OFFSET
1,2
COMMENTS
Number of digits in Fibonacci(k) is equal to floor(1 + k*log_10((1+sqrt(5))/2)-log_10(sqrt(5))).
MATHEMATICA
k = 0; Join[{1}, Table[While[d = IntegerDigits[Fibonacci[k]]; ! MemberQ[Partition[Differences[d], n - 1, 1], Table[0, {n - 1}]], k++]; Length[d], {n, 2, 8}]] (* T. D. Noe, Oct 02 2012 *)
PROG
(Python)
def A217191(n):
if n == 1:
return 1
else:
l, y, x = [str(d)*n for d in range(10)], 0, 1
for m in range(1, 10**9):
s = str(x)
for k in l:
if k in s:
return len(s)
y, x = x, y+x
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
CROSSREFS
KEYWORD
nonn,base,hard,more
AUTHOR
V. Raman, Sep 27 2012
EXTENSIONS
a(10)-a(11) from Chai Wah Wu, Dec 17 2014
a(12)-a(15) from Nick Hobson, Feb 15 2024
STATUS
approved
