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

Author

Emmanuel Vantieghem, Nov 15 2015

Keywords

• base
• easy
• nonn

Comments

Erdős conjectures that a(n) > 0 for n > 8.

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

Cf. A004642 (2^n in base 3), A020915 (number of terms), A036461 (number of 1's), A104320 (number of 0's).

Cf. A000108 (conjecture that A000108(n) is 6m+1 only for n = 0, 1 and 5 follows from Erdős's one).

Cf. A005836 (for numbers with no 2 in base 3).

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

Formula

a(n) = A020915(n) - A104320(n) - A036461(n). - Altug Alkan, Nov 15 2015

a(n) = A081603(A000079(n)). - Michel Marcus, Dec 03 2015