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.

1, 23, 111, 479, 1471, 6015, 24319, 28415, 490495, 6025215, 8122367, 98549759, 132104191, 1593769983, 1862205439, 29930291199, 479961546751, 514321285119, 8237743079423, 131872659079167, 136270705590271, 35461448750596095, 7998111458938322943, 9151032963545169919

Offset: 1

Author

Charles R Greathouse IV, Aug 06 2015

Keywords

• base
• nonn

Comments

This sequence is infinite, a result which follows from Stolarsky's Theorem 2.

a(22) > 2.4*10^13. - Giovanni Resta, Aug 07 2015

a(25) > 5.8*10^20. - Karl-Heinz Hofmann, Oct 14 2022

Examples

			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.
		

Links

• Kevin G. Hare, Shanta Laishram, and Thomas Stoll, Stolarsky's conjecture and the sum of digits of polynomial values, Proc. Amer. Math. Soc. 139:1 (2011), pp. 39-49.
• K. B. Stolarsky, The binary digits of a power, Proc. Amer. Math. Soc. 71 (1978), pp. 1-5.

Crossrefs

Cf. A000120, A159918, A230097.

Subsequence of A356877.

Programs

Mathematica

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 *)

PARI

r=2; forstep(n=1,1e9,2, t=hammingweight(n^2)/hammingweight(n); if(t
				

Extensions

a(16)-a(21) from Giovanni Resta, Aug 07 2015

a(22)-a(24) from Karl-Heinz Hofmann, Oct 14 2022