0,1

For odd n there is the Aurifeuillian factorization a(n) = Lucas[5n] = Lucas[n]*A[n]*B[n] = A000032[n]*A124296[n]*A124297[n], where A[n] = A124296[n] = 5*F(n)^2 - 5*F(n) + 1 and B[n] = A124297[n] = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci[n]. The largest prime divisors of a(n) for n>0 are listed in A121171[n] = {11, 41, 31, 2161, 151, 2521, 911, ...}. - Alexander Adamchuk (alex(AT)kolmogorov.com), Oct 25 2006

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures}, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 139.

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures}, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

Tanya Khovanova, Recursive Sequences

Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)

a(n) = Lucas(5n) = Fibonacci(5n-1) + Fibonacci(5n+1). - Alexander Adamchuk (alex(AT)kolmogorov.com), Oct 25 2006

a(n) = ((11 + 5sqrt(5))/2)^n + ((11 - 5sqrt(5))/2)^n. - Tanya Khovanova (tanyakh(AT)yahoo.com), Feb 06 2007

A001946:=(-2+11*z)/(-1+11*z+z**2); [Conjectured by S. Plouffe in his 1992 dissertation.]

Table[Fibonacci[5n-1]+Fibonacci[5n+1], {n, 0, 30}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Oct 25 2006

Cf. A000032, A000045, A121171, A124296, A124297.

Sequence in context: A130222 A057076 A118794 this_sequence A112864 A077391 A104087

Adjacent sequences: A001943 A001944 A001945 this_sequence A001947 A001948 A001949

easy,nonn

njas, Simon Plouffe (plouffe(AT)math.uqam.ca)