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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.
3
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
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 * A372627 A113312 A334954
KEYWORD
nonn,easy
AUTHOR
Don Reble, Nov 20 2003
STATUS
approved