|
| |
|
|
A159918
|
|
Number of ones in binary representation of n^2.
|
|
27
|
|
|
|
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.
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.
Index entries for sequences related to binary expansion of n
|
|
|
FORMULA
|
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
|
|
|
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
|
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
|
| |
|
|
