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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
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
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
KEYWORD
nonn,base
AUTHOR
Max Alekseyev, Feb 01 2021
STATUS
approved