A358268 a(n) is the least number k > 0 such that the binary weight of k^n is n times the binary weight of k.

1, 21, 5, 21, 17, 17, 9, 113, 17, 49, 665, 37, 149, 17, 275, 163, 33, 41, 97, 67, 141, 67, 135, 197, 49, 267, 81, 81, 69, 779, 1163, 69, 325, 49, 587, 837, 281, 197, 293, 49, 147, 677, 67, 651, 647, 67, 793, 277, 353, 49, 1233, 1177, 165, 775, 721, 353, 817, 69, 647, 709, 209, 1233, 69, 67, 263

Offset: 1

Author

Robert Israel, Nov 06 2022

Keywords

• nonn
• base
• look

Comments

a(n) is the least number k > 0 such that A000120(k^n) = n*A000120(k).

All terms are odd.

Conjecture: lim_{n -> infinity} a(n) = infinity.

Examples

			a(3) = 5 because 5 = 101_2 and 5^3 = 1111101_2 so A000120(5) = 2, A000120(5^3) = 6 and 6 = 3*2.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1004 from Robert Israel)

Crossrefs

Cf. A000120, A083567, A212314.

Programs

Maple

f:= proc(n) local k;
    for k from 1 by 2 do
      if convert(convert(k^n,base,2),`+`) = n*convert(convert(k,base,2),`+`) then return k
      fi
    od
end proc:
map(f, [$1..50]);

Mathematica

a[n_] := Module[{k = 1}, While[Divide @@ DigitCount[k^{n, 1}, 2, 1] != n, k += 2]; k]; Array[a, 65] (* Amiram Eldar, Nov 07 2022 *)

PARI

a(n) = my(k=1); while (hammingweight(k^n) != n*hammingweight(k), k++); k; \\ Michel Marcus, Nov 07 2022

Python

def a(n):
    k = 1
    while bin(k**n).count("1") != n*bin(k).count("1"): k += 2
    return k
print([a(n) for n in range(1, 66)]) # Michael S. Branicky, Nov 06 2022

Python

from itertools import count
def A358268(n): return next(filter(lambda k:k.bit_count()*n==(k**n).bit_count(),count(1,2))) # Chai Wah Wu, Nov 07 2022