%I #15 Apr 01 2022 21:37:44
%S 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,
%T 1,8,79,370,693,617,248,39,1,9,119,722,1821,2284,1458,450,53,1,10,173,
%U 1337,4338,7243,6553,3211,773,74,1
%N Triangle read by rows: A000012 * A047812 as infinite lower triangular matrices.
%C Row sums = A014138: (1, 3, 8, 22, 64, 196, 625, ...).
%C From _Petros Hadjicostas_, Jun 01 2020: (Start)
%C 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).
%C 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)
%H R. K. Guy, <a href="/A007042/a007042_1.pdf">Parker's permutation problem involves the Catalan numbers</a>, preprint, 1992. (Annotated scanned copy)
%H R. K. Guy, <a href="http://www.jstor.org/stable/2324467">Parker's permutation problem involves the Catalan numbers</a>, Amer. Math. Monthly 100 (1993), 287-289.
%F T(n,k) = Sum_{s=k+1..n} A047812(s,k) for n >= 1 and 0 <= k <= n-1. - _Petros Hadjicostas_, Jun 01 2020
%e Triangle T(n,k) (with rows n >= 1 and columns k = 0..n-1) begins:
%e 1;
%e 2, 1;
%e 3, 4, 1;
%e 4, 9, 8, 1;
%e 5, 18, 28, 12, 1;
%e 6, 31, 76, 63, 19, 1;
%e 7, 51, 176, 232, 131, 27, 1;
%e ...
%o (PARI) A(n, k) = polcoeff(prod(j=0, n-1, (1-q^(2*n-j))/(1-q^(j+1)) ), k*(n+1) );
%o T(n,k) = sum(s=k+1, n, A(s,k));
%o vector(15, n, vector(n, k, T(n, k-1))) \\ _Petros Hadjicostas_, Jun 01 2020
%Y Cf. A000012, A014138, A047812.
%K nonn,tabl
%O 0,2
%A _Gary W. Adamson_, Jan 30 2008