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
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 and 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.
Programs
-
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 *) -
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
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)). See Theorem 2.12 in Jones & Kiss. - Michel Marcus, Aug 02 2017
Comments