A137614 Triangle read by rows: A000012 * A047812 as infinite lower triangular matrices.

1, 2, 1, 3, 4, 1, 4, 9, 8, 1, 5, 18, 28, 12, 1, 6, 31, 76, 63, 19, 1, 7, 51, 176, 232, 131, 27, 1, 8, 79, 370, 693, 617, 248, 39, 1, 9, 119, 722, 1821, 2284, 1458, 450, 53, 1, 10, 173, 1337, 4338, 7243, 6553, 3211, 773, 74, 1

Offset

0,2

Comments

Row sums = A014138: (1, 3, 8, 22, 64, 196, 625, ...).

From Petros Hadjicostas, Jun 01 2020: (Start)

We prove the claim above. From Guy (1992, 1993), we know that A000108(n) = Sum_{k=0..n-1} A047812(k) (the row sums of Parker's triangle are Catalan numbers).

We then have Sum_{k=0..n-1} T(n,k) = Sum_{k=0..n-1} Sum_{s=k+1..n} A047812(s,k) = Sum_{s=1..n} Sum_{k=0..s-1} A047812(s,k) = Sum_{s=1..n} A000108(s) = A014138(n) because A014138 contains partial sums of the Catalan numbers. (End)

Links

Table of n, a(n) for n=0..54.

R. K. Guy, Parker's permutation problem involves the Catalan numbers, preprint, 1992. (Annotated scanned copy)

R. K. Guy, Parker's permutation problem involves the Catalan numbers, Amer. Math. Monthly 100 (1993), 287-289.

Formula

T(n,k) = Sum_{s=k+1..n} A047812(s,k) for n >= 1 and 0 <= k <= n-1. - Petros Hadjicostas, Jun 01 2020

Example

Triangle T(n,k) (with rows n >= 1 and columns k = 0..n-1) begins:
  1;
  2,  1;
  3,  4,   1;
  4,  9,   8,   1;
  5, 18,  28,  12,   1;
  6, 31,  76,  63,  19,  1;
  7, 51, 176, 232, 131, 27, 1;
  ...

Prog

(PARI) A(n, k) = polcoeff(prod(j=0, n-1, (1-q^(2*n-j))/(1-q^(j+1)) ), k*(n+1) );
T(n, k) = sum(s=k+1, n, A(s, k));
vector(15, n, vector(n, k, T(n, k-1))) \\ Petros Hadjicostas, Jun 01 2020

Crossrefs

Cf. A000012, A014138, A047812.

Sequence in context: A107616 A055208 A051128 * A204213 A143326 A327086

Adjacent sequences: A137611 A137612 A137613 * A137615 A137616 A137617

Keyword

nonn,tabl

Author

Gary W. Adamson, Jan 30 2008

Status

approved