A164344 Positive integers whose square contains the same number of 0's as 1's when represented in binary.

3, 7, 13, 15, 25, 29, 31, 54, 57, 61, 63, 103, 110, 113, 118, 121, 125, 127, 199, 203, 207, 212, 213, 214, 218, 230, 238, 241, 246, 249, 253, 255, 389, 393, 394, 395, 402, 404, 409, 421, 431, 433, 435, 439, 458, 468, 478, 481, 486, 494, 497, 502, 505, 509, 511

Offset

1,1

Comments

The squares must have an even number of binary digits, given by ceiling(log_2(a(n)^2)) = ceiling(2 log_2 a(n)), or equivalently, 2^(k-1/2) < a(n) < 2^k for some integer k > 0, which explains the jumps in the graph of the sequence. - M. F. Hasler, Jul 12 2022

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

James Propp et al., Perfectly balanced perfect squaresmath-fun mailing list (archive available to subscribers), Jul 12 2022.

Formula

{n | n^2 is in A031443} = {n | 2*A000120(n^2) = A070939(n^2)}, i.e., twice the Hamming weight must equal the number of binary digits, for the squares of the terms. - M. F. Hasler, Jul 12 2022

Mathematica

sn01Q[n_]:=Module[{idn2=IntegerDigits[n^2, 2]}, Count[idn2, 1] == Length[ idn2]/2]; Select[Range[600], sn01Q] (* Harvey P. Dale, Apr 03 2016 *)

Prog

(PARI) select( {is_A164344(n)=hammingweight(n^2)*2==exponent(n^2*2)}, [0..512]) \\ M. F. Hasler, Jul 12 2022
(Python)
def bal(n): return n and n.bit_length() == n.bit_count() * 2
print([k for k in range(512) if bal(k*k)]) # Michael S. Branicky, Jul 12 2022

Crossrefs

Cf. A031443 (digitally balanced numbers), A164343 (squares of the terms), A000120 (Hamming weight), A070939 (number of binary digits).

Sequence in context: A002236 A255682 A080565 * A002254 A356211 A331036

Adjacent sequences: A164341 A164342 A164343 * A164345 A164346 A164347

Keyword

base,nonn

Author

Leroy Quet, Aug 13 2009

Extensions

More terms from Sean A. Irvine, Oct 08 2009

Status

approved