A118354 Convolution triangle, read by rows, where diagonals are successive self-convolutions of A118351.

1, 1, 0, 1, 1, 0, 1, 2, 6, 0, 1, 3, 13, 42, 0, 1, 4, 21, 96, 325, 0, 1, 5, 30, 163, 770, 2688, 0, 1, 6, 40, 244, 1353, 6530, 23286, 0, 1, 7, 51, 340, 2093, 11760, 57612, 208659, 0, 1, 8, 63, 452, 3010, 18636, 105681, 523446, 1918314, 0, 1, 9, 76, 581, 4125, 27441, 170580, 973953, 4864795, 17994264, 0

Offset: 0

Author

Paul D. Hanna, Apr 26 2006

Keywords

• nonn
• tabl

Comments

A118351 equals the central terms of pendular triangle A118350 and the lower diagonals of this triangle form the semi-diagonals of the triangle A118350.

Examples

			Show: T(n,k) = T(n-1,k) - 3*T(n-1,k-1) + 3*T(n,k-1) + T(n+1,k-1)
at n=8,k=4: T(8,4) = T(7,4) - 3*T(7,3) + 3*T(8,3) + T(9,3)
or: 2093 = 1353 - 3*244 + 3*340 + 452.
Triangle begins:
  1;
  1, 0;
  1, 1,  0;
  1, 2,  6,   0;
  1, 3, 13,  42,    0;
  1, 4, 21,  96,  325,     0;
  1, 5, 30, 163,  770,  2688,      0;
  1, 6, 40, 244, 1353,  6530,  23286,      0;
  1, 7, 51, 340, 2093, 11760,  57612, 208659,       0;
  1, 8, 63, 452, 3010, 18636, 105681, 523446, 1918314,        0;
  1, 9, 76, 581, 4125, 27441, 170580, 973953, 4864795, 17994264, 0; ...
		

Links

• G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Crossrefs

Cf. A118350, A118351, A118352, A118353.

Row sums: A151616.

Programs

Mathematica

T[n_, k_]:= T[n, k]= If[k==0, 1, If[k==n, 0, T[n-1, k] -3*T[n-1, k-1] +3*T[n, k-1] +T[n+1, k-1]]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2021 *)

PARI

{T(n,k)=polcoeff((serreverse(x*(1-3*x+sqrt((1-3*x)*(1-7*x)+x*O(x^k)))/2/(1-3*x))/x)^(n-k),k)}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

Sage

@CachedFunction
def T(n, k):
    if (k==0): return 1
    elif (k==n): return 0
    else: return T(n-1, k) - 3*T(n-1, k-1) + 3*T(n, k-1) + T(n+1, k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 18 2021

Formula

Since g.f. G=G(x) of A118351 satisfies: G = 1 - 3*x*G + 3*x*G^2 + x*G^3 then

T(n,k) = T(n-1,k) - 3*T(n-1,k-1) + 3*T(n,k-1) + T(n+1,k-1).

Recurrence involving antidiagonals:

T(n,k) = T(n-1,k) + Sum_{j=1..k} [4*T(n-1+j,k-j) - 3*T(n-2+j,k-j)] for n>k>=0.