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%I #29 Jul 18 2023 02:27:55
%S 0,2,2,4,2,3,4,3,2,4,3,5,4,2,3,4,2,4,4,6,3,4,5,5,4,6,2,4,3,3,4,5,2,4,
%T 4,6,4,5,6,4,3,5,4,6,5,4,5,6,4,6,6,8,2,3,4,4,3,5,3,5,4,4,5,6,2,4,4,6,
%U 4,5,6,5,4,6,5,7,6,3,4,5,3,5,5,7,4,5,6,6,5,7,4,6,5
%N Number of 1's in binary expansion of 5n.
%C a(n) is also the largest integer such that 2^a(n) divides binomial(10n, 5n). - _Benoit Cloitre_, Mar 27 2002
%H Indranil Ghosh, <a href="/A046820/b046820.txt">Table of n, a(n) for n = 0..10000</a>
%H Michael Gilleland, <a href="/selfsimilar.html">Some Self-Similar Integer Sequences</a>.
%F a(n) = floor(log(gcd(binomial(10*n, 5*n), 2^floor(log(binomial(10*n, 5*n))/log(2))))/log(2)). - _Benoit Cloitre_, Mar 27 2002
%F a(n) = A000120(5*n). - _Indranil Ghosh_, Jan 18 2017
%e For n = 10, 5*n = 50 = 110010_2, having 3 1's. So, a(10) = 3. - _Indranil Ghosh_, Jan 18 2017
%t a[n_] := DigitCount[5*n, 2, 1]; Array[a, 100, 0] (* _Amiram Eldar_, Jul 18 2023 *)
%o (Python)
%o def A046820(n):
%o ....return bin(5*n)[2:].count("1") # _Indranil Ghosh_, Jan 18 2017
%o (PARI) a(n) = hammingweight(5*n); \\ _Michel Marcus_, Aug 19 2018
%Y Cf. A000120.
%K nonn,base
%O 0,2
%A _N. J. A. Sloane_
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