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A100234 G.f. A(x) satisfies: 6^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (6+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. 2

%I #8 Jun 14 2015 00:58:59

%S 1,4,5,-15,20,90,-695,1785,3895,-53985,196255,121635,-4907130,

%T 23332140,-13181145,-470127465,2866898820,-4455872910,-44776087145,

%U 356263904235,-873534120380,-3988869806010,44179467566755,-147200296896765,-293052319462105,5409366658571715

%N G.f. A(x) satisfies: 6^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (6+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 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 a(n)=-(3*(2*n-3)*a(n-1)+29*(n-3)*a(n-2))/n for n>2, with a(0)=1, a(1)=4, a(2)=5. G.f.: A(x) = (1+5*x+sqrt(1+6*x+29*x^2))/2.

%e From the table of powers of A(x) (A100235), we see that

%e 6^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:

%e A^1=[1,4],5,-15,20,90,-695,1785,...

%e A^2=[1,8,26],10,-55,190,-245,-1690,...

%e A^3=[1,12,63,139],15,-120,635,-2130,...

%e A^4=[1,16,116,436,726],20,-210,1480,...

%e A^5=[1,20,185,965,2830,3774],25,-325,...

%e A^6=[1,24,270,1790,7335,17634,19601],30,...

%o (PARI) a(n)=if(n==0,1,(6^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 (PARI) a(n)=if(n==0,1,if(n==1,4,if(n==2,5,-(3*(2*n-3)*a(n-1)+29*(n-3)*a(n-2))/n)))

%o (PARI) a(n)=polcoeff((1+5*x+sqrt(1+6*x+29*x^2+x^2*O(x^n)))/2,n)

%Y Cf. A100235, A100236, A100231.

%K sign

%O 0,2

%A _Paul D. Hanna_, Nov 29 2004

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