A361944 a(n) is the least k > 0 such that the binary expansion of k*n is an abelian square (A272653).

3, 5, 1, 9, 2, 6, 9, 17, 1, 1, 3, 3, 10, 11, 1, 33, 2, 2, 10, 26, 3, 6, 2, 22, 6, 5, 2, 21, 25, 5, 33, 65, 1, 1, 18, 1, 6, 5, 4, 13, 15, 14, 1, 3, 1, 1, 5, 11, 3, 3, 1, 3, 1, 1, 3, 41, 9, 37, 11, 10, 3, 39, 1, 129, 2, 2, 3, 2, 9, 9, 8, 11, 3, 3, 2, 27, 2, 2, 3

Offset: 1

Author

Rémy Sigrist, Mar 31 2023

Keywords

• nonn
• base

Examples

			a(8) = A361943(8)/8 = 136/8 = 17.
		

Links

• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A272653, A361943.

Programs

PARI

a(n) = { forstep (m = n, oo, n, my (w = #binary(m)); if (w%2==0 && hammingweight(m)==2*hammingweight(m % (2^(w/2))), return (m/n))) }

Python

from itertools import count
def a(n): return next(m//n for m in count(n, n) if not (w:= m.bit_length())&1 and m.bit_count() == ((m>>(w>>1)).bit_count())<<1)
print([a(n) for n in range(1, 80)]) # Michael S. Branicky, Mar 31 2023 after Rémy Sigrist

Formula

a(n) = A361943(n) / n.

a(n) = 1 iff n belongs to A272653.