A261593 Odd numbers n such that the sum of the binary digits of n and n^2 both equal 12.

4095, 10239, 11263, 12159, 12223, 12255, 12271, 12279, 12283, 14333, 15351, 15355, 15743, 15807, 18431, 19455, 19967, 20351, 20477, 22015, 22495, 22511, 24031, 24303, 24431, 24445, 25599, 26615, 26621, 27519, 27631, 27639, 28095, 28411, 28413, 28511, 28541, 28575

Offset: 1

Author

Charles R Greathouse IV, Aug 25 2015

Keywords

• nonn
• base

Comments

Hare, Laishram, & Stoll show that this sequence is infinite.

12 is the least constant for which the associated sequence is known to be infinite. 1 through 8 make finite sequences, the sequences with 9 and 10 are conjectured finite, and the sequence with 11 is unknown. (See Hare-Laishram-Stoll Theorem 1.3 and 1.4 plus section 5.)

Odd numbers n such that the sum of the binary digits of n-1 and n^2-1 both equal 11. - Chai Wah Wu, Aug 26 2015

Examples

			4095 = 111111111111_2 and 4095^2 = 111111111110000000000001_2 both have 12 1s in binary.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..1329
• K. G. Hare, S. Laishram, and T. Stoll, The sum of digits of n and n^2, International Journal of Number Theory 7:7 (2011), pp. 1737-1752.

Crossrefs

Subsequence of A261586 and hence of A077436.

Programs

Mathematica

Select[2 Range@ 15000 - 1, Total@ IntegerDigits[#, 2] == 12 && Total@ IntegerDigits[#^2, 2] == 12 &] (* Michael De Vlieger, Aug 27 2015 *)

PARI

is(n)=n%2 && hammingweight(n)==12 && hammingweight(n^2)==12

PARI

\\ List the elements below 2^(N+1).
go(N)=my(v=List(),n); forvec(u=vector(11,i,[1,N-11+i]), n=sum(i=1,11,2^u[i])+1; if(hammingweight(n^2)==12, listput(v,n)), 2); Set(v)

Python

from itertools import combinations
A261593_list = []
for c in combinations((2**x for x in range(15)),11):
    n = sum(c)
    if sum(int(d) for d in format(n*(n+1),'b')) == 11:
        A261593_list.append(2*n+1)
A261593_list = sorted(A261593_list) # Chai Wah Wu, Aug 26 2015