A088858 Define a Fibonacci-type sequence to be one of the form s(0) = s_1 >= 1, s(1) = s_2 >= 1, s(n+2) = s(n+1) + s(n); then a(n) = maximal m such that n is the m-th term in some Fibonacci-type sequence.

1, 2, 3, 3, 4, 3, 4, 5, 4, 4, 5, 4, 6, 5, 4, 5, 5, 6, 5, 5, 7, 5, 6, 5, 5, 6, 5, 6, 7, 5, 6, 5, 6, 8, 5, 6, 7, 6, 6, 5, 6, 7, 6, 6, 7, 6, 8, 6, 6, 7, 6, 6, 7, 6, 9, 6, 6, 7, 6, 8, 7, 6, 7, 6, 6, 7, 6, 8, 7, 6, 7, 6, 8, 7, 6, 9, 7, 6, 7, 6, 8, 7, 6, 7, 7, 8, 7, 6, 10, 7

Offset

1,2

Comments

A033192(k) is the number of integers m such that a(m) = k. - Michel Marcus, Aug 03 2017

Links

Michel Marcus, Table of n, a(n) for n = 1..1000

J. H. E. Cohn, Recurrent Sequences Including N, Fib. Quart., 29-1, 1991.

T. Denes, Problem 413, Discrete Math. 272 (2003), 302 (but there are several errors in the table given there).

James P. Jones, Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37.

Formula

For n>1, a(n) is the largest integer r>1 such that ceiling(((-1)^r*fibonacci(r-2)*n + 1)/fibonacci(r-1)) <= floor(((-1)^r*fibonacci(r-1)*n - 1)/fibonacci(r)), r++). See Theorem 2.12 in Jones & Kiss. - Michel Marcus, Aug 02 2017

Mathematica

max = 12; s[n_] := (1/2)*((3*s1 - s2)*Fibonacci[n] + (s2 - s1)*LucasL[n]); a[n_] := Reap[Do[If[s[m] == n, Sow[m - 1]], {m, 1, max}, {s1, 1, max}, {s2, 1, max}]][[2, 1]] // Max; Table[a[n], {n, 1, 90}] (* Jean-François Alcover, Jan 15 2013 *)

Prog

(PARI) a(n) = {if (n==1, return (1)); r = 2; while (ceil(((-1)^r*fibonacci(r-2)*n + 1)/fibonacci(r-1)) <= floor(((-1)^r*fibonacci(r-1)*n - 1)/fibonacci(r)), r++); r-1; } \\ Michel Marcus, Aug 02 2017

Crossrefs

Cf. A033192, A088527 (which is a(n)+1).

Sequence in context: A049108 A179846 A086925 * A113312 A334954 A053475

Adjacent sequences: A088855 A088856 A088857 * A088859 A088860 A088861

Keyword

nonn,easy

Author

Don Reble, Nov 20 2003

Status

approved