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 A159918 Number of ones in binary representation of n^2. 23
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS The binary weight (A000120) of n^2. LINKS Nathaniel Johnston, Table of n, a(n) for n = 0..10000 Bernt LindstrÃ¶m, On the binary digits of a power, Journal of Number Theory, Volume 65, Issue 2, August 1997, Pages 321-324. K. B. Stolarsky, The binary digits of a power, Proc. Amer. Math. Soc. 71 (1978), 1-5. FORMULA a(n) = A000120(A000290(n)); a(A077436(n)) = A000120(A077436(n)). LindstrÃ¶m 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 MAPLE A159918 := proc(n) return add(b, b=convert(n^2, base, 2)): end: seq(A159918(n), n=0..100); # Nathaniel Johnston, Jun 23 2011 PROG (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 CROSSREFS 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 KEYWORD nonn,base,easy AUTHOR Reinhard Zumkeller, Apr 25 2009 STATUS approved

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Last modified October 14 06:51 EDT 2019. Contains 327995 sequences. (Running on oeis4.)