A113506 Maximum element in the continued fraction expansion of F(n+1)^5/F(n)^5 where F=A000045.

1, 32, 7, 12, 17, 46, 3042, 319, 835, 2188, 5730, 375131, 39282, 102845, 269253, 704915, 46137317, 4831563, 12649196, 33116027, 86698886, 5674515856, 594243013, 1555748409, 4073002214, 10663258234, 697919312217, 73087059232

Offset

1,2

Links

Table of n, a(n) for n=1..28.

Formula

5*a(5n)=F(10n+1)-(-1)^n-5; 5*a(5n+1)=F(10n+3)-2*(-1)^n-5; a(5n+2)=5*F(10n+5)+7*(-1)^n-1; 5a(5n+3)=F(10n+7)-3*(-1)^n-5; 5a(5n+4)=F(10n+9)+(-1)^n-5.

Empirical g.f.: x*(6*x^19 -5*x^17 +x^16 -x^15 -732*x^14 -x^13 +593*x^12 -107*x^11 +118*x^10 -743*x^9 +94*x^8 -327*x^7 +786*x^6 +93*x^5 -5*x^4 -5*x^3 +25*x^2 -31*x -1) / ((x -1) * (x +1) * (x^2 -3*x +1) * (x^4 -x^3 +x^2 -x +1) * (x^4 -x^3 +6*x^2 +4*x +1) * (x^4 +4*x^3 +6*x^2 -x +1)). - Colin Barker, Jun 17 2013

Mathematica

Max[ContinuedFraction[#[[2]]/#[[1]]]]&/@Partition[Fibonacci[ Range[ 30]]^5, 2, 1] (* Harvey P. Dale, May 27 2018 *)

Prog

(PARI) a(n)=vecmax(contfrac(fibonacci(n+1)^5/fibonacci(n)^5))

Crossrefs

Cf. A113500.

Sequence in context: A298060 A298709 A240925 * A120315 A119362 A070630

Adjacent sequences: A113503 A113504 A113505 * A113507 A113508 A113509

Keyword

nonn

Author

Benoit Cloitre, Jan 11 2006

Status

approved