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G.f. satisfies A(x) = (1 + x*A(x)) * (1 + x*A(x)^5).
9

%I #15 Jul 19 2023 07:25:23

%S 1,2,13,130,1518,19358,261323,3670828,53100530,785657529,11834135909,

%T 180863294507,2797643204500,43715591710804,689030031494554,

%U 10941710269299893,174889301792724294,2811464199460768704,45426696813655278251

%N G.f. satisfies A(x) = (1 + x*A(x)) * (1 + x*A(x)^5).

%C The radius of convergence of g.f. A(x) is r = 0.05685644444171304880925020950930... with A(r) = 1.3208055627586104770123863310077013110788003146438630... where y=A(r) satisfies 4*y^9 - 5*y^8 + 25*y^6 - 40*y^5 + 10*y^4 - 5 = 0.

%H Seiichi Manyama, <a href="/A215624/b215624.txt">Table of n, a(n) for n = 0..807</a>

%F G.f. satisfies A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * A(x)^(4*k)).

%F The formal inverse of g.f. A(x) is (sqrt((1-x^4)^2 + 4*x^5) - (1+x^4))/(2*x^5).

%F a(n) = Sum_{k=0..n} binomial(n+4*k+1,k) * binomial(n+4*k+1,n-k) / (n+4*k+1). - _Seiichi Manyama_, Jul 19 2023

%e G.f.: A(x) = 1 + 2*x + 13*x^2 + 130*x^3 + 1518*x^4 + 19358*x^5 +...

%e Related expansions.

%e A(x)^5 = 1 + 10*x + 105*x^2 + 1250*x^3 + 16120*x^4 + 219162*x^5 +...

%e A(x)^6 = 1 + 12*x + 138*x^2 + 1720*x^3 + 22803*x^4 + 315840*x^5 +...

%e where A(x) = 1 + x*(A(x) + A(x)^5) + x^2*A(x)^6.

%e The logarithm of the g.f. equals the series:

%e log(A(x)) = (1 + A(x)^4)*x + (1 + 2^2*A(x)^4 + A(x)^8)*x^2/2 +

%e (1 + 3^2*A(x)^4 + 3^2*A(x)^8 + A(x)^12)*x^3/3 +

%e (1 + 4^2*A(x)^4 + 6^2*A(x)^8 + 4^2*A(x)^12 + A(x)^16)*x^4/4 +

%e (1 + 5^2*A(x)^4 + 10^2*A(x)^8 + 10^2*A(x)^12 + 5^2*A(x)^16 + A(x)^20)*x^5/5 +...

%e more explicitly,

%e log(A(x)) = 2*x + 22*x^2/2 + 320*x^3/3 + 4886*x^4/4 + 76962*x^5/5 + 1236784*x^6/6 + 20152260*x^7/7 +...

%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*(A+x*O(x^n))^(4*j))*x^m/m))); polcoeff(A, n)}

%o (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1+x*A)*(1+x*A^5)+x*O(x^n)); polcoeff(A, n)}

%o for(n=0,21,print1(a(n),", "))

%Y Cf. A198953, A215623, A036765, A198951, A181734.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Aug 17 2012