A212314 -id:A212314 - cp's OEIS frontend

A212315 Numbers m such that B(m) = B(triangular(m)), where B(m) is the binary weight of m (A000120).

0, 1, 3, 7, 15, 31, 39, 45, 63, 78, 79, 91, 93, 127, 139, 143, 158, 159, 175, 182, 187, 189, 222, 255, 286, 287, 318, 319, 351, 367, 375, 379, 381, 407, 446, 487, 511, 535, 543, 572, 574, 575, 607, 627, 638, 639, 703, 724, 727, 731, 747, 750, 759, 763, 765, 799, 823, 830

Offset: 1

Alex Ratushnyak, Oct 24 2013

Dumitru Damian, Table of n, a(n) for n = 1..10000

Cf. A000120, A077436, A083567, A118655, A212314.

def isa(n): return n.bit_count()==((n*(n+1))>>1).bit_count()
print(list(n for n in range(10**3) if isa(n)))  # Dumitru Damian, Mar 04 2023

A000120(a(n)) = A000120(a(n)*(a(n)+1)/2).

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.

Original entry on oeis.org

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

Robert Israel, Nov 06 2022

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.

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

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

Cf. A000120, A083567, A212314.

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]);

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 *)

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

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

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

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A212315 Numbers m such that B(m) = B(triangular(m)), where B(m) is the binary weight of m (A000120).

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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.

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