OFFSET
1,1
LINKS
MathOverflow, How to prove this polynomial always has integer values at all integers?, Jun 13 2015.
Wilberd van der Kallen, How to prove this polynomial always has integer values at all integers, arXiv:1509.08811 [math.NT], 2015.
FORMULA
T(0, 0) = 3/2, so sequence here as offset 1.
T(n, k) = 0 for k>n, so only the terms with k<=n are represented here.
EXAMPLE
Triangle starts:
-2;
0, 6;
0, 0, 24;
0, 0, 4, 118;
0, 0, 0, 60, 696;
0, 0, 0, 12, 720, 4824;
0, 0, 0, 0, 336, 8288, 38240;
0, 0, 0, 0, 60, 6516, 95928, 336822;
...
MATHEMATICA
Table[Sum[Sum[3 (-1)^(k + j) Binomial[2 k, k] Binomial[j, i] Binomial[n, i] Binomial[i, n - j]/(2 (2 i - 1) (2 j + 1) (2 n - 2 i - 1)), {j, k, n}], {i, 0, n}], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Oct 04 2015 *)
PROG
(PARI) d(n, k) = sum(i=0, n, sum(j=k, n, 3*(-1)^(k+j)*binomial(2*k, k)*binomial(j, i)*binomial(n, i)*binomial(i, n-j)/(2*(2*i-1)*(2*j+1)*(2*n-2*i-1))));
tabl(nn) = {for (n=1, nn, for (k=1, n, print1(d(n, k), ", "); ); print(); ); }
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Michel Marcus, Oct 04 2015
STATUS
approved