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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). 1
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 (list; graph; refs; listen; history; text; internal format)
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: A339906 A279430 A003714 * A010402 A010443 A035269

Adjacent sequences:  A340953 A340954 A340955 * A340957 A340958 A340959

KEYWORD

nonn,base

AUTHOR

Max Alekseyev, Feb 01 2021

STATUS

approved

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Last modified November 29 21:32 EST 2021. Contains 349416 sequences. (Running on oeis4.)