A185967 Inverse of Riordan array ((1-x)(1-x^2)(1-x^3)/(1-x^6), x(1-x)(1-x^2)(1-x^3)/(1-x^6)).

1, 1, 1, 3, 2, 1, 10, 7, 3, 1, 37, 26, 12, 4, 1, 146, 103, 49, 18, 5, 1, 602, 426, 207, 80, 25, 6, 1, 2563, 1818, 897, 359, 120, 33, 7, 1, 11181, 7946, 3966, 1628, 570, 170, 42, 8, 1, 49720, 35389, 17823, 7458, 2701, 852, 231, 52, 9, 1, 224540, 160024, 81177, 34484, 12815, 4212, 1218, 304, 63, 10, 1

Offset: 0

Author

Paul Barry, Feb 07 2011

Keywords

• nonn
• easy
• tabl

Comments

Riordan array (g(x),xg(x)) where x*g(x) = (x+2)/3 - 2*sqrt(1+x+x^2) * cos(arccos(-(2x^3+3x^2+24x-2) / (2(1+x+x^2)^(3/2)))/3)/3.

Examples

			Triangle begins
  1,
  1, 1,
  3, 2, 1,
  10, 7, 3, 1,
  37, 26, 12, 4, 1,
  146, 103, 49, 18, 5, 1,
  602, 426, 207, 80, 25, 6, 1,
  2563, 1818, 897, 359, 120, 33, 7, 1,
  11181, 7946, 3966, 1628, 570, 170, 42, 8, 1,
  49720, 35389, 17823, 7458, 2701, 852, 231, 52, 9, 1,
  224540, 160024, 81177, 34484, 12815, 4212, 1218, 304, 63, 10, 1
Production matrix is
  1, 1,
  2, 1, 1,
  3, 2, 1, 1,
  4, 3, 2, 1, 1,
  5, 4, 3, 2, 1, 1,
  6, 5, 4, 3, 2, 1, 1,
  7, 6, 5, 4, 3, 2, 1, 1,
  8, 7, 6, 5, 4, 3, 2, 1, 1,
  9, 8, 7, 6, 5, 4, 3, 2, 1, 1
  10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1
		

Links

• Emeric Deutsch, Luca Ferrari, and Simone Rinaldi, Production Matrices, Advances in Mathematics, 34 (2005) pp. 101-122.
• JiSun Huh, Sangwook Kim, Seunghyun Seo, and Heesung Shin, Bijections on pattern avoiding inversion sequences and related objects, arXiv:2404.04091 [math.CO], 2024. See p. 22.

Crossrefs

Inverse of number triangle A185962.

First column is A109081. Row sums are A106228(n+1).

Programs

Maple

T := (n, k) -> `if`(n=k, 1, (1 + k)*(n - k)*hypergeom([1 + k - n, -n, 1 - k + n], [3/2, 2], 1/4)):
seq(seq(simplify(T(n, k)), k=0..n),n=0..10); # Peter Luschny, Apr 02 2019

Mathematica

T[n_, k_] := (k+1)/(n+1) Sum[Binomial[n+1, i] Binomial[n-k+i-1, n-k-i], {i, 0, n-k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 25 2019, after Vladimir Kruchinin *)

Maxima

T(n,k):=(k+1)/(n+1)*sum(binomial(n+1,i)*binomial(n-k+i-1,n-k-i),i,0,n-k); /* Vladimir Kruchinin, Apr 02 2019 */

Formula

T(n, k) = (k + 1)/(n + 1)*Sum_{i=0..n-k} C(n+1, i)*C(n-k+i-1, n-k-i). - Vladimir Kruchinin, Apr 02 2019

T(n, k) = (1 + k)*(n - k)*hypergeom([1 + k - n, -n, 1 - k + n], [3/2, 2], 1/4) for k < n. - Peter Luschny, Apr 02 2019