OFFSET
0,4
COMMENTS
The binary weight (A000120) of n^2.
a(n) = 0 iff n = 0. a(n) = 1 iff n = 2^k for some k >= 0. a(n) = 2 iff n = 3*2^k for some k >= 0. Szalay proves that a(n) = 3 iff n = 7*2^k, 23*2^k, or 2^a + 2^b for k >= 0 and a > b >= 0. It seems that a(n) = 4 iff n = 13*2^k, 15*2^k, 47*2^k, or 111*2^k but this has not been proven! Any other n with a(n) = 4 are greater than 10^50, and there are finitely many odd solutions. - Charles R Greathouse IV, Jan 20 2022
REFERENCES
L. Szalay, The equations 2^n ± 2^m ± 2^l = z^2, Indagationes Mathematicae (N.S.) 13, no. 1 (2002), pp. 131-142.
LINKS
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.
FORMULA
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
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
MATHEMATICA
a[n_] := Total[IntegerDigits[n^2, 2]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Oct 27 2021 *)
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
KEYWORD
nonn,base,easy
AUTHOR
Reinhard Zumkeller, Apr 25 2009
STATUS
approved