OFFSET
0,3
COMMENTS
G. Kreweras explains that since the rows of A140136 are symmetric, they can be considered as linear combinations of the odd-indexed rows of the Pascal triangle. For instance, (1,1) = 1*(1,1) and (1,7,7,1) = 1*(1,3,3,1) + 4*(0,1,1,0) and (1,20,75,75,10,1) = 1*(1,5,10,10,5,1) + 15*(0,1,3,3,1) + 20*(0,0,1,1,0,0). These coefficients (1; 1, 4; 1, 15, 20;) are the rows of this triangle. - Michel Marcus, Nov 17 2014
LINKS
G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Germain Kreweras, Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, 6 (1965).
Germain Kreweras, page 93 of "Sur une classe de problèmes de dénombrement...", containing the defining formula for this sequence.
FORMULA
T(n,n) = A003645(n).
EXAMPLE
Triangle begins:
0: 1;
1: 1, 4;
2: 1, 15, 20;
3: 1, 35, 168, 112;
4: 1, 66, 714, 1680, 672;
5: 1, 110, 2178, 11352, 15840, 4224;
6: 1, 169, 5434, 51051, 156156, 144144, 27456;
7: 1, 245, 11830, 178035, 972400, 1953952, 1281280, 183040;
8: 1, 340, 23324, 520676, 4516798, 16102944, 22870848, 11202048, 1244672;
.....
MATHEMATICA
T[0, 0] := 1; T[0, k_] := 0; T[n_, k_] := T[n, k] = (2*n + 1)*(4*(2*n - 2*k + 1)*T[n - 1, k - 1] + (n + 2*k + 2)*T[n - 1, k])/((n + 2)*(2*n - 2*k + 1)); Table[If[k < 0, 0, T[n, k]], {n, 0, 5}, {k, 0, n}]//Flatten (* G. C. Greubel, Oct 13 2017 *)
PROG
(Sage)
@CachedFunction
def T(n, k):
if k < 0: return 0
if n < 0: return 0
if n == 0: return int( k==0 )
if k == 0: return 1
return ( (2*n+1)*( 4*(2*n-2*k+1)*T(n-1, k-1) + (n+2*k+2)*T(n-1, k) ) ) / ((n+2)*(2*n-2*k+1))
for n in [0..16]:
print([T(n, k) for k in range(0, n+1)])
# Joerg Arndt, Nov 21 2014
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Philippe Deléham, Oct 13 2006
EXTENSIONS
Corrected name, added more terms, Joerg Arndt, Nov 21 2014
STATUS
approved