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%I #25 Jul 27 2023 08:18:18
%S 1,2,3,6,5,6,7,8,12,13,11,15,13,14,17,20,20,20,24,19,26,29,25,27,30,
%T 19,31,33,29,36,37,33,39,34,42,40,44,42,38,46,53,54,49,52,52,53,50,49,
%U 54,60,58,60,54,64,58,74,61,67,74,65,61,77,74,81,86,78,87,85,82,89,83,79
%N Number of ones in binary expansion of 5^n.
%C Also binary weight of 10^n, which is verified easily enough: 10^n = 2^n * 5^n; it is obvious that 2^n in binary is a single 1 followed by n 0's, therefore, in the binary expansion of 2^n * 5^n, the 2^n contributes only the trailing zeros. - _Alonso del Arte_, Oct 28 2012
%H Robert Israel, <a href="/A118738/b118738.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) + A118737(n) = A061785(n) + 1 for n >= 1. - _Robert Israel_, Dec 24 2017 [corrected by _Amiram Eldar_, Jul 27 2023]
%e a(2) = 3 because 5^2 = 25 is 11001, which has 3 on bits.
%p seq(convert(convert(5^n,base,2),`+`),n=0..100); # _Robert Israel_, Dec 24 2017
%t Table[DigitCount[5^n, 2, 1], {n, 0, 71}] (* _Ray Chandler_, Sep 29 2006 *)
%o (PARI) a(n) = hammingweight(5^n) \\ _Iain Fox_, Dec 24 2017
%Y Cf. A000120, A061785, A118737.
%K base,nonn,easy
%O 0,2
%A _Zak Seidov_, May 22 2006
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