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

1,2

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.

Links

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

Example

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

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 *)

Prog

(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 3.10+)
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
(PARI) a(n) = my(k=1); while (hammingweight(k^n) != n*hammingweight(k), k++); k; \\ Michel Marcus, Nov 07 2022

Crossrefs

Cf. A000120, A083567, A212314.

Sequence in context: A034082 A040427 A070988 * A351321 A040426 A352137

Adjacent sequences: A358265 A358266 A358267 * A358269 A358270 A358271

Keyword

nonn,base,look

Author

Robert Israel, Nov 06 2022

Status

approved