A293433 a(n) is the number of the proper divisors of n that are Jacobsthal numbers (A001045).

0, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 2, 1, 1, 2, 2, 1, 3, 1, 2, 3, 1, 1, 2, 1, 2, 2, 1, 1, 2, 3, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 2, 1, 2, 2, 1, 1, 3, 2, 2, 2, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 1, 3, 2, 1, 2, 1, 1, 4

Offset

1,6

Links

Antti Karttunen, Table of n, a(n) for n = 1..21845

Formula

a(n) = Sum_{d|n, d<n} A147612(d).

a(n) = A293431(n) - A147612(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n>=2} 1/A001045(n) = 1.718591611927... . - Amiram Eldar, Jul 05 2025

Example

For n = 21, whose proper divisors are [1, 3, 7], both 1 and 3 are in A001045, thus a(21) = 2.

Mathematica

With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Table[DivisorSum[n, 1 &, And[MemberQ[s, #], # != n] &], {n, 105}]] (* Michael De Vlieger, Oct 09 2017 *)

Prog

(PARI)
A147612aux(n, i) = if(!(n%2), n, A147612aux((n+i)/2, -i));
A147612(n) = 0^(A147612aux(n, 1)*A147612aux(n, -1));
A293433(n) = sumdiv(n, d, (d<n)*A147612(d));
(Python)
from sympy import divisors
def A293433(n): return sum(1 for d in divisors(n, generator=True) if d<n and (m:=3*d+1).bit_length()>(m-3).bit_length()) # Chai Wah Wu, Apr 18 2025

Crossrefs

Cf. A001045, A147612, A293431, A293434.

Cf. also A293435.

Sequence in context: A350330 A154402 A210682 * A177025 A379552 A265210

Adjacent sequences: A293430 A293431 A293432 * A293434 A293435 A293436

Keyword

nonn

Author

Antti Karttunen, Oct 09 2017

Status

approved