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A100231 G.f. A(x) satisfies: 5^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (5+z)^n - (1+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.

%I A100231 #7 Jun 14 2015 00:59:29
%S A100231 1,3,4,-8,0,64,-192,-128,2816,-7680,-13312,157696,-352256,-1179648,
%T A100231 9748480,-16220160,-99614720,630456320,-651427840,-8218214400,
%U A100231 41481666560,-13191086080,-667334737920,2724661821440,1460876083200,-53446942130176,175607589634048,286761410363392
%N A100231 G.f. A(x) satisfies: 5^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (5+z)^n - (1+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.
%C A100231 More generally, if g.f. A(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]A(x)^n, then A(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z and A(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2.
%F A100231 a(n)=-((4*n-6)*a(n-1)+20*(n-3)*a(n-2))/n for n>2, with a(0)=1, a(1)=3, a(3)=4. G.f.: A(x) = (1+4*x+sqrt(1+4*x+20*x^2))/2.
%e A100231 From the table of powers of A(x) (A100232), we see that
%e A100231 5^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
%e A100231 A^1=[1,3],4,-8,0,64,-192,-128,...
%e A100231 A^2=[1,6,17],8,-32,64,64,-896,...
%e A100231 A^3=[1,9,39,75],12,-72,256,-384,...
%e A100231 A^4=[1,12,70,220,321],16,-128,640,...
%e A100231 A^5=[1,15,110,470,1165,1363],20,-200,...
%e A100231 A^6=[1,18,159,852,2895,5922,5777],24,...
%o A100231 (PARI) a(n)=if(n==0,1,(5^n-1-sum(k=0,n,polcoeff(sum(j=0,min(k,n-1),a(j)*x^j)^n+x*O(x^k),k)))/n)
%o A100231 (PARI) a(n)=if(n==0,1,if(n==1,3,if(n==2,4,-((4*n-6)*a(n-1)+20*(n-3)*a(n-2))/n)))
%o A100231 (PARI) a(n)=polcoeff((1+4*x+sqrt(1+4*x+20*x^2+x^2*O(x^n)))/2,n)
%Y A100231 Cf. A100232, A100233, A100228.
%K A100231 sign
%O A100231 0,2
%A A100231 _Paul D. Hanna_, Nov 29 2004