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%I #26 Oct 13 2017 00:18:08
%S 0,0,0,1,1,0,1,1,0,3,2,3,3,2,2,5,5,4,3,6,4,6,5,3,1,4,2,7,8,6,9,8,8,6,
%T 7,7,4,5,8,8,11,10,7,10,10,7,9,7,8,9,11,15,13,9,9,11,13,15,12,12,15,
%U 14,11,14,16,13,14,11,14,14,14,15
%N Number of 2's in the base-3 representation of 2^n - 1.
%H Indranil Ghosh, <a href="/A265157/b265157.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = A081603(A000225(n)). - _Michel Marcus_, Dec 03 2015
%F a(n) = A260683(n) - A000035(n). - _Robert Israel_, Dec 03 2015
%t n = -1; While[n < 71, n++; s = IntegerDigits[2^n - 1, 3]; k = Count[s, 2]; AppendTo[a, k]]; a (* after _Emmanuel Vantieghem_ (see A260683)*)
%t Table[DigitCount[2^n-1,3,2],{n,0,80}] (* _Harvey P. Dale_, Dec 04 2015 *)
%o (PARI) a(n) = {my(d = digits(n, 3)); sum(k=1, #d, d[k]==2);} \\ _Michel Marcus_, Dec 03 2015
%Y Cf. A000079 (2^n), A000225 (2*n - 1).
%Y Cf. A260683 (Number of 2's in the expansion of 2^n in base 3).
%Y Cf. A000035, A081603.
%K nonn,base
%O 0,10
%A _L. Edson Jeffery_, Dec 02 2015