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 A002894 a(n) = binomial(2n, n)^2. (Formerly M3664 N1490) 108
 1, 4, 36, 400, 4900, 63504, 853776, 11778624, 165636900, 2363904400, 34134779536, 497634306624, 7312459672336, 108172480360000, 1609341595560000, 24061445010950400, 361297635242552100, 5445717990022688400, 82358080713306090000, 1249287673091590440000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 REFERENCES 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). LINKS 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. 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, 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, 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. FORMULA 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 EXAMPLE 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/ MAPLE A002894 := n-> binomial(2*n, n)^2. MATHEMATICA 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 *) PROG (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 CROSSREFS 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). Sequence in context: A019999 A291274 A239112 * A202828 A131765 A244559 Adjacent sequences:  A002891 A002892 A002893 * A002895 A002896 A002897 KEYWORD nonn,nice,easy AUTHOR EXTENSIONS Edited by N. J. A. Sloane, Feb 18 2016 STATUS approved

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Last modified April 21 12:18 EDT 2021. Contains 343150 sequences. (Running on oeis4.)