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G.f.: exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n^2, k*(n-k))*x^k ).
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%I #8 Sep 07 2013 20:34:34

%S 1,1,2,4,17,171,3171,101741,7181615,1274607729,428568152553,

%T 223160743256395,185627109707405932,320952534083059792786,

%U 1367454166673309618606950,11078799748881429582280609036,137939599816546528357634500253053,2679390013936303204526656964298150849

%N G.f.: exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n^2, k*(n-k))*x^k ).

%C The logarithmic derivative yields A207138.

%C Equals the antidiagonal sums of triangle A228900.

%e G.f.: A(x) = 1 + x + 2*x^2 + 4*x^3 + 17*x^4 + 171*x^5 + 3171*x^6 +...

%e where the logarithm of the g.f. equals the l.g.f. of A207138:

%e log(A(x)) = x + 3*x^2/2 + 7*x^3/3 + 51*x^4/4 + 761*x^5/5 + 17913*x^6/6 +...

%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n,x^m/m*sum(k=0,m,binomial(m^2,k*(m-k))*x^k))+x*O(x^n)),n)}

%o for(n=0,25,print1(a(n),", "))

%Y Cf. A207138 (log), A207135, A228900, A206850, A206830, A167006.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 15 2012