Search: a002897 -id:a002897
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1, 2, 4, 9, 23, 65, 197, 626, 2056, 6918, 23714, 82500, 290512, 1033412, 3707852, 13402697, 48760367, 178405157, 656043857, 2423307047, 8987427467, 33453694487, 124936258127, 467995871777, 1757900019101, 6619846420553, 24987199492705, 94520750408709, 358268702159069
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OFFSET
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0,2
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COMMENTS
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This is also the result of applying the transformation on generating functions A -> 1/((1 - x)*(1 - x*A)) to the g.f. for the Catalan numbers.
p divides a(p) - 3 for prime p = 3 and p = {7, 13, 19, 31, 37, 43, ...} = A002476 (Primes of the form 6*n + 1). p^2 divides a(p^2) - 3 for prime p > 3. - Alexander Adamchuk, Jul 11 2006
Prime p divides a(p) for p = {2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, ...} = A045309 (Primes congruent to {0, 2} mod 3); and A045309 (Primes p such that x^3 = n (integer) has only one solution mod p). Nonprime numbers n such that n divides a(n) are listed in A128287 = {1, 8, 133, ...}. - Alexander Adamchuk, Feb 23 2007
For p prime >= 5, a(p-1) = 1 or -2 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). For example, with p=5, a(p-1) = 23 = -2 (mod p). - David Callan, Nov 29 2007
Hankel transform is A010892(n+1). - Paul Barry, Apr 24 2009
Equals INVERTi transform of A000245: (1, 3, 9, 28, ...). - Gary W. Adamson, May 15 2009
The subsequence of prime partial sums of Catalan numbers begins: a(1) = 2, a(4) = 23, a(6) = 197, a(16) = 48760367; see A121852. - Jonathan Vos Post, Feb 10 2010
Number of lattice paths from (0,0) to (n,n) which do not go above the diagonal x=y using steps (1,k), (k,1) with k >= 1 including two kinds of (1,1). - Alois P. Heinz, Oct 14 2015
Binomial transform of A086246(n+1) = [1, 1, 1, 2, 4, 9, ...], or, equivalently, of A001006 (Motzkin numbers) with 1 prepended.
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LINKS
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G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0 to 200 by T. D. Noe)
G. Alvarez, J. E. Bergner, and R. Lopez, Action Graphs and Catalan Numbers, J. Int. Seq. 18 (2015), 15.7.2.
Maciej Bendkowski and Pierre Lescanne, Combinatorics of explicit substitutions, arXiv:1804.03862 [cs.LO], 2018.
W. Chammam, F. Marcellán and R. Sfaxi, Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation, Linear Algebra Appl. 436(7) (2012), 2105-2116.
Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, 18 (2015), Article 15.5.8.
Nancy S. S. Gu, Nelson Y. Li, and Toufik Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Ângela Mestre and José Agapito, A Family of Riordan Group Automorphisms, J. Int. Seq., Vol. 22 (2019), Article 19.8.5.
I. Pak, Partition identities and geometric bijections, Proc. Amer. Math. Soc. 132 (2004), 3457-3462.
Hao Pan and Zhi-Wei Sun, A combinatorial identity with application to Catalan numbers, arXiv:math/0509648 [math.CO], 2005-2006.
Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016. See Appendix B2.
Kevin Topley, Computationally Efficient Bounds for the Sum of Catalan Numbers, arXiv:1601.04223 [math.CO], 2016.
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FORMULA
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a(n) = A014138(n-1) + 1.
G.f.: (1 - (1 - 4*x)^(1/2))/(2*x*(1 - x)).
Sum_{i=1..n} c(i) = Sum_{i=1..n} binomial(2*i-2, i-1)/i = 1/(n-1)! * [n^(n-2) + binomial(n, 2)*n^(n-3) + {8*binomial(n-4, 0) + 19*binomial(n-4, 1) + 24*binomial(n-4, 2) + 14*binomial(n-4, 3) + 3*binomial(n-4, 4)}*n^(n-4) +{18*binomial(n-5, 0) + 82*binomial(n-5, 1) + 229*binomial(n-5, 2) + 323*binomial(n-5, 3) + 244*binomial(n-5, 4) + 95*binomial(n-5, 5) + 15*binomial(n-5, 6)}*n^(n-5) + ... + binomial(n-3, 0)*(n-1)! ] (where c() = Catalan numbers A000108). - André F. Labossière, May 17 2004
a(n) = Sum_{k=0..n} (2k)!/(k!)^2/(k+1). - Alexander Adamchuk, Jul 11 2006
D-finite with recurrence: (n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011
Mathar's formula reduces to 2*(2*n-1)*C(n-1) = (n+1)*C(n), which is a known recurrence of the Catalan numbers, so the conjecture is true. - Peter J. Taylor, Mar 23 2015
Let C(n+1) = binomial(2*n+2,n+1)/(n+2) and H(n) = hypergeometric([1,n+3/2],[n+3],4) then A014137(n) = -(-1)^(2/3) - C(n+1)*H(n) and A014138(n) = -I^(2/3) - C(n+1)*H(n). - Peter Luschny, Aug 09 2012
G.f. (conjecture): Q(0)/(1-x), where Q(k)= 1 + (4*k + 1)*x/(k + 1 - 2*x*(k + 1)*(4*k + 3)/(2*x*(4*k + 3) + (2*k + 3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
a(n) ~ 2^(2*n + 2)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 10 2013
0 = a(n)*(16*a(n+1) - 26*a(n+2) + 10*a(n+3)) + a(n+1)*(-14*a(n+1) + 23*a(n+2) - 11*a(n+3)) + a(n+2)*(a(n+2) + a(n+3)) if n >= 0. - Michael Somos, Oct 24 2015
a(n) = (1 + A000108(n)*(3*(n+1)*hypergeom([1,-n], [1/2-n], 1/4) - 4*n - 2))/2. - Vladimir Reshetnikov, Oct 03 2016
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EXAMPLE
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G.f. = 1 + 2*x + 4*x^2 + 9*x^3 + 23*x^4 + 65*x^5 + 197*x^6 + 626*x^7 + 2056*x^8 + ...
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MAPLE
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a:= proc(n) option remember; `if`(n<2, n+1,
((5*n-1)*a(n-1)-(4*n-2)*a(n-2))/(n+1))
end:
seq(a(n), n=0..30); # Alois P. Heinz, May 18 2013
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MATHEMATICA
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Table[Sum[(2k)!/(k!)^2/(k+1), {k, 0, n}], {n, 0, 30}] (* Alexander Adamchuk, Jul 11 2006 *)
Accumulate[CatalanNumber[Range[0, 30]]] (* Harvey P. Dale, May 08 2012 *)
a[ n_] := SeriesCoefficient[ (1 - (1 - 4 x)^(1/2)) / (2 x (1 - x)), {x, 0, n}]; (* Michael Somos, Oct 24 2015 *)
Table[(1 + CatalanNumber[n] (3 (n + 1) Hypergeometric2F1[1, -n, 1/2 - n, 1/4] - 4 n - 2))/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)
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PROG
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(PARI) Vec((1-(1-4*x)^(1/2))/(2*x*(1-x))+O(x^99)) \\ Charles R Greathouse IV, Feb 11 2011
(PARI)
sm(v)={my(s=vector(#v)); s[1]=v[1]; for(n=2, #v, s[n]=v[n]+s[n-1]); s; }
C(n)=binomial(2*n, n)/(n+1);
sm(vector(66, n, C(n-1)))
/* Joerg Arndt, May 04 2013 */
(Python)
from __future__ import division
A014137_list, b, s = [], 1, 0
for n in range(10**2):
s += b
A014137_list.append(s)
b = b*(4*n+2)//(n+2) # Chai Wah Wu, Jan 28 2016
(Sage)
def A014137():
f, c, n = 1, 1, 1
while True:
yield f
n += 1
c = c * (4*n - 6) // n
f = c + f
a = A014137()
print([next(a) for _ in range(29)]) # Peter Luschny, Nov 30 2016
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CROSSREFS
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Cf. A000108, A000245, A000984, A001246, A002476, A002897, A006134, A033536, A045309, A079727, A082894, A094638, A094639, A128287.
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KEYWORD
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nonn,nice
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AUTHOR
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N. J. A. Sloane
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STATUS
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approved
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A002894
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a(n) = binomial(2n, n)^2.
(Formerly M3664 N1490)
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+10
108
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1, 4, 36, 400, 4900, 63504, 853776, 11778624, 165636900, 2363904400, 34134779536, 497634306624, 7312459672336, 108172480360000, 1609341595560000, 24061445010950400, 361297635242552100, 5445717990022688400, 82358080713306090000, 1249287673091590440000
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OFFSET
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0,2
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COMMENTS
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a(n) is the number of monotonic paths (only moving N and E) in the lattice [0..2n] X [0..2n] that contain the points (0,0), (n,n) and (2n,2n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Expansion of K(k) / (Pi/2) in powers of m/16 = (k/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at k. - Michael Somos, Mar 04 2003
Square lattice walks that start and end at origin after 2n steps. - Gareth McCaughan (gareth.mccaughan(AT)pobox.com) and Michael Somos, Jun 12 2004
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic) then a(n)=E[(tr(A^k))^{2n}] for any k > 4. - Andrew V. Sutherland, Apr 01 2008
From R. H. Hardin, Feb 03 2016 and R. J. Mathar, Feb 18 2016: (Start)
Also, number of 2 X (2n) arrays of permutations of 2n copies of 0 or 1 with row sums equal.
For example, some solutions for n=3:
0 1 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 1 1 1 0 0 0
1 0 0 0 1 1 1 1 0 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1
There is a simple combinatorial argument to show that this is a(n): We have 2n copies of 0's and 1's and need equal row sums. Therefore there must be n 1's in each of the two rows. Otherwise there are no constraints, so there are C(2n,n) ways of placing the 1's in the first row and independently C(2n,n) ways of placing the 1's in the second. The product is clearly C(2n,n)^2. (End)
Also the even part of the bisection of A241530. One half of the odd part is given in A000894. - Wolfdieter Lang, Sep 06 2016
From Peter Bala, Jan 26 2018: (Start)
Let S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} be a set of four column vectors. Then a(n) equals the number of 3 X k arrays whose columns belong to the set S and whose row sums are all equal to n (apply Eger, Theorem 3). An example is given below. Equivalently, a(n) equals the number of lattice paths from (0,0,0) to (n,n,n) using steps (1,0,0), (0,1,0), (1,0,1) and (0,1,1).
The o.g.f. for the sequence equals the diagonal of the rational function 1/(1 - (x + y + x*z + y*z)).
Row sums of A069466. (End)
Also, the constant term in the expansion of (x + 1/x + y + 1/y)^(2n). - Christopher J. Smyth, Sep 26 2018
Number of ways to place 2n^2 nonattacking pawns on a 2n x 2n board. - Tricia Muldoon Brown, Dec 12 2018
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 591,828.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.
Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
Lipshitz, Leonard, and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339-358.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..100
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
R. Bacher, Meander algebras, Institut Fourier, 1999.
E. Barcucci, A. Frosini and S. Rinaldi, On directed-convex polyominoes in a rectangle, Discr. Math., 298 (2005). 62-78.
Arnaud Beauville, Les familles stables de courbes elliptiques sur P^1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, no. 294, May 24 1982, 657-660. MR0664643 (83h:14008)
Tricia Muldoon Brown, The Problem of Pawns, The Electronic Journal of Combinatorics (2019) Vol. 26, Issue 3, #P3.21. Also arXiv:1811.09606, [math.CO], 2018.
John Maxwell Campbell, New series involving harmonic numbers and squared central binomial coefficients, Rocky Mountain J. Math., 49 (2019), 2513-2544.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
Murray Elder, Cogrowth, 2011.
M. Elder, A. Rechnitzer, E. J. Janse van Rensburg, and T. Wong, The cogrowth series for BS(N,N) is D-finite, arXiv:1309.4184 [math.GR], 2013.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 90
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2010.
L. Lipshitz and A. J. van der Poorten, Rational functions, diagonals, automata and arithmetic
Eric M. Rains, High powers of random elements of compact Lie groups, Probability Theory and Related Fields 107 (1997), 219-241.
Grzegorz Siudem and Agata Fronczak, Bell polynomials in the series expansions of the Ising model, arXiv:2007.16132 [math-ph], 2020.
Eric Weisstein's World of Mathematics, Lattice Path.
D. Zagier, Integral solutions of Apéry-like recurrence equations. See line G in sporadic solutions table of page 5.
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FORMULA
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D-finite with recurrence: (n + 1)^2*a(n+1) = 4*(2*n + 1)^2*a(n). - Matthijs Coster, Apr 28 2004
a(n) ~ Pi^-1*n^-1*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
G.f.: F(1/2, 1/2; 1; 16*x) = 1 / AGM(1, (1 - 16*x)^(1/2)) = K(4*sqrt(x)) / (Pi/2), where AGM(x, y) is the arithmetic-geometric mean of Gauss and Legendre. - Michael Somos, Mar 04 2003
G.f.: 2*EllipticK(4*sqrt(x))/Pi, using Maple's convention for elliptic integrals.
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2x)^2.
a(n) = A000984(n)^2 = ((2*n)!/(n!)^2)^2 = (((2*n)!)^2)/((n!)^4). a(n) = A000984(n)^2 = ((((2^n)*(2*n-1)!!)/(n!)))^2 = (((2^(2*n))*(2*n-1)!!)^2)/(n!)^2). - Jonathan Vos Post, Jun 17 2007
E.g.f.: (BesselI(0, 2x))^2=1+(2*x^2)/(U(0)-2*x^2); U(k)=(2*x^2)*(2*k+1)+(k+1)^3-(2*x^2)*(2*k+3)*((k+1)^3)/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
In generally, for (BesselI(b, 2x))^2=((x^(2*b))/(GAMMA(b+1))^2)*(1+(2*x^2)*(2*b+1)/(Q(0)-(2*x^2)*(2*b+1)); Q(k)=(2*x^2)*(2*k+2*b+1)+(k+1)*(k+b+1)*(k+2*b+1)-(2*x^2)*(k+1)*(k+b+1)*(k+2*b+1)*(2*k+2*b+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 4*(2*k+1)^2*x*(1+4*x)^2/(4*(2*k+1)^2*x*(1+4*x)^2 + (k+1)^2*(1+4*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
0 = +a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for all n in Z. - Michael Somos, Aug 06 2014
Integral representation as the n-th moment of a positive function W(x) on (0,16), in Maple notation, W(x)=EllipticK(sqrt(1-x/16)/(2*Pi^2*sqrt(x)); a(n)=int(x^n* W(x), x=0..16), n>=0. The function W(x) is singular at x=0 and W(16)=1/(16*Pi). This representation is unique since W(x) is the solution of the Hausdorff moment problem. - Stanley Smith and Karol A. Penson, Jun 19 2015
a(n) ~ 16^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)^2/((4*n+1)* Pi). - Peter Luschny, Oct 14 2015
a(n) = binomial(2*n,n)*binomial(2*n,n) = ( [x^n](1 + x)^(2*n) ) *( [x^n](1 + x)^(2*n) ) = [x^n](F(x)^(4*n)), where F(x) = 1 + x + x^2 + 4*x^3 + 20*x^4 + 120*x^5 + 798*x^6 + 5697*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) = Sum_{k = 0..n} binomial(2*n + k,k)*binomial(n,k)^2. Cf. A005258(n) = Sum_{k = 0..n} binomial(n + k,k)*binomial(n,k)^2. - Peter Bala, Jul 27 2016
a(n) = A241530(2*n), n >= 0. - Wolfdieter Lang, Sep 06 2016
E.g.f.: 2F2(1/2,1/2; 1,1; 16*x). - Ilya Gutkovskiy, Jan 23 2018
a(n) = 16^n*hypergeom([1/2, -2*n, 2*n + 1], [1, 1], 1). - Peter Luschny, Mar 14 2018
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-4)^(n-k) = a(n). - Peter Bala, Mar 16 2018
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EXAMPLE
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G.f. = 1 + 4*x + 36*x^2 + 400*x^3 + 4900*x^4 + 63504*x^5 + 853776*x^6 + ... - Michael Somos, Aug 06 2014
From Peter Bala, Jan 26 2018: (Start)
a(2) = 36: The thirty six 3 x k arrays with columns belonging to the set of column vectors S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} and having all row sums equal to 2 are the 6 distinct arrays obtained by permuting the columns of
/1 1 0 0\
|0 0 1 1|,
\0 0 1 1/
the 6 distinct arrays obtained by permuting the columns of
/0 0 1 1\
|1 1 0 0|
\0 0 1 1/
and the 24 arrays obtained by permuting the columns of
/1 0 1 0\
|0 1 0 1|. (End)
\0 0 1 1/
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MAPLE
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A002894 := n-> binomial(2*n, n)^2.
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MATHEMATICA
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CoefficientList[Series[Hypergeometric2F1[1/2, 1/2, 1, 16x], {x, 0, 20}], x]
Table[Binomial[2n, n]^2, {n, 0, 20}] (* Harvey P. Dale, Jul 06 2011 *)
a[ n_] := SeriesCoefficient[ EllipticK[16 x] / (Pi/2), {x, 0, n}]; (* Michael Somos, Aug 06 2014 *)
a[n_] := 16^n HypergeometricPFQ[{1/2, -2 n, 2 n + 1}, {1, 1}, 1];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 14 2018 *)
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PROG
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(PARI) {a(n) = binomial(2*n, n)^2};
(PARI) {a(n) = if( n<0, 0, polcoeff( polcoeff( polcoeff( 1 / (1 - x * (y + z + 1/y + 1/z)) + x * O(x^(2*n)), 2*n), 0), 0))}; /* Michael Somos, Jun 12 2004 */
(Sage) [binomial(2*n, n)**2 for n in range(17)] # Zerinvary Lajos, Apr 21 2009
(MAGMA) [Binomial(2*n, n)^2: n in [0..20]]; // Vincenzo Librandi, Aug 07 2014
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CROSSREFS
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Row sums of A069466.
Row 2 of A268367 (even terms).
Equals 4*A060150.
Cf. A000984, A000515, A010370, A054474 (INVERTi transform), A172390, A000897, A002897, A006480, A008977, A186420, A188662, A000894, A241530, A002898 (walks hex lattice).
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KEYWORD
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nonn,nice,easy
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AUTHOR
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N. J. A. Sloane
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EXTENSIONS
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Edited by N. J. A. Sloane, Feb 18 2016
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STATUS
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approved
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A094638
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Triangle read by rows: T(n,k) = |s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind A008276 (1 <= k <= n; in other words, the unsigned Stirling numbers of the first kind in reverse order).
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+10
63
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1, 1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24, 1, 15, 85, 225, 274, 120, 1, 21, 175, 735, 1624, 1764, 720, 1, 28, 322, 1960, 6769, 13132, 13068, 5040, 1, 36, 546, 4536, 22449, 67284, 118124, 109584, 40320, 1, 45, 870, 9450, 63273, 269325, 723680, 1172700, 1026576, 362880
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OFFSET
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1,5
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COMMENTS
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Triangle of coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in decreasing powers of x. - T. D. Noe, Feb 22 2008
Row n also gives the number of permutation of 1..n with complexity 0,1,...,n-1. See the comments in A008275. - N. J. A. Sloane, Feb 08 2019
T(n,k) is the number of deco polyominoes of height n and having k columns. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: T(2,1)=1 and T(2,2)=1 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, having, respectively, 1 and 2 columns. - Emeric Deutsch, Aug 14 2006
Sum_{k=1..n} k*T(n,k) = A121586. - Emeric Deutsch, Aug 14 2006
Let the triangle U(n,k), 0 <= k <= n, read by rows, be given by [1,0,1,0,1,0,1,0,1,0,1,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,...] where DELTA is the operator defined in A084938; then T(n,k) = U(n-1,k-1). - Philippe Deléham, Jan 06 2007
From Tom Copeland, Dec 15 2007: (Start)
Consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5, ...).
Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...). And T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n > 0. Call this sequence the right factorial (n+)!.
Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1, 0, 0, 0, ...), the left factorial (n-)!.
Sampling every other integer to the right, we obtain (1,3,5,7,9,...). T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n > 0, the right double factorial, (n+)!!.
Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7,...). T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left double factorial, (n-)!!.
Sampling every 3 steps to the right, we obtain (1,4,7,10,...). T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...), giving A007559 for n > 0, the right triple factorial, (n+)!!!.
Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...) = signed A008544, the left triple factorial, (n-)!!!.
The list partition transform A133314 of [1,T * c(t)] gives [1,T * c(-t)] with all odd terms negated; e.g., LPT[1,T*c(2)] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147). And e.g.f. for [1,T * c(t)] = (1-xt)^(-1/t).
The above results hold for t any real or complex number. (End)
Let R_n(x) be the real and I_n(x) the imaginary part of Product_{k=0..n} (x + I*k). Then, for n=1,2,..., we have R_n(x) = Sum_{k=0..floor((n+1)/2)}(-1)^k*stirling1(n+1,n+1-2*k)*x^(n+1-2*k), I_n(x) = Sum_{k=0..floor(n/2)}(-1)^(k+1)*stirling1(n+1,n-2*k)*x^(n-2*k). - Milan Janjic, May 11 2008
T(n,k) is also the number of permutations of n with "reflection length" k (i.e., obtained from 12..n by k not necessarily adjacent transpositions). For example, when n=3, 132, 213, 321 are obtained by one transposition, while 231 and 312 require two transpositions. - Kyle Petersen, Oct 15 2008
From Tom Copeland, Nov 02 2010: (Start)
[x^(y+1) D]^n = x^(n*y) [T(n,1)(xD)^n + T(n,2)y (xD)^(n-1) + ... + T(n,n)y^(n-1)(xD)], with D the derivative w.r.t. x.
E.g., [x^(y+1) D]^4 = x^(4*y) [(xD)^4 + 6 y(xD)^3 + 11 y^2(xD)^2 + 6 y^3(xD)].
(xD)^m can be further expanded in terms of the Stirling numbers of the second kind and operators of the form x^j D^j. (End)
With offset 0, 0 <= k <= n: T[n,k) is the sum of products of each size k subset of {1,2,...,n}. For example, T(3,2) = 11 because there are three subsets of size two: {1,2},{1,3},{2,3}. 1*2 + 1*3 + 2*3 = 11. - Geoffrey Critzer, Feb 04 2011
The Kn11, Fi1 and Fi2 triangle sums link this triangle with two sequences, see the crossrefs. For the definitions of these triangle sums see A180662. The mirror image of this triangle is A130534. - Johannes W. Meijer, Apr 20 2011
T(n+1,k+1) is the elementary symmetric function a_k(1,2,...,n), n >= 0, k >= 0, (a_0(0):=1). See the T. D. Noe and Geoffrey Critzer comments given above. For a proof see the Stanley reference, p. 19, Second Proof. - Wolfdieter Lang, Oct 24 2011
Let g(t) = 1/d(log(P(j+1,-t)))/dt (see Tom Copeland's 2007 formulas). The Mellin transform (t to s) of t*Dirac[g(t)] gives Sum_{n=1..j} n^(-s), which as j tends to infinity gives the Riemann zeta function for Re(s) > 1. Dirac(x) is the Dirac delta function. The complex contour integral along a circle of radius 1 centered at z=1 of z^s/g(z) gives the same result. - Tom Copeland, Dec 02 2011
Rows are coefficients of the polynomial expansions of the Pochhammer symbol, or rising factorial, Pch(n,x) = (x+n-1)!/(x-1)!. Expansion of Pch(n,xD) = Pch(n,Bell(.,:xD:)) in a polynomial with terms :xD:^k=x^k*D^k gives the Lah numbers A008297. Bell(n,x) are the unsigned Bell polynomials or Stirling polynomials of the second kind A008277. - Tom Copeland, Mar 01 2014
From Tom Copeland, Dec 09 2016: (Start)
The Betti numbers, or dimension, of the pure braid group cohomology. See pp. 12 and 13 of the Hyde and Lagarias link.
Row polynomials and their products appear in presentation of the Jack symmetric functions of R. Stanley. See Copeland link on the Witt differential generator.
(End)
From Tom Copeland, Dec 16 2019: (Start)
The e.g.f. given by Copeland in the formula section appears in a combinatorial Dyson-Schwinger equation of quantum field theory in Yeats in Thm. 2 on p. 62 related to a Hopf algebra of rooted trees. See also the Green function on p. 70.
Per comments above, this array contains the coefficients in the expansion in polynomials of the Euler, or state number, operator xD of the rising factorials Pch(n,xD) = (xD+n-1)!/(xD-1)! = x [:Dx:^n/n!]x^{-1} = L_n^{-1}(-:xD:), where :Dx:^n = D^n x^n and :xD:^n = x^n D^n. The polynomials L_n^{-1} are the Laguerre polynomials of order -1, i.e., normalized Lah polynomials.
The Witt differential operators L_n = x^(n+1) D and the row e.g.f.s appear in Hopf and dual Hopf algebra relations presented by Foissy. The Witt operators satisfy L_n L_k - L_k L_n = (k-n) L_(n+k), as for the dual Hopf algebra. (End)
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REFERENCES
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M. Miyata and J. W. Son, On the complexity of permutations and the metric space of bijections, Tensor, 60 (1998), No. 1, 109-116 (MR1768839).
R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.
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LINKS
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T. D. Noe, Rows n=1..51 of triangle, flattened
E. Barcucci, A. Del Lungo and R. Pinzani, "Deco" polyominoes, permutations and random generation, Theoretical Computer Science, 159, 1996, 29-42.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48. Added Mar 01 2014
F. Bergeron, Philippe Flajolet and Bruno Salvy, Varieties of increasing trees, HAL, Rapport De Recherche Inria. Added Mar 01 2014
T. Copeland, Addendum to Mathemagical Forests
T. Copeland, A Class of Differential Operators and the Stirling Numbers
T. Copeland, Mellin Interpolation of Differential Ops and Associated Infinigens and Appell Polynomials: The Ordered, Laguerre, and Scherk-Witt-Lie Diff Ops
T. Copeland, Witt differential generator for special Jack symmetric functions / polynomials
FindStat - Combinatorial Statistic Finder, The absolute length of a permutation
L. Foissy, Faa di Bruno subalgebras of the Hopf algebra of planar trees from combinatorial Dyson-Schwinger equations, arXiv:0707.1204 [math.RA], (2007).
O. Furdui, T. Trif, On the Summation of Certain Iterated Series, J. Int. Seq. 14 (2011) #11.6.1
F. Hivert, J.-C. Novelli and J.-Y. Thibon, The Algebra of Binary Search Trees, Theoretical Computer Science, 339 (2005), 129-165.
T. Hyde and J. Lagarias Polynomial splitting measures and cohomology of the pure braid group, arXiv preprint arXiv:1604.05359 [math.RT], 2016.
MathOverflow, Motivation of Virasoro algebra, an answer by Tom Copeland to an MO question posed in 2012.
R. Mestrovic, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820 [math.NT], 2014.
Robert E. Moritz, On the sum of products of n consecutive integers, Univ. Washington Publications in Math., 1 (No. 3, 1926), 44-49 [Annotated scanned copy]
M. D. Schmidt, Generalized j-Factorial Functions, Polynomials, and Applications , J. Int. Seq. 13 (2010), 10.6.7.
M. Z. Spivey, On Solutions to a General Combinatorial Recurrence, J. Int. Seq. 14 (2011) # 11.9.7.
K. Yeats, A Combinatorial Perspective on Quantum Field Theory, SpringerBriefs in Mathematical Physics, Vol. 15, 2017.
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FORMULA
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With P(n,t) = Sum_{k=0..n-1} T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n-1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1-tx)^(-1/t). T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] evaluated at t=x=0. (1-tx)^(-1/t) - 1 is the e.g.f. for a plane m-ary tree when t=m-1. See Bergeron et al. in "Varieties of Increasing Trees". - Tom Copeland, Dec 09 2007
First comment and formula above rephrased as o.g.f. for row n: Product_{i=0...n} (1+i*x). - Geoffrey Critzer, Feb 04 2011
n-th row polynomials with alternate signs are the characteristic polynomials of the (n-1)x(n-1) matrices with 1's in the superdiagonal, (1,2,3,...) in the main diagonal, and the rest zeros. For example, the characteristic polynomial of [1,1,0; 0,2,1; 0,0,3] is x^3 - 6*x^2 + 11*x - 6. - Gary W. Adamson, Jun 28 2011
E.g.f.: A(x,y) = x*y/(1 - x*y)^(1 + 1/y) = Sum_{n>=1, k=1..n} T(n,k)*x^n*y^k/(n-1)!. - Paul D. Hanna, Jul 21 2011
With F(x,t) = (1-t*x)^(-1/t) - 1 an e.g.f. for the row polynomials P(n,t) of A094638 with P(0,t)=0, G(x,t)= [1-(1+x)^(-t)]/t is the comp. inverse in x. Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+x)^(t+1),
P(n,t) = [(H(x,t)*d/dx)^n] x evaluated at x=0; i.e.,
F(x,t) = exp[x*P(.,t)] = exp[x*H(u,t)*d/du] u, evaluated at u = 0.
Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 20 2011
T(n,k) = |A008276(n,k)|. - R. J. Mathar, May 19 2016
The row polynomials of this entry are the reversed row polynomials of A143491 multiplied by (1+x). E.g., (1+x)(1 + 5x + 6x^2) = (1 + 6x + 11x^2 + 6x^3). - Tom Copeland, Dec 11 2016
Regarding the row e.g.f.s in Copeland's 2007 formulas, e.g.f.s for A001710, A001715, and A001720 give the compositional inverses of the e.g.f. here for t = 2, 3, and 4 respectively. - Tom Copeland, Dec 28 2019
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EXAMPLE
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Triangle starts:
1;
1, 1;
1, 3, 2;
1, 6, 11, 6;
1, 10, 35, 50, 24;
...
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MAPLE
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T:=(n, k)->abs(Stirling1(n, n+1-k)): for n from 1 to 10 do seq(T(n, k), k=1..n) od; # yields sequence in triangular form. # Emeric Deutsch, Aug 14 2006
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MATHEMATICA
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Table[CoefficientList[Series[Product[1 + i x, {i, n}], {x, 0, 20}], x], {n, 0, 6}] (* Geoffrey Critzer, Feb 04 2011 *)
Table[Abs@StirlingS1[n, n-k+1], {n, 10}, {k, n}]//Flatten (* Michael De Vlieger, Aug 29 2015 *)
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PROG
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(PARI) {T(n, k)=if(n<1 || k>n, 0, (n-1)!*polcoeff(polcoeff(x*y/(1 - x*y+x*O(x^n))^(1 + 1/y), n, x), k, y))} /* Paul D. Hanna, Jul 21 2011 */
(Maxima) create_list(abs(stirling1(n+1, n-k+1)), n, 0, 10, k, 0, n); / * Emanuele Munarini, Jun 01 2012 */
(Haskell)
a094638 n k = a094638_tabl !! (n-1) !! (k-1)
a094638_row n = a094638_tabl !! (n-1)
a094638_tabl = map reverse a130534_tabl
-- Reinhard Zumkeller, Aug 01 2014
(MAGMA) [(-1)^(k+1)*StirlingFirst(n, n-k+1): k in [1..n], n in [1..10]]; // G. C. Greubel, Dec 29 2019
(Sage) [[stirling_number1(n, n-k+1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, Dec 29 2019
(GAP) Flat(List([1..10], n-> List([1..n], k-> Stirling1(n, n-k+1) ))); # G. C. Greubel, Dec 29 2019
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CROSSREFS
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A008276 gives the (signed) Stirling numbers of the first kind.
Cf. A000108, A014137, A001246, A033536, A000984, A094639, A006134, A082894, A002897, A079727, A000217 (2nd column), A000914 (3rd column), A001303 (4th column), A000915 (5th column), A053567 (6th column), A000142 (row sums).
Triangle sums (see the comments): A124380 (Kn11), A001710 (Fi1, Fi2). - Johannes W. Meijer, Apr 20 2011
Cf. A121586, A130534, A143491.
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KEYWORD
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easy,nonn,tabl
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AUTHOR
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André F. Labossière, May 17 2004
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EXTENSIONS
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Edited by Emeric Deutsch, Aug 14 2006
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STATUS
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approved
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A008977
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a(n) = (4*n)!/(n!)^4.
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+10
35
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1, 24, 2520, 369600, 63063000, 11732745024, 2308743493056, 472518347558400, 99561092450391000, 21452752266265320000, 4705360871073570227520, 1047071828879079131681280, 235809301462142612780721600
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,2
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COMMENTS
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Number of paths of length 4*n in an n X n X n X n grid from (0,0,0,0) to (n,n,n,n).
a(n) occurs in Ramanujan's formula 1/Pi = (sqrt(8)/9801) * Sum_{n>=0} (4*n)!/(n!)^4 * (1103 + 26390*n)/396^(4*n) ). - Susanne Wienand, Jan 05 2013
a(n) is the number of ballot results that lead to a 4-way tie when 4*n voters each cast three votes for three out of four candidates vying for 3 slots on a county commission; each of these ballot results give 3*n votes to each of the four candidates. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X + Y + Z + 1/(X*Y*Z))^(4*n). - Mark van Hoeij, May 07 2013
In Narumiya and Shiga on page 158 the g.f. is given as a hypergeometric function. - Michael Somos, Aug 12 2014
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w+x+y+z)). - Gheorghe Coserea, Jul 15 2016
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..100
R. M. Dickau, Paths through a 4-D lattice
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Michaël Moortgat, The Tamari order for D^3 and derivability in semi-associative Lambek-Grishin Calculus, 15th Workshop: Computational Logic and Applications (CLA 2020).
N. Narumiya and H. Shiga, The mirror map for a family of K3 surfaces induced from the simplest 3-dimensional reflexive polytope, Proceedings on Moonshine and related topics (Montréal, QC, 1999), 139-161, CRM Proc. Lecture Notes, 30, Amer. Math. Soc., Providence, RI, 2001. MR1877764 (2002m:14030).
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FORMULA
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a(n) = A139541(n)*(A001316(n)/A049606(n))^3. - Reinhard Zumkeller, Apr 28 2008
Self-convolution of A178529, where A178529(n) = (4^n/n!^2) * Product_{k=0..n-1} (8*k + 1)*(8*k + 3).
G.f.: hypergeom([1/8, 3/8], [1], 256*x)^2. - Mark van Hoeij, Nov 16 2011
a(n) ~ 2^(8*n - 1/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Mar 07 2014
G.f.: hypergeom([1/4, 2/4, 3/4], [1, 1], 256*x). - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(24*n)), where F(x) = 1 + x + 29*x^2 + 2246*x^3 + 239500*x^4 + 30318701*x^5 + 4271201506*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008978, A008979, A186420 and A188662. (End)
0 = (x^2-256*x^3)*y''' + (3*x-1152*x^2)*y'' + (1-816*x)*y' - 24*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..3*n} (-1)^(n+k)*binomial(4*n,n + k)* binomial(n + k,k)^4.
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(4*n,k)*binomial(n + k,k)^4. (End)
E.g.f.: 3F3(1/4,1/2,3/4; 1,1,1; 256*x). - Ilya Gutkovskiy, Jan 23 2018
From Peter Bala, Feb 16 2020: (Start)
Supercongruences: a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(4*n). (End)
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EXAMPLE
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a(13)=52!/(13!)^4=53644737765488792839237440000 is the number of ways of dealing the four hands in Bridge or Whist. - Henry Bottomley, Oct 06 2000
a(1)=24 since, in a 4-voter 3-vote election that ends in a four-way tie for candidates A, B, C, and D, there are 4! ways to arrange the needed vote sets {A,B,C}, {A,B,D}, {A,C,D}, and {B,C,D} among the 4 voters. - Dennis P. Walsh, May 02 2013
G.f. = 1 + 24*x + 2520*x^2 + 369600*x^3 + 63063000*x^4 + 11732745024*x^5 + ...
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MAPLE
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A008977 := n->(4*n)!/(n!)^4;
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MATHEMATICA
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Table[(4n)!/(n!)^4, {n, 0, 16}] (* Harvey P. Dale, Oct 24 2011 *)
a[ n_] := If[ n < 0, 0, (4 n)! / n!^4]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/4, 2/4, 3/4}, {1, 1}, 256 x], {x, 0, n}]; (* Michael Somos, Aug 12 2014 *)
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PROG
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(Maxima) A008977(n):=(4*n)!/(n!)^4$ makelist(A008977(n), n, 0, 20); /* Martin Ettl, Nov 15 2012 */
(MAGMA) [Factorial(4*n)/Factorial(n)^4: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014
(PARI) a(n) = (4*n)!/n!^4; \\ Gheorghe Coserea, Jul 15 2016
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CROSSREFS
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Cf. A000984, A006480, A008978, A178529, A000897, A002894, A002897, A006480, A008979, A186420, A188662.
Related to diagonal of rational functions: A268545-A268555.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane
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STATUS
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approved
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A000897
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a(n) = (4*n)! / ((2*n)!*n!^2).
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+10
19
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1, 12, 420, 18480, 900900, 46558512, 2498640144, 137680171200, 7735904619300, 441233078286000, 25467973278667920, 1484298740174927040, 87202550985276963600, 5157850293780050462400, 306839461354466267304000
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listen;
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internal format)
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OFFSET
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0,2
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COMMENTS
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Appears in Ramanujan's theory of elliptic functions of signature 4.
H. A. Verrill proves that a(n) = Sum_{p + q + r = 3n} w^(p-q) * {(3n)!/(p! q! r!)}^2, with p, q, r >= 0 and w = primitive 3rd root of unity.
The family of elliptic curves "x=2*H1=p^2+q^2-(1/4)*q^4, 0<x<1" generates these a_n as the coefficients of the period-energy function "T(x)=2*Pi*2F1(1/4,3/4;1;x)". Applying complex transformation "q->sqrt(-1)*q" to H1 produces "x=2*H2=p^2-q^2-(1/4)*q^4, 0<x<1", with "T(x)=sqrt(2)*Pi*2F1(1/4,3/4;1;1-x)". This explains the appearance of factor sqrt(2)/2 in Ramanujan's nome q_1. - Bradley Klee, Feb 25 2018
Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z)). - Gheorghe Coserea, Aug 09 2018
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REFERENCES
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E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..200
H. J. Brothers, Pascal's Prism: Supplementary Material
B. Klee, Geometric G.F. for Ramanujan Periods, seqfan mailing list, 2017.
S. Ramanujan, Modular Equations and Approximations to Pi, Quarterly Journal of Mathematics, XLV (1914), 350-372.
L. C. Shen, A note on Ramanujan’s identities involving the hypergeometric function 2F1(1/6,5/6;1;z), The Ramanujan Journal, 30.2 (2013), 211-222.
H. A. Verrill, Sums of squares of binomial coefficients, with applications to Picard-Fuchs equations, arXiv:math/0407327v1 [math.CO], 2004.
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FORMULA
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E.g.f.: Sum_{k>=0} (-1)^k * a(k) * x^(4*k) / (4*k)! = BesselI(0, 2x) * BesselJ(0, 2x).
G.f.: F(1/4, 3/4; 1; 64*x). - Michael Somos, Oct 31 2005
a(n) = A008977(n)/A000984(n) - Zerinvary Lajos, Jun 28 2007
Sum_{k>=0} a(k) * x^(3k)/(3k)!^2 = f(x)*f(x*w)*f(x/w) where f(x) = BesselI(0, 2*sqrt(x)) and w = primitive 3rd root of unity. - Michael Somos, Jul 25 2007
In general, for (BesselI(b, 2x))*(BesselJ(b, 2x))=((x^(2*b))/((GAMMA(b+1))^2)*(1-(x^4)/(Q(0)+(x^4))); Q(k)=(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)-(x^4)+(x^4)*(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)/Q(k+1)) ; (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
D-finite with recurrence 0 = a(n)*4*(4*n + 1)*(4*n + 3) - a(n+1)*(n + 1)^2 for all n in Z. - Michael Somos, Aug 12 2014
0 = a(n)*(-4026531840*a(n+2) +2005401600*a(n+3) -103896576*a(n+4) +1251948*a(n+5)) + a(n+1)*(+41418752*a(n+2) -30435328*a(n+3) +1863228*a(n+4) -24604*a(n+5)) + a(n+2)*(-16896*a(n+2) +75608*a(n+3) -6740*a(n+4) +105*a(n+5)) for all n in Z. - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(3*n,n)*binomial(4*n,n) = A005809(n)*A005810(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(12*n)), where F(x) = 1 + x + 6*x^2 + 105*x^3 + 2448*x^4 + 67043*x^5 + 2028307*x^6 + ... appears to have integer coefficients. Cf. A002894, A002897, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 2^(6*n-1/2)/(Pi*n). - Ilya Gutkovskiy, Jul 12 2016
G.f.: 2*EllipticK(sqrt((sqrt(1-64*x)-1)/(2*sqrt(1-64*x))))/(Pi*(1-64*x)^(1/4)) where EllipticK is the complete elliptic integral of the first kind (in Maple's notation). - Robert Israel, Jul 12 2016
a(n) = Sum_{k = 0..3*n} (-1)^k*C(3*n,k)*C(6*n-k,3*n)*C(2*k,k). - Peter Bala, Feb 10 2018
From Bradley Klee, Feb 27 2018: (Start)
a(n) = A000984(n)*A001448(n).
G.f.: (1/(sqrt(2)*Pi))*Integral_{q=-oo..oo} 1/sqrt(q^2+(1/4)*q^4+(1-64*x)) dq.
G.f.: (1/(2*Pi))*Integral_{phi=0..2*Pi} 1/sqrt(1-64*x*sin^4(phi)) dphi. (End)
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|
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EXAMPLE
|
G.f.: 1 + 12*x + 420*x^2 + 18480*x^3 + 900900*x^4 + 46558512*x^5 + 2498640144*x^6 + ...
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MAPLE
|
seq((4*n)!/(n!)^4/binomial(2*n, n), n=0..14); # Zerinvary Lajos, Jun 28 2007
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MATHEMATICA
|
Table[(4n)!/((2n)! n!^2), {n, 0, 30}] (* Stefan Steinerberger, Apr 14 2006 *)
a[ n_] := Binomial[ 4 n, 2 n] Binomial[ 2 n, n]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 3/4, 1, 64 x], {x, 0, n}]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := If[ n < 0, 0, With[{m = 4 n}, (-1)^n m! SeriesCoefficient[ BesselI[ 0, 2 x] BesselJ[ 0, 2 x], {x, 0, m}]]]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := 64^n Pochhammer[1/4, n] Pochhammer[3/4, n] / n!^2; (* Michael Somos, Aug 12 2014 *)
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PROG
|
(PARI) {a(n) = if( n<0, 0, (4*n)! / ((2*n)! * n!^2))}; /* Michael Somos, Oct 31 2005 */
(GAP) a:=n->Sum([0..3*n], k->(-1)^k*Binomial(3*n, k)*Binomial(6*n-k, 3*n)*
Binomial(2*k, k));;
A000897:=List([0..14], n->a(n)); # Muniru A Asiru, Feb 11 2018
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CROSSREFS
|
Cf. A002897, A008977, A186420, A188662. Elliptic Integrals: A002894, A113424, A006480. Factors: A005809, A005810, A000984, A001448.
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KEYWORD
|
nonn,easy
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AUTHOR
|
N. J. A. Sloane
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STATUS
|
approved
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| |
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A004981
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|
a(n) = (2^n/n!) * Product_{k=0..n-1} (4*k + 1).
|
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+10
18
|
|
|
|
1, 2, 10, 60, 390, 2652, 18564, 132600, 961350, 7049900, 52169260, 388898120, 2916735900, 21987701400, 166478310600, 1265235160560, 9647418099270, 73774373700300, 565603531702300, 4346216612028200, 33465867912617140, 258165266754475080, 1994913424920943800
(list;
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OFFSET
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0,2
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COMMENTS
|
The convolution of this sequence with itself yields A059304. - T. D. Noe, Jun 11 2002
Conjecture: a(p*n) = a(n) (mod p^2) for prime p = 1 (mod 4) and all positive integers n. Cf. A004982 and A298799. - Peter Bala, Dec 22 2019
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LINKS
|
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Vogt, Resummation of small-x double logarithms in QCD: semi-inclusive electron-positron annihilation, arXiv preprint arXiv:1108.2993 [hep-ph], 2011.
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FORMULA
|
a(n) ~ Gamma(1/4)^-1*n^(-3/4)*2^(3*n)*{1 - 3/32*n^-1 - ...}
G.f.: (1-8*x)^(-1/4).
A002897(n) = Sum_{k=0..n} a(k)^2*a(n-k)^2. - Michael Somos, Jan 31 2007
a(n) = (Sum_{m=1..n} m*Sum_{k=m..n} binomial(-m+2*k-1,k-1)*2^(n+m-k)*binomial(2*n-k-1,n-1))/n, n>0, a(0)=1. - Vladimir Kruchinin, Dec 26 2011
D-finite with recurrence: n*a(n) = 2*(4*n-3)*a(n-1). - R. J. Mathar, Mar 14 2014
From Karol A. Penson, Dec 19 2015: (Start)
a(n) = (-8)^n*binomial(-1/4,n).
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([1/4], [1], 8*x).
Representation as n-th moment of a positive function on (0, 8): a(n)=int(x^n*(sqrt(2)/(16*Pi*(x/8)^(3/4)*(1-x/8)^(1/4))), x=0..8), n=0, 1, ... . This function is the solution of the Hausdorff moment problem on (0, 8) with moments equal to a(n). As a consequence this representation is unique. (End)
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MAPLE
|
A004981 := n -> (-8)^n*binomial(-1/4, n):
seq(A004981(n), n=0..25); # Peter Luschny, Oct 23 2018
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MATHEMATICA
|
CoefficientList[Series[(1-8x)^(-1/4), {x, 0, 25}], x] (* Vincenzo Librandi, Mar 16 2014 *)
Table[8^n*Pochhammer[1/4, n]/n!, {n, 0, 25}] (* G. C. Greubel, Aug 22 2019 *)
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PROG
|
(PARI) a(n)=if(n<0, 0, prod(k=1, n, (8*k-6)/k))
(PARI) {a(n)=if(n<0, 0, polcoeff( (1-8*x+x*O(x^n))^(-1/4), n))} /* Michael Somos, Jan 31 2007 */
(Maxima) a(n):=if n=0 then 1 else (sum(m*sum(binomial(-m+2*k-1, k-1) *2^(n+m-k)*binomial(2*n-k-1, n-1), k, m, n), m, 1, n))/(n); /* Vladimir Kruchinin, Dec 26 2011 */
(MAGMA) [1] cat [2^n*&*[4*k+1: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019
(Sage) [8^n*rising_factorial(1/4, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019
(GAP) List([0..25], n-> 2^n*Product([0..n-1], k-> 4*k+1)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
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CROSSREFS
|
Cf. A002897, A004982, A059304, A127776, A298799.
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KEYWORD
|
nonn,easy
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AUTHOR
|
Joe Keane (jgk(AT)jgk.org)
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EXTENSIONS
|
More terms from James A. Sellers, May 01 2000
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STATUS
|
approved
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A008978
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|
a(n) = (5*n)!/(n!)^5.
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|
+10
16
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|
1, 120, 113400, 168168000, 305540235000, 623360743125120, 1370874167589326400, 3177459078523411968000, 7656714453153197981835000, 19010638202652030712978200000, 48334775757901219912115629238400
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,2
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|
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COMMENTS
|
Number of paths of length 5n in Z^5 from (0,0,0,0,0) to (n,n,n,n,n).
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LINKS
|
Vincenzo Librandi, Table of n, a(n) for n = 0..100
V. Batyrev, Review of "Mirror Symmetry and Algebraic Geometry", by D. A. Cox and S. Katz, Bull. Amer. Math. Soc., 37 (No. 4, 2000), 473-476.
R. M. Dickau, 5-D shortest path diagrams
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
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FORMULA
|
a(n) ~ 5^(5*n+1/2) / (4 * Pi^2 * n^2). - Vaclav Kotesovec, Mar 07 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n]( F(x)^(120*n) ), where F(x) = 1 + x + 353*x^2 + 318986*x^3 + 408941594*x^4 + 633438203535*x^5 + 1105336091531052*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. (End)
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(5*n,n + k)*binomial(n + k,k)^5.
a(n) = Sum_{k = 0..5*n} (-1)^(n+k)*binomial(5*n,k)*binomial(n + k,k)^5. (End)
From Ilya Gutkovskiy, Nov 23 2017: (Start)
O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1,1,1; 3125*x).
E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1,1,1,1; 3125*x). (End)
From Peter Bala, Feb 16 2020: (Start)
Supercongruences: a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z*u)^n] (1 + x + y + z + u )^(5*n). (End)
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MATHEMATICA
|
Table[(5 n)!/(n)!^5, {n, 0, 20}] (* Vincenzo Librandi, Mar 08 2014 *)
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PROG
|
(MAGMA) [Factorial(5*n)/Factorial(n)^5: n in [0..10]]; // Vincenzo Librandi, Mar 08 2014
(PARI) a(n) = (5*n)!/(n!)^5; \\ Michel Marcus, Mar 08 2014
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CROSSREFS
|
Cf. A000984, A006480, A008977, A002894, A002897, A186420, A188662, A001460.
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KEYWORD
|
nonn,easy
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AUTHOR
|
N. J. A. Sloane.
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STATUS
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approved
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A181543
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Triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, read by rows.
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+10
16
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1, 1, 1, 1, 8, 1, 1, 27, 27, 1, 1, 64, 216, 64, 1, 1, 125, 1000, 1000, 125, 1, 1, 216, 3375, 8000, 3375, 216, 1, 1, 343, 9261, 42875, 42875, 9261, 343, 1, 1, 512, 21952, 175616, 343000, 175616, 21952, 512, 1, 1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1
(list;
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OFFSET
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0,5
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COMMENTS
|
Diagonal of rational function R(x,y,z,t) = 1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z) with respect to x, y, z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t). - Gheorghe Coserea, Jul 01 2018
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LINKS
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Indranil Ghosh, Rows 0..120 of triangle, flattened
Jeffrey S. Geronimo, Hugo J. Woerdeman, and Chung Y. Wong, The autoregressive filter problem for multivariable degree one symmetric polynomials, arXiv:2101.00525 [math.CA], 2021.
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FORMULA
|
Row sums equal A000172, the Franel numbers.
Central terms are A002897(n) = C(2n,n)^3.
Antidiagonal sums equal A181545;
The g.f. of the antidiagonal sums is Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).
G.f. for column k: [Sum_{j=0..2k} A181544(k,j)*x^j]/(1-x)^(3k+1), where the row sums of A181544 equals De Bruijn's s(3,n) = (3n)!/(n!)^3.
G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1). - Paul D. Hanna, Nov 04 2010
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EXAMPLE
|
Triangle begins:
1;
1, 1;
1, 8, 1;
1, 27, 27, 1;
1, 64, 216, 64, 1;
1, 125, 1000, 1000, 125, 1;
1, 216, 3375, 8000, 3375, 216, 1;
1, 343, 9261, 42875, 42875, 9261, 343, 1;
1, 512, 21952, 175616, 343000, 175616, 21952, 512, 1;
1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1;
...
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MAPLE
|
T:= (n, k)-> binomial(n, k)^3:
seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Jan 06 2021
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MATHEMATICA
|
Flatten[Table[Binomial[n, k]^3, {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, May 23 2011 *)
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PROG
|
(PARI) T(n, k)=binomial(n, k)^3
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print())
(PARI) T(n, k)=polcoeff(polcoeff(sum(m=0, n, (3*m)!/m!^3*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(3*m+1)), n, x), k, y)
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Nov 04 2010
(PARI)
diag(expr, N=22, var=variables(expr)) = {
my(a = vector(N));
for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
for (n = 1, N, a[n] = expr;
for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
return(a);
};
x='x; y='y; z='z; t='t;
concat(apply(Vec, diag(1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z), 10, [x, y, z]))) \\ Gheorghe Coserea, Jul 01 2018
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CROSSREFS
|
Cf. A000172 (row sums), A181545 (antidiagonal sums), A002897, A181544, A248658.
Variants: A008459, A007318.
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KEYWORD
|
nonn,tabl
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AUTHOR
|
Paul D. Hanna, Oct 30 2010
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STATUS
|
approved
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A008979
|
|
a(n) = (6n)!/(n!)^6.
|
|
+10
12
|
| |
|
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OFFSET
|
0,2
|
|
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LINKS
|
Vincenzo Librandi, Table of n, a(n) for n = 0..100
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
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FORMULA
|
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*
binomial(5*n,n)*binomial(6*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) * ( [x^n](1 + x)^(6*n) ) = [x^n](F(x)^(720*n)), where F(x) = 1 + x + 4478*x^2 + 53085611*x^3 + 926072057094*x^4 + 19977558181209910*x^5 + 493286693783478576177*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A008978, A186420 and A188662. (End)
a(n) ~ 3^(6*n+1/2)*4^(3*n-1)/(Pi*n)^(5/2). - Ilya Gutkovskiy, Jul 12 2016
From Peter Bala, Feb 14 2020: (Start)
Supercongruences: a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z*u*v)^n] (1 + x + y + z + u + v)^(6*n). (End)
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MAPLE
|
seq( (6*n)!/(n!)^6, n=0..20); # G. C. Greubel, Feb 17 2020
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MATHEMATICA
|
Table[(6 n)!/(n)!^6, {n, 0, 20}] (* Vincenzo Librandi, Aug 13 2014 *)
|
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PROG
|
(MAGMA) [Factorial(6*n)/Factorial(n)^6: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014
(PARI) vector(21, n, my(m=n-1); (6*m)!/(m!)^6 ) \\ G. C. Greubel, Feb 17 2020
(Sage) [factorial(6*n)/factorial(n)^6 for n in (0..20)] # G. C. Greubel, Feb 17 2020
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CROSSREFS
|
Cf. A000897, A000984, A002894, A002897, A006480, A008977, A008978, A071549, A071550, A071551, A071552, A186420, A188662.
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KEYWORD
|
nonn,easy
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|
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AUTHOR
|
N. J. A. Sloane
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STATUS
|
approved
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A186420
|
|
a(n) = binomial(2n,n)^4.
|
|
+10
11
|
|
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|
1, 16, 1296, 160000, 24010000, 4032758016, 728933458176, 138735983333376, 27435582641610000, 5588044012339360000, 1165183173971324375296, 247639903129149250277376, 53472066459540320483696896, 11701285507234585729600000000, 2589980371199606611713600000000
(list;
graph;
refs;
listen;
history;
text;
internal format)
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|
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OFFSET
|
0,2
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|
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LINKS
|
Seiichi Manyama, Table of n, a(n) for n = 0..417
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FORMULA
|
a(n) = A000984(n)^4 = A002894(n)^2.
a(n) = binomial(2*n,n)^4 = ( [x^n](1 + x)^(2*n) )^4 = [x^n](F(x)^(16*n)), where F(x) = 1 + x + 25*x^2 + 1798*x^3 + 183442*x^4 + 22623769*x^5 + 3142959012*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 256^n/(Pi*n)^2. - Ilya Gutkovskiy, Jul 13 2016
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EXAMPLE
|
G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,1},256x) where 4F3 is a hypergeometric series.
|
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MATHEMATICA
|
Table[Binomial[2n, n]^4, {n, 0, 20}]
Table[Coefficient[Series[HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {1, 1, 1}, 256 x], {x, 0, n}], x, n], {n, 0, 14}] (* Michael De Vlieger, Jul 13 2016 *)
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|
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PROG
|
(Maxima)
makelist(binomial(2*n, n)^4, n, 0, 40);
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|
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CROSSREFS
|
Cf. binomial(2n,n)^k: A000984 (k=1), A002894 (k=2), A002897 (k=3), this sequence (k=4).
Cf. A000108, A000888, A186414, A186415, A186416, A186418, A186419, A000897, A006480, A008977, A188662.
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|
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KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
Emanuele Munarini, Feb 21 2011
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STATUS
|
approved
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