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%I #4 Sep 07 2013 20:16:25
%S 1,1,1,1,3,1,1,6,12,1,1,10,71,76,1,1,15,281,2153,701,1,1,21,861,29166,
%T 129509,8477,1,1,28,2212,244725,7664343,12391414,126126,1,1,36,4998,
%U 1477391,218030412,3875325345,1699148352,2223278,1,1,45,10242,7017577,3748460115,448713017405,3284369541969,315158247170,45269999,1
%N Triangle defined by g.f. A(x,y) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n, k)^(k+1) * y^k ), as read by rows.
%C Note that the following g.f. does NOT yield an integer triangle: exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n, k)^k * y^k ).
%e This triangle begins:
%e 1;
%e 1, 1;
%e 1, 3, 1;
%e 1, 6, 12, 1;
%e 1, 10, 71, 76, 1;
%e 1, 15, 281, 2153, 701, 1;
%e 1, 21, 861, 29166, 129509, 8477, 1;
%e 1, 28, 2212, 244725, 7664343, 12391414, 126126, 1;
%e 1, 36, 4998, 1477391, 218030412, 3875325345, 1699148352, 2223278, 1;
%e 1, 45, 10242, 7017577, 3748460115, 448713017405, 3284369541969, 315158247170, 45269999, 1; ...
%e ...
%e G.f.: A(x,y) = 1 + (1+y)*x + (1+3*y+y^2)*x^2 + (1+6*y+12*y^2+y^3)*x^3 + (1+10*y+71*y^2+76*y^3+y^4)*x^4 + (1+15*y+281*y^2+2153*y^3+701*y^4+y^5)*x^5 +...
%e The logarithm of the g.f. equals the series:
%e log(A(x)) = (1 + x)*x
%e + (1 + 2^2*x + x^2)*x^2/2
%e + (1+ 3^2*y + 3^3*y^2 + y^3)*x^3/3
%e + (1+ 4^2*y + 6^3*y^2 + 4^4*y^3 + x^4)*x^4/4
%e + (1+ 5^2*y + 10^3*y^2 + 10^4*y^3 + 5^5*y^4 + y^5)*x^5/5
%e + (1+ 6^2*y + 15^3*y^2 + 20^4*y^3 + 15^5*y^4 + 6^6*y^5 + y^6)*x^6/6 +...
%e in which the coefficients form A219207(n,k) = binomial(n, k)^(k+1).
%o (PARI) {T(n, k)=polcoeff(polcoeff(exp(sum(m=1, n, x^m/m*sum(j=0, m, binomial(m, j)^(j+1)*y^j))+x*O(x^n)), n, x), k, y)}
%o for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
%Y Cf. A184730 (row sums), A181070 (antidiagonal sums), A060946 (diagonal).
%Y Cf. related triangles: A219207, A209424, A228904.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Sep 07 2013
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