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A203838 a(n) = sigma_3(n)*Fibonacci(n), where sigma_3(n) = A001158(n), the sum of cubes of divisors of n. 4

%I #8 Jan 06 2018 13:24:50

%S 1,9,56,219,630,2016,4472,12285,25738,62370,118548,294336,512134,

%T 1167192,2152080,4620147,7847658,17604792,28681660,62224470,105431872,

%U 212319468,348698376,759507840,1181718775,2401396326,4014783920,7980869832,12542045310

%N a(n) = sigma_3(n)*Fibonacci(n), where sigma_3(n) = A001158(n), the sum of cubes of divisors of n.

%C Compare g.f. to the Lambert series identity: Sum_{n>=1} n^3*x^n/(1-x^n) = Sum_{n>=1} sigma_3(n)*x^n.

%F G.f.: Sum_{n>=1} n^3*fibonacci(n)*x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) = Sum_{n>=1} sigma_3(n)*fibonacci(n)*x^n, where Lucas(n) = A000204(n).

%e G.f.: A(x) = x + 9*x^2 + 56*x^3 + 219*x^4 + 630*x^5 + 2016*x^6 +...

%e where A(x) = x/(1-x-x^2) + 2^3*1*x^2/(1-3*x^2+x^4) + 3^3*2*x^3/(1-4*x^3-x^6) + 4^3*3*x^4/(1-7*x^4+x^8) + 5^3*5*x^5/(1-11*x^5-x^10) + 6^3*8*x^6/(1-18*x^6+x^12) +...+ n^3*fibonacci(n)*x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) +...

%o (PARI) {a(n)=sigma(n,3)*fibonacci(n)}

%o (PARI) {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}

%o {a(n)=polcoeff(sum(m=1,n,m^3*fibonacci(m)*x^m/(1-Lucas(m)*x^m+(-1)^m*x^(2*m)+x*O(x^n))),n)}

%Y Cf. A203847, A203848, A203849, A001158 (sigma_3), A000204 (Lucas), A000045.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Jan 12 2012

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Last modified April 23 02:14 EDT 2024. Contains 371906 sequences. (Running on oeis4.)