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A118738
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Number of ones in binary expansion of 5^n.
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4
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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, 19, 31, 33, 29, 36, 37, 33, 39, 34, 42, 40, 44, 42, 38, 46, 53, 54, 49, 52, 52, 53, 50, 49, 54, 60, 58, 60, 54, 64, 58, 74, 61, 67, 74, 65, 61, 77, 74, 81, 86, 78, 87, 85, 82, 89, 83, 79
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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a(2) = 3 because 5^2 = 25 is 11001, which has 3 on bits.
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MAPLE
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seq(convert(convert(5^n, base, 2), `+`), n=0..100); # Robert Israel, Dec 24 2017
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MATHEMATICA
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Table[DigitCount[5^n, 2, 1], {n, 0, 71}] (* Ray Chandler, Sep 29 2006 *)
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PROG
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(PARI) a(n) = hammingweight(5^n) \\ Iain Fox, Dec 24 2017
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CROSSREFS
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KEYWORD
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base,nonn,easy
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AUTHOR
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STATUS
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approved
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