A260683 Number of 2's in the expansion of 2^n in base 3.
0, 1, 0, 2, 1, 1, 1, 2, 0, 4, 2, 4, 3, 3, 2, 6, 5, 5, 3, 7, 4, 7, 5, 4, 1, 5, 2, 8, 8, 7, 9, 9, 8, 7, 7, 8, 4, 6, 8, 9, 11, 11, 7, 11, 10, 8, 9, 8, 8, 10, 11, 16, 13, 10, 9, 12, 13, 16, 12, 13, 15, 15, 11, 15, 16, 14, 14, 12, 14, 15, 14, 16, 11, 18, 11, 17, 10
Offset: 0
Examples
For n=5, the expansion of 2^n in number base 3 is 1012, thus: a(n)=1 For n=10, the expansion of 2^n in number base 3 is 1101221, thus: a(n)=2
References
- R. K. Guy, Unsolved Problems in Number Theory, B33. [Does not seem to be in section B33.]
Links
- Robert Israel, Table of n, a(n) for n = 0..10000
- Paul Erdős, Some unconventional problems in number theory, Mathematics Magazine, Vol. 52, No. 2 (1979), pp. 67-70.
Crossrefs
Programs
-
Maple
seq(numboccur(2, convert(2^n,base,3)),n=0..100); # Robert Israel, Nov 15 2015
-
Mathematica
S={};n=-1;While[n<150,n++;A=IntegerDigits[2^n,3];k=Count[A,2];AppendTo[S, k]];S -
PARI
c(k, d, b) = {my(c=0, f); while (k>b-1, f=k-b*(k\b); if (f==d, c++); k\=b); if (k==d, c++); return(c)} for(n=0, 300, print1(c(2^n, 2, 3)", ")) \\ Altug Alkan, Nov 15 2015 -
PARI
a(n) = #select(x->(x==2), digits(2^n, 3)); \\ Michel Marcus, Nov 28 2018
-
PARI
a(n) = hammingweight(digits(2^n, 3)\2); \\ Ruud H.G. van Tol, May 09 2024
-
Perl
use ntheory ":all"; sub a260683 { scalar grep { $==2 } todigits(vecprod((2) x shift), 3) } # _Dana Jacobsen, Aug 16 2016
Comments