A384252 The number of integers k from 1 to n such that the greatest divisor of k that is an infinitary divisor of n is a power of 2.

1, 2, 2, 4, 4, 4, 6, 8, 8, 8, 10, 8, 12, 12, 8, 16, 16, 16, 18, 16, 12, 20, 22, 16, 24, 24, 18, 24, 28, 16, 30, 32, 20, 32, 24, 32, 36, 36, 24, 32, 40, 24, 42, 40, 32, 44, 46, 32, 48, 48, 32, 48, 52, 36, 40, 48, 36, 56, 58, 32, 60, 60, 48, 64, 48, 40, 66, 64, 44

Offset: 1

Author

Amiram Eldar, May 23 2025

Keywords

• nonn
• easy
• mult

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000079, A006519, A077609.

Analogous sequences: A062570, A384056.

The number of integers k from 1 to n such that the greatest divisor of k that is an infinitary divisor of n is: A384247(1), A384249 (squarefree), A384250 (powerful), A384251 (odd), this sequence (power of 2).

Programs

Mathematica

f[p_, e_] := p^e*(1 - 1/p^(2^(IntegerExponent[e, 2]))); f[2, e_] := 2^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n)); n * prod(i = 1, #f~, if(f[i,1] == 2, 1, (1 - 1/f[i,1]^(1 << valuation(f[i,2], 2)))));}

Formula

Multiplicative with a(2^e) = 2^e, and a(p^e) = p^e * (1 - 1/p^A006519(e)) if p is an odd prime.

a(n) = A384247(n)/A384251(n).

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime >= 3} f(1/p) = 0.83603370570658499764..., and f(x) = 1 - (1-x)*Sum_{k>=1} x^(2^k)/(1-x^(2^k)).