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

Author

David W. Wilson

Keywords

• sign
• base
• look

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

Crossrefs

Cf. A000045, A023184, A020344, A020345.

Programs

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 *)

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