OFFSET
1,3
LINKS
G. C. Greubel, Rows n = 1..50 of the triangle, flattened
G. E. Andrews, W. Gawronski and L. L. Littlejohn, The Legendre-Stirling Numbers
G. E. Andrews et al., The Legendre-Stirling numbers, Discrete Math., 311 (2011), 1255-1272.
FORMULA
From G. C. Greubel, Jun 06 2021: (Start)
T(n, k) = Ps(n, n-k+1), where Ps(n, k) = Sum_{j=0..k} (-1)^(j+k)*(2*j+1)*j^n*(1 + j)^n/((j+k+1)!*(k-j)!).
Sum_{k=1..n} T(n, k) = A135921(n). (End)
EXAMPLE
Triangle begins:
1;
1 2;
1 8 4;
1 20 52 8;
1 40 292 320 16;
1 70 1092 3824 1936 32;
1 112 3192 25664 47824 11648 64;
1 168 7896 121424 561104 585536 69952 128;
...
MATHEMATICA
Ps[n_, k_]:= Sum[(-1)^(j+k)*(2*j+1)*j^n*(1+j)^n/((j+k+1)!*(k-j)!), {j, 0, k}];
Table[Ps[n, n-k+1], {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jun 06 2021 *)
PROG
(PARI) T071951(n, k) = sum(i=0, k, (-1)^(i+k) * (2*i + 1) * (i*i + i)^n / (k-i)! / (k+i+1)! );
for (n=1, 10, for (k=1, n, print1(T071951(n, n-k+1), ", ")); print); \\ Michel Marcus, Nov 24 2019
(Sage)
def Ps(n, k): return sum( (-1)^(j+k)*(2*j+1)*j^n*(1+j)^n/(factorial(j+k+1) * factorial(k-j)) for j in (0..k) )
flatten([[Ps(n, n-k+1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Jun 06 2021
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
N. J. A. Sloane, Jun 19 2011
EXTENSIONS
More terms from Omar E. Pol, Jan 10 2012
More terms from Michel Marcus, Nov 24 2019
STATUS
approved