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A020345
Smallest Fibonacci number beginning with n.
6
0, 1, 2, 3, 4181, 5, 610, 75025, 8, 987, 10946, 1134903170, 121393, 13, 144, 1597, 165580141, 17711, 1836311903, 196418, 20365011074, 21, 225851433717, 233, 24157817, 2584, 267914296, 27777890035288, 28657, 2971215073, 308061521170129, 317811
OFFSET
0,3
COMMENTS
The graph of the indices A020344 is much more interesting. - T. D. Noe, Apr 02 2014
a(1382) is the first term with > 1000 digits (1004). - Michael S. Branicky, Jul 08 2022
LINKS
Michael S. Branicky, Table of n, a(n) for n = 0..1389 (terms 1..400 from T. D. Noe)
Ron Knott, Every number starts some Fibonacci Number, The Mathematical Magic of the Fibonacci Numbers.
FORMULA
a(n) = A000045(A020344(n)).
EXAMPLE
a(4) = 4181 is a Fibonacci number starting with 4.
MATHEMATICA
nn = 31; t = tn = Table[0, {nn}]; found = 0; n = 0; While[found < nn, n++; f = Fibonacci[n]; d = IntegerDigits[f]; i = 1; While[i <= Length[d], k = FromDigits[Take[d, i]]; If[k > nn, Break[]]; If[t[[k]] == 0, t[[k]] = f; tn[[k]] = n; found++]; i++]]; t = Join[{0}, t] (* T. D. Noe, Apr 02 2014 *)
PROG
(Python)
def aupton(nn):
ans, f, g, k = dict(), 0, 1, 0
while len(ans) < nn+1:
sf = str(f)
for i in range(1, len(sf)+1):
if int(sf[:i]) > nn:
break
if sf[:i] not in ans:
ans[sf[:i]] = f
f, g, k = g, f+g, k+1
return [int(ans[str(i)]) for i in range(nn+1)]
print(aupton(31)) # Michael S. Branicky, Jul 08 2022
CROSSREFS
KEYWORD
nonn,base
STATUS
approved