%I #8 Nov 09 2022 19:32:57

%S 1,1,2,1,9,9,1,34,96,64,1,125,750,1250,625,1,461,5265,16470,19440,

%T 7776,1,1715,35329,184877,386561,352947,117649,1,6434,232288,1913408,

%U 6307840,9863168,7340032,2097152,1,24309,1513656,18921924,92365758,220016574

%N Triangle, read by rows, such that row n equals the inverse binomial transform of row n of table A060543, where A060543(n,k) = C(n+n*k+k, n*k+k).

%C Row sums form A108292. Main diagonal is A000169(n) = (n+1)^n. Triangle with rows reversed is A108291.

%C The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (1+n*x)*(2+n*x)*...*(n-1+n*x)/(n-1)! in the basis made of the binomial(x+i,i). - _F. Chapoton_, Nov 01 2022

%e BINOMIAL[1, 9, 9] = {1, 10, 28, 55, 91, 136, 190, 253, ...}.

%e BINOMIAL[1, 34, 96, 64] = {1, 35, 165, 455, 969, 1771, 2925, ...}.

%e BINOMIAL[1, 125, 750, 1250, 625] = {1, 126, 1001, 3876, 10626, ...}.

%e Triangle begins:

%e 1;

%e 1, 2;

%e 1, 9, 9;

%e 1, 34, 96, 64;

%e 1, 125, 750, 1250, 625;

%e 1, 461, 5265, 16470, 19440, 7776;

%e 1, 1715, 35329, 184877, 386561, 352947, 117649;

%e 1, 6434, 232288, 1913408, 6307840, 9863168, 7340032, 2097152; ...

%o (PARI) {T(n,k)=local(X=x+x*O(x^k)); polcoeff(sum(j=0,n,binomial(n+n*j+j,n*j+j)*(x/(1+X))^j)/(1+X),k)}

%Y Cf. A108267, A060543, A108290, A000169.

%K nonn,tabl

%O 0,3

%A _Paul D. Hanna_, May 31 2005