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A262886 Regular triangle read by rows: T(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)))). 0
-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, 0, 0, 0, 0, 0, 2520, 109872, 1131732, 3215544, 0, 0, 0, 0, 0, 392, 67904, 1735320, 13647840, 32651544 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,1

LINKS

Table of n, a(n) for n=1..55.

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

Sequence in context: A047918 A321981 A322481 * A138701 A332400 A274878

Adjacent sequences:  A262883 A262884 A262885 * A262887 A262888 A262889

KEYWORD

sign,tabl

AUTHOR

Michel Marcus, Oct 04 2015

STATUS

approved

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Last modified September 21 22:33 EDT 2020. Contains 337274 sequences. (Running on oeis4.)