cp's OEIS Frontend

A260986 Numbers n such that H(n)/H(n^2) is a new record, where H(n) = A000120(n) is the sum of the binary digits of n.

%I A260986 #26 Sep 21 2023 13:26:12
%S A260986 1,23,111,479,1471,6015,24319,28415,490495,6025215,8122367,98549759,
%T A260986 132104191,1593769983,1862205439,29930291199,479961546751,
%U A260986 514321285119,8237743079423,131872659079167,136270705590271,35461448750596095,7998111458938322943,9151032963545169919
%N A260986 Numbers n such that H(n)/H(n^2) is a new record, where H(n) = A000120(n) is the sum of the binary digits of n.
%C A260986 This sequence is infinite, a result which follows from Stolarsky's Theorem 2.
%C A260986 a(22) > 2.4*10^13. - _Giovanni Resta_, Aug 07 2015
%C A260986 a(25) > 5.8*10^20. - _Karl-Heinz Hofmann_, Oct 14 2022
%H A260986 Kevin G. Hare, Shanta Laishram, and Thomas Stoll, <a href="https://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P34_Stolarsky.pdf">Stolarsky's conjecture and the sum of digits of polynomial values</a>, Proc. Amer. Math. Soc. 139:1 (2011), pp. 39-49.
%H A260986 K. B. Stolarsky, <a href="http://dx.doi.org/10.1090/S0002-9939-1978-0495823-5">The binary digits of a power</a>, Proc. Amer. Math. Soc. 71 (1978), pp. 1-5.
%e A260986 23 is 10111 in binary and 23^2 = 529 is 1000010001 in binary. Each smaller number has H(n)/H(n^2) <= 1, but H(23)/H(529) = 4/3 > 1, so 23 is in this sequence.
%t A260986 DeleteDuplicates[Table[{n,Total[IntegerDigits[n,2]]/Total[IntegerDigits[n^2,2]]},{n,500000}],GreaterEqual[ #1[[2]],#2[[2]]]&][[;;,1]] (* The program generates the first 9 terms of the sequence. *) (* _Harvey P. Dale_, Sep 21 2023 *)
%o A260986 (PARI) r=2; forstep(n=1,1e9,2, t=hammingweight(n^2)/hammingweight(n); if(t<r, r=t; print1(n", ")))
%Y A260986 Cf. A000120, A159918, A230097.
%Y A260986 Subsequence of A356877.
%K A260986 base,nonn
%O A260986 1,2
%A A260986 _Charles R Greathouse IV_, Aug 06 2015
%E A260986 a(16)-a(21) from _Giovanni Resta_, Aug 07 2015
%E A260986 a(22)-a(24) from _Karl-Heinz Hofmann_, Oct 14 2022