A113500 Maximum element in the continued fraction for F(5n+3)^5/F(5n+2)^5 where F=A000045 are Fibonacci numbers.

32, 3042, 375131, 46137317, 5674515856, 697919312217, 85838400887831, 10557425389890242, 1298477484555612931, 159702173174950499517, 19642068823034355828656, 2415814763060050816424417

Offset: 0

Author

Benoit Cloitre, Jan 10 2006

Keywords

• nonn

References

• B. Cloitre, On rational sequences yielding continued fractions with unbounded coefficients, in preparation

Links

• G. C. Greubel, Table of n, a(n) for n = 0..475

Crossrefs

Cf. A000045.

Programs

Mathematica

Table[2*LucasL[10*n + 4] + LucasL[10*n + 5] + 7*(-1)^n - 1, {n,0,50}] (* G. C. Greubel, Mar 13 2017 *)

PARI

a(n)=vecmax(contfrac(fibonacci(5*n+3)^5/fibonacci(5*n+2)^5))

Formula

a(n) = 2*L(10*n+4) + L(10*n+5) + (-1)^n*7 - 1, where L(k) denotes the k-th Lucas number L(k) = F(k-1) + F(k+1), for n >= 0.

Empirical g.f.: (x^4-140*x^3-965*x^2+894*x-32) / ((x-1)*(x+1)*(x^2-123*x+1)). - Colin Barker, Jun 17 2013