login
a(n) = Sum_{k=0..[n/2]} binomial((n-k)*k, k^2).
3

%I #9 Jan 27 2019 05:50:07

%S 1,1,2,3,5,20,77,437,5509,54475,1031232,31874836,789351469,

%T 47552777430,3302430043985,223753995897916,39177880844093733,

%U 5954060239110086680,1226026438114057710320,551315671593483499670137,188615011023291125237647365,124995445742889226418307452940

%N a(n) = Sum_{k=0..[n/2]} binomial((n-k)*k, k^2).

%C Equals antidiagonal sums of triangle A228832.

%F Limit n->infinity a(n)^(1/n^2) = ((1-r)/(1-2*r))^(r/2) = 1.171233876693210503..., where r = A323773 = 0.366320150305283... is the root of the equation (1-2*r)^(4*r-1) * (1-r)^(1-2*r) = r^(2*r). - _Vaclav Kotesovec_, Sep 06 2013

%t Table[Sum[Binomial[(n-k)*k, k^2],{k,0,Floor[n/2]}],{n,0,15}] (* _Vaclav Kotesovec_, Sep 06 2013 *)

%o (PARI) {a(n)=sum(k=0,n\2,binomial(n*k-k^2, k^2))}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A228832.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 04 2013