This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A352086 #42 May 28 2025 11:11:35
%S A352086 1,21,2697,4736533,14244123157,4804953862344753
%N A352086 a(n) is the smallest positive integer k such that wt(k^2) / wt(k) = n where wt(k) = A000120(k) is the binary weight of k.
%C A352086 Theorem (proofs in Diophante link):
%C A352086 For any n and any base b, there exists m such that sod_b(m^2) / sod_b(m) = n, where sod_b(m) = sum of digits of m in base b (A280012 for base 10).
%C A352086 a(n) is odd. Proof: a(n) exists. Furthermore, if a(n) is even then wt(a(n)) = wt(a(n)/2) and wt(a(n)^2) = wt((a(n)/2)^2) so then a(n)/2 so that a(n)/2 is a lesser candidate, a contradiction. - _David A. Corneth_, Mar 06 2022
%H A352086 Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/4786-a1730-des-chiffres-a-sommer-pour-un-entier">A1730 - Des chiffres à sommer pour un entier</a> (in French).
%H A352086 Wojciech Muła, Nathan Kurz and Daniel Lemire, <a href="https://arxiv.org/abs/1611.07612">Faster Population Counts using AVX2 Instructions</a>, arXiv:1611.07612 [cs.DS], Sep 05 2018.
%F A352086 a(n) > 2^(n^2/2) for n > 1. - _Charles R Greathouse IV_, Mar 16 2022
%e A352086 We have 21_10 = 10101_2, so wt(21) = 3 ones; then 21^2 = 441_10 = 110111001_2, so wt(21^2) = 6 ones; as 6/3 = 2 and 21 is the smallest integer k such that wt(k^2) / wt(k) = 2, hence a(2) = 21.
%t A352086 r[n_] := Total[IntegerDigits[n^2, 2]]/Total[IntegerDigits[n, 2]]; seq[max_, nmax_] := Module[{s = Table[0, {max}], c = 0, n = 1, i}, While[c < max && n < nmax, i = r[n]; If[IntegerQ[i] && s[[i]] == 0, c++; s[[i]] = n]; n+=2]; TakeWhile[s, # > 0 &]]; seq[4, 5*10^6] (* _Amiram Eldar_, Mar 06 2022 *)
%o A352086 (Python)
%o A352086 from gmpy2 import popcount
%o A352086 aDict=dict()
%o A352086 for k in range(1, 10**11, 2):
%o A352086 if popcount(k*k)%popcount(k)==0:
%o A352086 n=popcount(k*k)//popcount(k)
%o A352086 if not n in aDict:
%o A352086 print(n, k); aDict[n]=k # _Martin Ehrenstein_, Mar 16 2022
%Y A352086 Cf. A000120, A077436, A083567, A159918, A280012, A352084, A352085.
%K A352086 nonn,base,more
%O A352086 1,2
%A A352086 _Bernard Schott_, Mar 06 2022
%E A352086 a(3)-a(5) from _David A. Corneth_, Mar 06 2022
%E A352086 a(6) -- using the Muła et al. Faster Population Counts algorithm -- from _Martin Ehrenstein_, Mar 15 2022