

A039598


Triangle formed from oddnumbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.


67



1, 2, 1, 5, 4, 1, 14, 14, 6, 1, 42, 48, 27, 8, 1, 132, 165, 110, 44, 10, 1, 429, 572, 429, 208, 65, 12, 1, 1430, 2002, 1638, 910, 350, 90, 14, 1, 4862, 7072, 6188, 3808, 1700, 544, 119, 16, 1, 16796, 25194, 23256, 15504, 7752, 2907, 798, 152, 18, 1
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OFFSET

0,2


COMMENTS

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges.  Emeric Deutsch, Jan 15 2005
Riordan array ((12xsqrt(14x))/(2x^2),(12xsqrt(14x))/(2x)). Inverse array is A053122.  Paul Barry, Mar 17 2005
T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper halfplane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW.  Emeric Deutsch, Apr 15 2005
Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n1,0) + T(n1,1), T(n,k) = T(n1,k1) + 2*T(n1,k) + T(n1,k+1) for k>=1.  Philippe Deléham, Mar 30 2007
Number of (2n+1)step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu.  Philippe Deléham, Apr 16 2007, Apr 18 2007
Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths.  Philippe Deléham, Sep 25 2007
This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = x*T(n1,0) + T(n1,1), T(n,k) = T(n1,k1) + y*T(n1,k) + T(n1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) > A053121; (0,1) > A089942; (0,2) > A126093; (0,3) > A126970; (1,0) > A061554; (1,1) > A064189; (1,2) > A039599; (1,3) > A110877; (1,4) > A124576; (2,0) > A126075; (2,1) > A038622; (2,2) > A039598; (2,3) > A124733; (2,4) > A124575; (3,0) > A126953; (3,1) > A126954; (3,2) > A111418; (3,3) > A091965; (3,4) > A124574; (4,3) > A126791; (4,4) > A052179; (4,5) > A126331; (5,5) > A125906.  Philippe Deléham, Sep 25 2007
With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.
T(n, k) is also the number of orderpreserving full transformations (of an nchain) with exactly k fixed points.  Abdullahi Umar, Oct 02 2008
T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3.  Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008
The signed triangle S(n,k) := (1)^(nk)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1  5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n and ksummation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1  c(x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n)  5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(x) = 1 + 2*x  (x*c(x))^2.  Wolfdieter Lang, Aug 27 2012
O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:
((1+x)  C(z))/(x  (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)1)/x,C(x)1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section.  Wolfdieter Lang, Nov 13 2012
The Asequence for this Riordan triangle is [1,2,1] and the Zsequence is [2,1]. See a W. Lang link under A006232 with details and references.  Wolfdieter Lang, Nov 13 2012
T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular Ngon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)
The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122).  Tom Copeland, Dec 11 2015 (revised Dec 28 2015)
T(n,k) is the number of pairs of nonintersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976.  Peter Bala, Apr 12 2017


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. 796.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5648007388.


LINKS

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].
José Agapito, Ângela Mestre, Maria M. Torres, and Pasquale Petrullo, On OneParameter Catalan Arrays, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.1.
L. W. Shapiro, A Catalan triangle, Discrete Math. 14 (1976), no. 1, 8390. [Annotated scanned copy]


FORMULA

Row n: C(2n, nk)  C(2n, nk2).
a(n, k) = C(2n+1, nk)*2*(k+1)/(n+k+2) = A050166(n, nk) = a(n1, k1) + 2*a(n1, k)+ a (n1, k+1) [with a(0, 0) = 1 and a(n, k) = 0 if n<0 or n<k].  Henry Bottomley, Sep 24 2001
T(n, 0) = A000108(n+1), T(n, k) = 0 if n<k; for k>0, T(n, k) = Sum_{j=1..n} T(nj, k1)*A000108(j).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+2) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
Sum_{k>=0} T(m, k)*T(n, k) = A000108(m+n+1). (End)
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [2,2,2,...] in the main diagonal.  Gary W. Adamson, Dec 17 2006
G.f.: G(t,x) = C^2/(1txC^2), where C = (1sqrt(14x))/(2x) is the Catalan function. From here G(1,x)=C, i.e., the alternating row sums are the Catalan numbers (A000108).  Emeric Deutsch, Jan 20 2007
Sum_{k=0..n} T(n,k)*x^k = A000957(n+1), A000108(n), A000108(n+1), A001700(n), A049027(n+1), A076025(n+1), A076026(n+1) for x=2,1,0,1,2,3,4 respectively (see square array in A067345).  Philippe Deléham, Mar 21 2007, Nov 04 2011
G.f.: 1/(1xy2xx^2/(12xx^2/(12xx^2/(12xx^2/(12xx^2/(1.... (continued fraction).
Sum_{k=0..n} T(n,k)*x^(nk) = A000012(n), A001700(n), A194723(n+1), A194724(n+1), A194725(n+1), A194726(n+1), A194727(n+1), A194728(n+1), A194729(n+1), A194730(n+1) for x = 0,1,2,3,4,5,6,7,8,9 respectively.  Philippe Deléham, Nov 03 2011
This triangle factorizes in the Riordan group as ( C(x), x*C(x) ) * ( 1/(1  x), x/(1  x) ) = A033184 * A007318, where C(x) = (1  sqrt(1  4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Let U denote the lower unit triangular array with 1's on or below the main diagonal and zeros elsewhere. For k = 0,1,2,... define U(k) to be the lower unit triangular block array
/I_k 0\
\ 0 U/ having the k X k identity matrix I_k as the upper left block; in particular, U(0) = U. Then this array equals the biinfinite product (...*U(2)*U(1)*U(0))*(U(0)*U(1)*U(2)*...). (End)
O.g.f. G(x,t) = (1/x) * series reversion of ( x/f(x,t) ), where f(x,t) = ( 1 + (1 + t)*x )^2/( 1 + t*x ).
1 + x*d/dx(G(x,t))/G(x,t) = 1 + (2 + t)*x + (6 + 4*t + t^2)*x^2 + ... is the o.g.f for A094527. (End)
Conjecture: Sum_{k=0..n} T(n,k)/(k+1)^2 = H(n+1)*A000108(n)*(2*n+1)/(n+1), where H(n+1) = Sum_{k=0..n} 1/(k+1).  Werner Schulte, Jul 23 2015
Sum_{k=0..n} T(n,k)*(k+1)^2 = (2*n+1)*binomial(2*n,n). (A002457)
Sum_{k=0..n} T(n,k)*(k+1)^3 = 4^n*(3*n+2)/2.
Sum_{k=0..n} T(n,k)*(k+1)^4 = (2*n+1)^2*binomial(2*n,n).
Sum_{k=0..n} T(n,k)*(k+1)^5 = 4^n*(15*n^2+15*n+4)/4. (End)
The o.g.f. G(x,t) is such that G(x,t+1) is the o.g.f. for A035324, but with an offset of 0, and G(x,t1) is the o.g.f. for A033184, again with an offset of 0.  Peter Bala, Sep 20 2015


EXAMPLE

Triangle T(n,k) starts:
n\k 0 1 2 3 4 5 6 7 8 9 10
0: 1
1: 2 1
2: 5 4 1
3: 14 14 6 1
4: 42 48 27 8 1
5: 132 165 110 44 10 1
6: 429 572 429 208 65 12 1
7: 1430 2002 1638 910 350 90 14 1
8: 4862 7072 6188 3808 1700 544 119 16 1
9: 16796 25194 23256 15504 7752 2907 798 152 18 1
10: 58786 90440 87210 62016 33915 14364 4655 1120 189 20 1
Production matrix begins:
2, 1
1, 2, 1
0, 1, 2, 1
0, 0, 1, 2, 1
0, 0, 0, 1, 2, 1
0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 0, 1, 2, 1
Recurrence: T(5,1) = 165 = 1*42 + 2*48 +1*27. The Riordan Asequence is [1,2,1].
Recurrence from Riordan Zsequence [2,1]: T(5,0) = 132 = 2*42 + 1*48. (End)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^5 = 5*R(N, 2) + 4*R(N, 4) + 1*R(N, 6) = 5*S(1, rho(N)) + 4*S(3, rho(N)) + 1*S(5, rho(N)), identical in N >= 1. For N=5 (the pentagon with only one distinct diagonal) the degree delta(5) = 2, hence R(5, 4) and R(5, 6) can be reduced, namely to R(5, 1) = 1 and R(5, 6) = R(5,1) = 1, respectively. Thus rho(5)^5 = 5*R(N, 2) + 4*1 + 1*(1) = 3 + 5*R(N, 2) = 3 + 5*rho(5), with the golden section rho(5). (End)


MAPLE

T:=(n, k)>binomial(2*n, nk)  binomial(2*n, nk2); # N. J. A. Sloane, Aug 26 2013
# Uses function PMatrix from A357368. Adds row and column above and to the left.
PMatrix(10, n > binomial(2*n, n) / (n + 1)); # Peter Luschny, Oct 07 2022


MATHEMATICA

Flatten[Table[Binomial[2n, nk]  Binomial[2n, nk2], {n, 0, 9}, {k, 0, n}]] (* JeanFrançois Alcover, May 03 2011 *)


PROG

(Sage) # Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle.
D = [0]*(n+2); D[1] = 1
b = True; h = 1
for i in range(2*n) :
if b :
for k in range(h, 0, 1) : D[k] += D[k1]
h += 1
else :
for k in range(1, h, 1) : D[k] += D[k+1]
b = not b
if b : print([D[z] for z in (1..h1) ])
(Magma) /* As triangle: */ [[Binomial(2*n, nk)  Binomial(2*n, nk2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2015


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STATUS

approved



