|
%I #20 Jul 13 2022 14:49:50
%S 3,7,13,15,25,29,31,54,57,61,63,103,110,113,118,121,125,127,199,203,
%T 207,212,213,214,218,230,238,241,246,249,253,255,389,393,394,395,402,
%U 404,409,421,431,433,435,439,458,468,478,481,486,494,497,502,505,509,511
%N Positive integers whose square contains the same number of 0's as 1's when represented in binary.
%C 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
%H Harvey P. Dale, <a href="/A164344/b164344.txt">Table of n, a(n) for n = 1..1000</a>
%H James Propp et al., <a href="https://mailman.xmission.com/hyperkitty/list/math-fun@mailman.xmission.com/thread/ABJYBY4XWNAXLLMEIEKTE622UBEBHITA/">Perfectly balanced perfect squares</a>math-fun mailing list (archive available to subscribers), Jul 12 2022.
%F {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
%t sn01Q[n_]:=Module[{idn2=IntegerDigits[n^2,2]},Count[idn2,1] == Length[ idn2]/2]; Select[Range[600],sn01Q] (* _Harvey P. Dale_, Apr 03 2016 *)
%o (PARI) select( {is_A164344(n)=hammingweight(n^2)*2==exponent(n^2*2)}, [0..512]) \\ _M. F. Hasler_, Jul 12 2022
%o (Python)
%o def bal(n): return n and n.bit_length() == n.bit_count() * 2
%o print([k for k in range(512) if bal(k*k)]) # _Michael S. Branicky_, Jul 12 2022
%Y Cf. A031443 (digitally balanced numbers), A164343 (squares of the terms), A000120 (Hamming weight), A070939 (number of binary digits).
%K base,nonn
%O 1,1
%A _Leroy Quet_, Aug 13 2009
%E More terms from _Sean A. Irvine_, Oct 08 2009
|
