%I #30 May 03 2018 05:11:45

%S 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,

%T 20,20,22,24,26,24,22,22,22,18,20,26,28,28,28,26,30,30,30,26,26,26,32,

%U 38,40,38,38,28,34,40,42,38,40,46,40,38,42,48,44,42,40,42,48,48,44

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

%C 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}).

%C 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

%H Giovanni Resta, <a href="/A261009/b261009.txt">Table of n, a(n) for n = 0..10000</a>

%H Cernenoks J., Iraids J., Opmanis M., Opmanis R., Podnieks K., <a href="http://arxiv.org/abs/1409.0446">Integer complexity: experimental and analytical results II</a>, arXiv:1409.0446 [math.NT] (September 2014)

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

%H Sebastian Tim Holdum, Frederik Ravn Klausen, Peter Michael Reichstein Rasmussen, <a href="http://www.integers-ejcnt.org/p43/p43.mail.html">Powers in prime bases and a problem on central coefficients</a>, Integers 15 (2015), #A43

%H K. Podnieks, <a href="http://arxiv.org/abs/1411.3911">Digits of pi: limits to the seeming randomness</a>, arXiv:1411.3911 [math.NT], 2014.

%F a(n) = A053735(A000079(n)). - _Michel Marcus_, Aug 14 2015

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

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

%t Table[Total@ IntegerDigits[2^n, 3], {n, 0, 100}] (* _Giovanni Resta_, Aug 14 2015 *)

%o (PARI) a(n) = vecsum(digits(2^n, 3)); \\ _Michel Marcus_, Aug 14 2015

%o (Haskell)

%o a261009 = a053735 . a000079 -- _Reinhard Zumkeller_, Aug 14 2015

%Y Cf. A000079, A000984, A053735, A007089.

%Y Sum of digits of k^n in base b for various pairs (k,b): A001370 (2,10), A011754 (3,2), A261009 (2,3), A261010 (5,3).

%K nonn,base

%O 0,2

%A _N. J. A. Sloane_, Aug 14 2015