A352084 Integers m such that wt(m) divides wt(m^2) where wt(m) = A000120(m) is the binary weight of m.

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

Martin Ehrenstein, Table of n, a(n) for n = 1..10000

Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

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. A000120, A159918, A268415, A351650, A352085, A352086.

Cf. A351650 (similar for base 10).

Subsequences: A000079, A023758, A077436, A083567.

Sequence in context: A032900 A370684 A268415 * A277779 A166935 A114391

Adjacent sequences: A352081 A352082 A352083 * A352085 A352086 A352087

Keyword

nonn,base

Author

Bernard Schott, Mar 03 2022

Extensions

More terms from Amiram Eldar, Mar 03 2022

Status

approved