A237654 G.f.: exp( Sum_{n>=1} 5*Fibonacci(n-1)*Fibonacci(n+1) * x^n/n ).
1, 0, 5, 5, 25, 49, 150, 365, 990, 2550, 6726, 17550, 46015, 120390, 315275, 825299, 2160775, 5656855, 14809980, 38772875, 101508876, 265753500, 695751900, 1821501900, 4768754125, 12484760124, 32685526625, 85571819345, 224029931845, 586517975725, 1535523995826, 4020054011225, 10524638038410
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 5*x^2 + 5*x^3 + 25*x^4 + 49*x^5 + 150*x^6 + 365*x^7 + ... where the logarithm begins: log(A(x)) = 5*1*2*x^2/2 + 5*1*3*x^3/3 + 5*2*5*x^4/4 + 5*3*8*x^5/5 + 5*5*13*x^6/6 + 5*8*21*x^7/7 + 5*13*34*x^8/8 + ...
Links
- Index entries for linear recurrences with constant coefficients, signature (0,5,5,0,-1).
Programs
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Mathematica
LinearRecurrence[{0,5,5,0,-1},{1,0,5,5,25},40] (* Harvey P. Dale, Apr 17 2025 *) -
PARI
{a(n)=polcoeff(exp(sum(m=1, n, 5*fibonacci(m-1)*fibonacci(m+1)*x^m/m) + x*O(x^n)), n)} for(n=0,36,print1(a(n),", "))
Formula
G.f.: 1 / ( (1-3*x+x^2) * (1+x)^3 ).
a(n) = (2*Lucas(2*n+5) + (28+25*n+5*n^2)*(-1)^(n))/50 where Lucas = A000032. - Greg Dresden, Jan 01 2021
Comments