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A261593
Odd numbers n such that the sum of the binary digits of n and n^2 both equal 12.
1
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
1,1
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
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.
EXAMPLE
4095 = 111111111111_2 and 4095^2 = 111111111110000000000001_2 both have 12 1s in binary.
MATHEMATICA
Select[2 Range@ 15000 - 1, Total@ IntegerDigits[#, 2] == 12 && Total@ IntegerDigits[#^2, 2] == 12 &] (* Michael De Vlieger, Aug 27 2015 *)
PROG
(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
CROSSREFS
Subsequence of A261586 and hence of A077436.
Sequence in context: A038463 A359084 A043452 * A290869 A075952 A075957
KEYWORD
nonn,base
AUTHOR
STATUS
approved