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A181144 G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^4*y^k] * x^n/n ) = Sum_{n>=0,k=0..n} T(n,k)*x^n*y^k, as a triangle of coefficients T(n,k) read by rows.

%I A181144 #7 Oct 22 2012 12:41:13
%S A181144 1,1,1,1,9,1,1,36,36,1,1,100,419,100,1,1,225,2699,2699,225,1,1,441,
%T A181144 12138,35052,12138,441,1,1,784,42865,286206,286206,42865,784,1,1,1296,
%U A181144 127191,1696820,3932898,1696820,127191,1296,1,1,2025,330903,7958563
%N A181144 G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^4*y^k] * x^n/n ) = Sum_{n>=0,k=0..n} T(n,k)*x^n*y^k, as a triangle of coefficients T(n,k) read by rows.
%C A181144 Compare g.f. to that of the following triangle variants:
%C A181144 * Pascal's: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)*y^k] * x^n/n );
%C A181144 * Narayana: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n );
%C A181144 * A181143: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^3*y^k] * x^n/n );
%C A181144 * A218115: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^5*y^k] * x^n/n );
%C A181144 * A218116: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^6*y^k] * x^n/n ).
%e A181144 G.f.: A(x,y) = 1 + (1+y)*x + (1+9*y+y^2)*x^2 + (1+36*y+36*y^2+y^3)*x^3 + (1+100*y+419*y^2+100*y^3+y^4)*x^4 +...
%e A181144 The logarithm of the g.f. equals the series:
%e A181144 log(A(x,y)) = (1 + y)*x
%e A181144 + (1 + 2^4*y + y^2)*x^2/2
%e A181144 + (1 + 3^4*y + 3^4*y^2 + y^3)*x^3/3
%e A181144 + (1 + 4^4*y + 6^4*y^2 + 4^4*y^3 + y^4)*x^4/4
%e A181144 + (1 + 5^4*y + 10^4*y^2 + 10^4*y^3 + 5^4*y^4 + y^5)*x^5/5 +...
%e A181144 Triangle begins:
%e A181144 1;
%e A181144 1, 1;
%e A181144 1, 9, 1;
%e A181144 1, 36, 36, 1;
%e A181144 1, 100, 419, 100, 1;
%e A181144 1, 225, 2699, 2699, 225, 1;
%e A181144 1, 441, 12138, 35052, 12138, 441, 1;
%e A181144 1, 784, 42865, 286206, 286206, 42865, 784, 1;
%e A181144 1, 1296, 127191, 1696820, 3932898, 1696820, 127191, 1296, 1;
%e A181144 1, 2025, 330903, 7958563, 36955542, 36955542, 7958563, 330903, 2025, 1;
%e A181144 1, 3025, 776688, 31205941, 261852055, 525079969, 261852055, 31205941, 776688, 3025, 1; ...
%e A181144 Note that column 1 forms the sum of cubes (A000537), and forms the squares of the triangular numbers.
%e A181144 Inverse binomial transform of columns begins:
%e A181144 [1];
%e A181144 [1, 8, 19, 18, 6];
%e A181144 [1, 35, 348, 1549, 3713, 5154, 4161, 1818, 333];
%e A181144 [1, 99, 2500, 27254, 161793, 589819, 1409579, 2282850, 2529900, 1893972, 917349, 259854, 32726]; ...
%o A181144 (PARI) {T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^4*y^j)*x^m/m)+O(x^(n+1))),n,x),k,y)}
%o A181144 for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
%Y A181144 Cf. A000537 (column 1), A166992 (row sums), A166898 (antidiagonal sums), A218140.
%Y A181144 Cf. variants: A001263 (Narayana), A181143, A218115, A218116.
%K A181144 nonn,tabl
%O A181144 0,5
%A A181144 _Paul D. Hanna_, Oct 13 2010