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%I #6 Oct 21 2012 15:02:00
%S 1,1,1,1,17,1,1,98,98,1,1,354,2251,354,1,1,979,23803,23803,979,1,1,
%T 2275,158367,617036,158367,2275,1,1,4676,773842,8763293,8763293,
%U 773842,4676,1,1,8772,3031668,82498785,241082026,82498785,3031668,8772,1,1,15333
%N G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^5 * 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 Compare g.f. to that of the following triangle variants:
%C * Pascal's: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)*y^k] * x^n/n );
%C * Narayana: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n );
%C * A181143: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^3*y^k] * x^n/n );
%C * A181144: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^4*y^k] * x^n/n );
%C * A218116: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^6*y^k] * x^n/n ).
%e G.f.: A(x,y) = 1 + (1+y)*x + (1+17*y+y^2)*x^2 + (1+98*y+98*y^2+y^3)*x^3 + (1+354*y+2251*y^2+354*y^3+y^4)*x^4 +...
%e The logarithm of the g.f. equals the series:
%e log(A(x,y)) = (1 + y)*x
%e + (1 + 2^5*y + y^2)*x^2/2
%e + (1 + 3^5*y + 3^5*y^2 + y^3)*x^3/3
%e + (1 + 4^5*y + 6^5*y^2 + 4^5*y^3 + y^4)*x^4/4
%e + (1 + 5^5*y + 10^5*y^2 + 10^5*y^3 + 5^5*y^4 + y^5)*x^5/5 +...
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 17, 1;
%e 1, 98, 98, 1;
%e 1, 354, 2251, 354, 1;
%e 1, 979, 23803, 23803, 979, 1;
%e 1, 2275, 158367, 617036, 158367, 2275, 1;
%e 1, 4676, 773842, 8763293, 8763293, 773842, 4676, 1;
%e 1, 8772, 3031668, 82498785, 241082026, 82498785, 3031668, 8772, 1;
%e 1, 15333, 10057620, 575963523, 4066874561, 4066874561, 575963523, 10057620, 15333, 1; ...
%e Note that column 1 forms the sum of fourth powers (A000538).
%o (PARI) {T(n, k)=polcoeff(polcoeff(exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^5*y^j)*x^m/m)+O(x^(n+1))), n, x), k, y)}
%o for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
%Y Cf. A000538 (column 1), A218117 (row sums).
%Y Cf. variants: A001263 (Narayana), A181143, A181144, A218116.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Oct 21 2012
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