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A352084
Integers m such that wt(m) divides wt(m^2) where wt(m) = A000120(m) is the binary weight of m.
4
1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 21, 24, 28, 30, 31, 32, 37, 42, 45, 48, 53, 56, 60, 62, 63, 64, 69, 73, 74, 79, 81, 83, 84, 90, 91, 96, 106, 112, 120, 124, 126, 127, 128, 133, 137, 138, 141, 146, 148, 155, 157, 158, 159, 161, 162, 165, 166, 168, 177, 180
OFFSET
1,2
COMMENTS
Integers m such that A000120(m) divides A159918(m).
This is a problem proposed by the French site Diophante in the links section.
The first 18 terms are the same as A268415, then A268415(19) = 41 while a(19) = 42.
The corresponding quotients are in A352085.
The smallest term k such that the corresponding quotient = n is A352086(n).
Some subsequences:
-> wt(m^2) = wt(m) iff m is in A077436.
-> wt(m^2) / wt(m) = 2 iff m is in A083567.
-> When m is a power of 2 (A000079): wt(2^k) = wt((2^k)^2) = wt(2^(2k)) = 1.
LINKS
EXAMPLE
37_10 = 100101_2, digsum_2(37) = 1+1+1 = 3; then 37^2 = 1369_10 = 10101011001_2, digsum_2(1369) = 1+1+1+1+1+1 = 6; as 3 divides 6, 37 is a term.
MATHEMATICA
Select[Range[180], Divisible[Total[IntegerDigits[#^2, 2]], Total[IntegerDigits[#, 2]]] &] (* Amiram Eldar, Mar 03 2022 *)
PROG
(Python)
def ok(n): return n > 0 and bin(n**2).count('1')%bin(n).count('1') == 0
print([m for m in range(1, 200) if ok(m)]) # Michael S. Branicky, Mar 03 2022
(PARI) isok(m) = !(hammingweight(m^2) % hammingweight(m)); \\ Michel Marcus, Mar 03 2022
CROSSREFS
Cf. A351650 (similar for base 10).
Subsequences: A000079, A023758, A077436, A083567.
Sequence in context: A032900 A370684 A268415 * A277779 A166935 A114391
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Mar 03 2022
EXTENSIONS
More terms from Amiram Eldar, Mar 03 2022
STATUS
approved