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G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^4 * x^n/n ).
7

%I #16 Oct 18 2020 05:11:13

%S 1,5,15,60,295,1625,9430,56465,345010,2139595,13419500,84926105,

%T 541398665,3472389210,22385362895,144945232375,942089445030,

%U 6143582084115,40181143112035,263482860974570,1731780213622125,11406235045261205,75268685723935940

%N G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^4 * x^n/n ).

%C Conjecture: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k) * x^n/n ) is an integer series for integers k>=0.

%C Note that exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k+1) * x^n/n ) is not an integer series for integers k.

%C Note that exp( Sum_{n>=1} Fibonacci(n)^(2*k) * x^n/n ) is not an integer series for integers k.

%F The o.g.f. A(x) = 1 + 5*x + 15*x^2 + 60*x^3 + ... is an algebraic function: A(x)^5 = (1 + 3*x + x^2)^4/( (1 - 7*x + x^2)*(1 - 2*x + x^2)^3 ). Cf. A203804. - _Peter Bala_, Apr 03 2014

%F a(n) ~ 2^(4/5) * 5^(1/10) * phi^(4*n) / (Gamma(1/5) * 3^(1/5) * n^(4/5)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Oct 18 2020

%e G.f.: A(x) = 1 + 5*x + 15*x^2 + 60*x^3 + 295*x^4 + 1625*x^5 + 9430*x^6 +...

%e such that

%e log(A(x))/5 = x + x^2/2 + 2^4*x^3/3 + 3^4*x^4/4 + 5^4*x^5/5 + 8^4*x^6/6 + 13^4*x^7/7 +...+ Fibonacci(n)^4*x^n/n +...

%o (PARI) {a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k)^4*x^k/k)+x*O(x^n)),n)}

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

%Y Cf. A054888, A207970, A207971, A207972, A207834, A207835.

%Y A077916, A203804.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 22 2012