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A164344
Positive integers whose square contains the same number of 0's as 1's when represented in binary.
4
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
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
KEYWORD
base,nonn
AUTHOR
Leroy Quet, Aug 13 2009
EXTENSIONS
More terms from Sean A. Irvine, Oct 08 2009
STATUS
approved