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 A046820 Number of 1's in binary expansion of 5n. 2
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n) is also the largest integer such that 2^a(n) divides binomial(10n, 5n). - Benoit Cloitre, Mar 27 2002 LINKS Indranil Ghosh, Table of n, a(n) for n = 0..10000 Michael Gilleland, Some Self-Similar Integer Sequences. FORMULA 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 a(n) = A000120(5*n). - Indranil Ghosh, Jan 18 2017 EXAMPLE For n = 10, 5*n = 50 = 110010_2, having 3 1's. So, a(10) = 3. - Indranil Ghosh, Jan 18 2017 MATHEMATICA a[n_] := DigitCount[5*n, 2, 1]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *) PROG (Python) def A046820(n): ....return bin(5*n)[2:].count("1") # Indranil Ghosh, Jan 18 2017 (PARI) a(n) = hammingweight(5*n); \\ Michel Marcus, Aug 19 2018 CROSSREFS Cf. A000120. Sequence in context: A061338 A135714 A103274 * A356878 A043262 A130860 Adjacent sequences: A046817 A046818 A046819 * A046821 A046822 A046823 KEYWORD nonn,base AUTHOR N. J. A. Sloane STATUS approved

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Last modified August 15 10:38 EDT 2024. Contains 375173 sequences. (Running on oeis4.)