A065710 Number of 2's in the decimal expansion of 2^n.
0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 2, 2, 0, 2, 0, 0, 1, 1, 0, 2, 1, 1, 1, 1, 2, 1, 0, 0, 0, 1, 1, 0, 1, 4, 0, 3, 1, 2, 0, 1, 1, 3, 3, 3, 1, 2, 0, 1, 2, 1, 2, 2, 2, 3, 1, 3, 0, 2, 2, 3, 3, 2, 2, 4, 4, 4, 0, 1, 2, 4, 3, 1, 3, 6, 2, 0, 2, 4, 4, 4, 2, 3, 6, 2, 1, 5, 1, 2, 4, 4, 1, 2, 6
Offset: 0
Examples
2^31 = 2147483648 so a(31) = 1.
Links
- M. F. Hasler, Table of n, a(n) for n = 0..10000 (first 1001 terms from Harry J. Smith), Feb 10 2023
Crossrefs
Programs
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Maple
seq(numboccur(2, convert(2^n,base,10)),n=0..100); # Robert Israel, Jul 09 2025
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Mathematica
Table[ Count[ IntegerDigits[2^n], 2], {n, 0, 100} ]
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PARI
a(n) = #select(x->(x==2), digits(2^n)); \\ Michel Marcus, Jun 15 2018
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Python
def A065710(n): return str(2**n).count('2') # Chai Wah Wu, Feb 14 2020
Formula
a(n) = a(floor(n/10)) + [n == 2 (mod 10)], where [...] is the Iverson bracket. - M. F. Hasler, Feb 10 2023
Extensions
More terms from Robert G. Wilson v, Dec 07 2001
Comments