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A159918
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Number of ones in binary representation of n^2.
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24
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0, 1, 1, 2, 1, 3, 2, 3, 1, 3, 3, 5, 2, 4, 3, 4, 1, 3, 3, 5, 3, 6, 5, 3, 2, 5, 4, 6, 3, 5, 4, 5, 1, 3, 3, 5, 3, 6, 5, 7, 3, 5, 6, 7, 5, 8, 3, 4, 2, 5, 5, 5, 4, 8, 6, 7, 3, 6, 5, 7, 4, 6, 5, 6, 1, 3, 3, 5, 3, 6, 5, 7, 3, 6, 6, 9, 5, 7, 7, 5, 3, 6, 5, 8, 6, 7, 7, 7, 5, 9, 8, 5, 3, 6, 4, 5, 2, 5, 5, 6, 5, 9, 5, 7, 4
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OFFSET
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0,4
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COMMENTS
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The binary weight (A000120) of n^2.
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LINKS
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Nathaniel Johnston, Table of n, a(n) for n = 0..10000
Bernt Lindström, On the binary digits of a power, Journal of Number Theory, Volume 65, Issue 2, August 1997, Pages 321-324.
Nick MacKinnon, Problems and Solutions #12140, The American Mathematical Monthly, 126:9 (2019), 850.
K. B. Stolarsky, The binary digits of a power, Proc. Amer. Math. Soc. 71 (1978), 1-5.
Index entries for sequences related to binary expansion of n
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FORMULA
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a(n) = A000120(A000290(n)); a(A077436(n)) = A000120(A077436(n)).
Lindström shows that lim sup wt(m^2)/log_2 m = 2. - N. J. A. Sloane, Oct 11 2013
a(n) = [x^(n^2)] (1/(1 - x))*Sum_{k>=0} x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Mar 27 2018
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MAPLE
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A159918 := proc(n) return add(b, b=convert(n^2, base, 2)): end: seq(A159918(n), n=0..100); # Nathaniel Johnston, Jun 23 2011
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PROG
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(Haskell)
a159918 = a000120 . a000290 -- Reinhard Zumkeller, Oct 12 2013
(Python)
def A159918(n):
return bin(n*n).count('1') # Chai Wah Wu, Sep 03 2014
(PARI) a(n)=hammingweight(n^2) \\ Charles R Greathouse IV, Aug 06 2015
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CROSSREFS
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Cf. A000120, A007088, A192085, A004159, A214560, A231897, A231898. For records see A230097.
Sequence in context: A066376 A151682 A318928 * A278573 A108663 A307314
Adjacent sequences: A159915 A159916 A159917 * A159919 A159920 A159921
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KEYWORD
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nonn,base,easy
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AUTHOR
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Reinhard Zumkeller, Apr 25 2009
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STATUS
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approved
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