A023183 a(n) = least k such that Fibonacci(k) ends with n, or -1 if there are none.

0, 1, 3, 4, 9, 5, 21, 14, 6, 11, 15, 22, 216, 7, 111, 130, 168, 37, 27, 112, 60, 8, 117, 64, 198, 25, 99, 136, 204, 29, 105, 88, 174, 13, 9, 70, 222, 43, 93, 172, 30, 41, 63, 124, 12, 55, 21, 154, 186, 49, 75, 148, 36, 67, 129, 10, 162, 23, 87, 118, 180, 61, 57, 166, 72, 20

Offset

0,3

Comments

It appears that if n is greater than 99 and congruent to 4 or 6 (mod 8) then there is no Fibonacci number ending in that n. - Jason Earls, Jun 19 2004

This is because there is no Fibonacci number == 4 or 6 (mod 8). - Robert Israel, Sep 11 2020

Links

Robert Israel, Table of n, a(n) for n = 0..9999

Maple

V:= Array(0..999, -1):
V[0]:= 0: u:= 1: v:= 0:
for n from 1 to 1500 do
  t:= v;
  v:= u+v mod 1000;
  u:= t;
  if V[v] = -1 then V[v]:= n fi;
  if V[v mod 100] = -1 then V[v mod 100] := n fi;
  if V[v mod 10] = -1 then V[v mod 10]:= n fi;
od:
seq(V[i], i=0..999); # Robert Israel, Sep 11 2020

Mathematica

d[n_]:=IntegerDigits[n]; Table[j=0; While[Length[d[Fibonacci[j]]]<(le=Length[y=d[n]]), j++]; i=j; While[Take[d[Fibonacci[i]], -le]!=y, i++]; i, {n, 0, 65}] (* Jayanta Basu, May 18 2013 *)

Prog

(Python)
from itertools import count
def A023183(n):
    if n < 2: return n
    if n > 99 and n%8 in {4, 6}: return -1
    k, f, g, s = 3, 1, 2, str(n)
    pow10, seen = 10**len(s), set()
    while (f, g) not in seen:
        seen.add((f, g))
        if g%pow10 == n:
            return k
        f, g, k = g, (f+g)%pow10, k+1
    return -1
print([A023183(n) for n in range(66)]) # Michael S. Branicky, Jun 27 2024

Crossrefs

Cf. A000045, A023184, A020344, A020345.

Sequence in context: A256469 A192334 A140439 * A102320 A227717 A021290

Adjacent sequences: A023180 A023181 A023182 * A023184 A023185 A023186

Keyword

sign,base,look

Author

David W. Wilson

Status

approved