OFFSET
1,1
COMMENTS
a(n+1)/a(n) = ~5.
FORMULA
t = sorted union of 2^k (mod 10^n), 0<k<=5^n; then count only those which have n digits.
Conjecture: a(n) = 18*5^(n-2) for n>1. - Mohammed Yaseen, Jul 19 2022
MATHEMATICA
Do[t = Union[ Table[ PowerMod[2, i, 10^n], {i, 5^n}]]; Print[ Length[ Select[t, Floor[ Log[10, # ] + 1] == n &]]], {n, 10}] (* Robert G. Wilson v, Aug 27 2004 *)
PROG
(Python)
def a(n):
b, m, t = 5**n+1, 10**n, 10**(n-1)
return len(set(p for p in (pow(2, k, m) for k in range(b)) if p >= t))
print([a(n) for n in range(1, 10)]) # Michael S. Branicky, Jul 25 2022
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Paul D. Hanna and Robert G. Wilson v, Aug 27 2004
EXTENSIONS
a(11)-a(12) from Michael S. Branicky, Jul 25 2022
STATUS
approved