A261009 - OEIS

A261009

Write 2^n in base 3, add up the "digits".

1, 2, 2, 4, 4, 4, 4, 6, 4, 8, 8, 10, 10, 8, 10, 16, 12, 14, 12, 16, 14, 18, 16, 12, 10, 12, 14, 20, 20, 22, 24, 26, 24, 22, 22, 22, 18, 20, 26, 28, 28, 28, 26, 30, 30, 30, 26, 26, 26, 32, 38, 40, 38, 38, 28, 34, 40, 42, 38, 40, 46, 40, 38, 42, 48, 44, 42, 40, 42, 48, 48, 44

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OFFSET

0,2

COMMENTS

Comment from Jean-Paul Allouche, Oct 25 2015: As mentioned by Holdum et al. (2015) the following problem, cited in "Concrete Mathematics" by Graham, Knuth, and Patashnik (1994), is still open: prove that for all n > 256, binomial(2n,n) is either divisible by 4 or by 9 (cf. A000984). This can be easily reduced to show that, for all k >= 9, 2*a(k) - a(k+1) >= 4. This has been proved up to huge values of k (Holdum et al. mention k = 10^{13}).

For additional information about the divisibility of binomial(2n,n) by squares see the comments and references in A000984, - N. J. A. Sloane, Oct 29 2015

LINKS

Giovanni Resta, Table of n, a(n) for n = 0..10000

Cernenoks J., Iraids J., Opmanis M., Opmanis R., Podnieks K., Integer complexity: experimental and analytical results II, arXiv:1409.0446 [math.NT] (September 2014)

Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994

Sebastian Tim Holdum, Frederik Ravn Klausen, Peter Michael Reichstein Rasmussen, Powers in prime bases and a problem on central coefficients, Integers 15 (2015), #A43

K. Podnieks, Digits of pi: limits to the seeming randomness, arXiv:1411.3911 [math.NT], 2014.

FORMULA

a(n) = A053735(A000079(n)). - Michel Marcus, Aug 14 2015

EXAMPLE

2^7 = 128_10 = 11202_3, so a(7) = 1+1+2+0+2 = 6.

MAPLE

S:=n->add(i, i in convert(2^n, base, 3)); [seq(S(n), n=0..100)];

MATHEMATICA

Table[Total@ IntegerDigits[2^n, 3], {n, 0, 100}] (* Giovanni Resta, Aug 14 2015 *)

PROG

(PARI) a(n) = vecsum(digits(2^n, 3)); \\ Michel Marcus, Aug 14 2015

(Haskell)