A294903 Number of divisors of n that are in A257691.

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

Offset: 1

Author

Antti Karttunen, Nov 10 2017

Keywords

• nonn
• base

Comments

Number of terms of A257691 that divide n.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..25000

Crossrefs

Cf. A000120, A257691, A292257, A294901, A294902, A294904, A294905.

Cf. also A294893.

Programs

Mathematica

q[n_] := DivisorSum[n, DigitCount[#, 2, 1] &] <= 2*DigitCount[n, 2, 1]; a[n_] := DivisorSum[n, 1 &, q[#] &]; Array[a, 100] (* Amiram Eldar, Jul 20 2023 *)

PARI

A292257(n) = sumdiv(n,d,(dA294905(n) = (A292257(n) <= hammingweight(n));
A294903(n) = sumdiv(n,d,A294905(d));

Formula

a(n) = Sum_{d|n} A294905(d).

a(n) = A294901(n) + A294905(n).

a(n) + A294904(n) = A000005(n).