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

Author

Bernard Schott, Mar 03 2022

Keywords

• nonn
• base

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.

Examples

			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.
		

Links

• Martin Ehrenstein, Table of n, a(n) for n = 1..10000
• Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

Crossrefs

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

Cf. A351650 (similar for base 10).

Subsequences: A000079, A023758, A077436, A083567.

Programs

Mathematica

Select[Range[180], Divisible[Total[IntegerDigits[#^2, 2]], Total[IntegerDigits[#, 2]]] &] (* Amiram Eldar, Mar 03 2022 *)

PARI

isok(m) = !(hammingweight(m^2) % hammingweight(m)); \\ Michel Marcus, Mar 03 2022

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

Extensions

More terms from Amiram Eldar, Mar 03 2022