A340956 Integers n such that s(n^2) = s(n)*(s(n)+1)/2, where s(n) is the sum of binary digits of n (A000120).

0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37, 40, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 84, 128, 129, 130, 132, 133, 136, 137, 138, 144, 146, 148, 160, 161, 162, 168, 256, 257, 258, 260, 261, 264, 265, 266, 272, 273, 274, 276, 277, 288, 289, 292, 293, 296, 320, 321, 322, 324, 336, 337, 512, 513, 514, 516

Offset

1,3

Comments

Also, integers n such that A159918(n) = A000217(A000120(n)).

For any n, s(n^2) <= s(n)*(s(n)+1)/2. This sequence gives n for which the equality is achieved.

Links

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Maple

q:= n-> (s-> is(s(n^2)=(t->t(s(n)))(h->h*(h+1)/2)))(
         k-> add(i, i=Bits[Split](k))):
select(q, [$0..555])[];  # Alois P. Heinz, Feb 01 2021

Mathematica

s[n_] := DigitCount[n, 2, 1]; t[n_] := n*(n + 1)/2; Select[Range[0, 500], s[#^2] == t[s[#]] &] (* Amiram Eldar, Feb 01 2021 *)

Prog

(PARI) isok(n) = my(hn=hammingweight(n)); hammingweight(n^2) == hn*(hn+1)/2; \\ Michel Marcus, Feb 01 2021

Crossrefs

Cf. A000120, A000217, A000290, A159918.

Sequence in context: A359267 A279430 A003714 * A010402 A010443 A035269

Adjacent sequences: A340953 A340954 A340955 * A340957 A340958 A340959

Keyword

nonn,base

Author

Max Alekseyev, Feb 01 2021

Status

approved