A089942 Inverse binomial matrix applied to A039599.

1, 0, 1, 1, 1, 1, 1, 3, 2, 1, 3, 6, 6, 3, 1, 6, 15, 15, 10, 4, 1, 15, 36, 40, 29, 15, 5, 1, 36, 91, 105, 84, 49, 21, 6, 1, 91, 232, 280, 238, 154, 76, 28, 7, 1, 232, 603, 750, 672, 468, 258, 111, 36, 8, 1, 603, 1585, 2025, 1890, 1398, 837, 405, 155, 45, 9, 1, 1585, 4213, 5500

Offset: 0

Author

Paul Barry, Nov 16 2003

Keywords

• nonn
• tabl
• changed

Comments

Reverse of A071947 - related to lattice paths. First column is A005043.

Triangle T(n,k), 0 <= k <= n, defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Feb 27 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(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k >= 1. Other triangles arise from 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

Riordan array (f(x),x*g(x)), where f(x)is the o.g.f. of A005043 and g(x)is the o.g.f. of A001006. - Philippe Deléham, Nov 22 2009

Riordan array ((1+x-sqrt(1-2x-3x^2))/(2x(1+x)), (1-x-sqrt(1-2x-3x^2))/(2x)). Inverse of Riordan array ((1+x)/(1+x+x^2),x/(1+x+x^2)). E.g.f. of column k is exp(x)*(Bessel_I(k,2x)-Bessel_I(k+1,2x)).

Diagonal sums are A187306.

Simultaneous equations using the first n rows solve for diagonal lengths of odd N = (2n+1) regular polygons, with constants c^0, c^1, c^2, ...; where c = 1 + 2*cos( 2*Pi/N) = sin(3*Pi/N)/sin(Pi/N) = the third longest diagonal of N>5. By way of example, take the first 4 rows relating to the 9-gon (nonagon), N=(2*4 + 1), with c = 1 + 2*cos(2*Pi/9) = 2.5320888.... The simultaneous equations are (1,0,0,0) = 1; (0,1,0,0) = c; (1,1,1,0) = c^2, (1,3,2,1) = c^3. The answers are 1, 2.532..., 2.879..., and 1.879...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. - Gary W. Adamson, Sep 07 2011

Number of linearly independent irreducible representations of a given weight j in a tensor of given rank n. - Mikkel N. Schmidt, Aug 20 2025

Examples

			Triangle begins
   1,
   0,   1,
   1,   1,   1,
   1,   3,   2,   1,
   3,   6,   6,   3,   1,
   6,  15,  15,  10,   4,  1,
  15,  36,  40,  29,  15,  5,  1,
  36,  91, 105,  84,  49, 21,  6, 1,
  91, 232, 280, 238, 154, 76, 28, 7, 1
Production matrix is
  0, 1,
  1, 1, 1,
  0, 1, 1, 1,
  0, 0, 1, 1, 1,
  0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 0, 1, 1, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 1, 1
		

Links

• G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
• Paul Barry and Aoife Hennessy, Four-term Recurrences, Orthogonal Polynomials and Riordan Arrays, Journal of Integer Sequences, 2012, article 12.4.2. - From _N. J. A. Sloane_, Sep 21 2012
• J. A. R. Coope, R. F. Snider, and F. R. McCourt, Irreducible Cartesian Tensors, The Journal of Chemical Physics, vol 43, no 7, 1965.
• Emeric Deutsch, Luca Ferrari, and Simone Rinaldi, Production Matrices, Advances in Applied Mathematics, 34 (2005) pp. 101-122.
• Toufik Mansour and Mark Shattuck, Enumeration of Catalan and smooth words according to capacity, Integers (2025) Vol. 25, Art. No. A5. See p. 29.
• Donatella Merlini, Douglas G. Rogers, Renzo Sprugnoli, and M. Cecilia Verri, On some alternative characterizations of Riordan arrays, Canad J. Math., 49 (1997), 301-320.
• Yidong Sun and Luping Ma, Minors of a class of Riordan arrays related to weighted partial Motzkin paths, Eur. J. Comb. (2014) Vol. 39, 157-169. See Table 2.2.

Crossrefs

Row sums give A002426 (central trinomial coefficients).

Programs

Maple

T:= (n,k) -> simplify(GegenbauerC(n-k,-n+1,-1/2)-GegenbauerC(n-k-1,-n+1,-1/2)): for n from 1 to 9 do seq(T(n,k), k=1..n) od; # Peter Luschny, May 12 2016
# Or by recurrence:
T := proc(n, k) option remember;
if n = k then 1 elif k < 0 or n < 0 or k > n then 0
elif k = 0 then T(n-1, 1) else T(n-1, k-1) + T(n-1, k) + T(n-1, k+1) fi end:
for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # Peter Luschny, May 25 2021

Mathematica

T[n_, k_] := GegenbauerC[n - k, -n + 1, -1/2] - GegenbauerC[n - k - 1, -n + 1, -1/2]; Table[T[n, k], {n,1,10}, {k,1,n}] // Flatten (* G. C. Greubel, Feb 28 2017 *)

Formula

G.f.: (1+z-q)/[(1+z)(2z-t+tz+tq)], where q = sqrt(1-2z-3z^2).

Sum_{k>=0} T(m,k)*T(n,k) = T(m+n,0) = A005043(m+n). - Philippe Deléham, Mar 22 2007

Sum_{k=0..n} T(n,k)*(2k+1) = 3^n. - Philippe Deléham, Mar 22 2007

Sum_{k=0..n} T(n,k)*2^k = A112657(n). - Philippe Deléham, Apr 01 2007

T(n,2k) + T(n,2k+1) = A109195(n,k). - Philippe Deléham, Nov 11 2008

T(n,k) = GegenbauerC(n-k,-n+1,-1/2) - GegenbauerC(n-k-1,-n+1,-1/2) for 1 <= k <= n. - Peter Luschny, May 12 2016

Extensions

Edited by Emeric Deutsch, Mar 04 2004