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A113500 Maximum element in the continued fraction for F(5n+3)^5/F(5n+2)^5 where F=A000045 are Fibonacci numbers. 3
32, 3042, 375131, 46137317, 5674515856, 697919312217, 85838400887831, 10557425389890242, 1298477484555612931, 159702173174950499517, 19642068823034355828656, 2415814763060050816424417 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

REFERENCES

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

LINKS

Table of n, a(n) for n=0..11.

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

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

MATHEMATICA

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

PROG

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

CROSSREFS

Cf. A000045.

Sequence in context: A248074 A220299 A264115 * A064018 A067321 A104652

Adjacent sequences:  A113497 A113498 A113499 * A113501 A113502 A113503

KEYWORD

nonn

AUTHOR

Benoit Cloitre, Jan 10 2006

STATUS

approved

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Last modified October 27 06:31 EDT 2021. Contains 348271 sequences. (Running on oeis4.)