A380402 Number of proper prime powers (in A246547) that do not exceed primorial A002110(n).

0, 0, 1, 6, 14, 34, 75, 187, 551, 1954, 8317, 38582, 200978, 1125541, 6562122, 40444003, 266832233, 1870169623, 13424553758, 101495825622, 793832121165, 6325729776075, 52616754936494, 450157758564742, 3999323787879764, 37180986240914714, 353667558431662474

Offset: 0

Author

Michael De Vlieger, Jan 23 2025

Keywords

• nonn
• hard

Examples

			Let s = A246547.
a(0) = a(1) = 0 since P(0) = 1 and P(1) = 2, and the smallest number in s is 4.
a(2) = 1 since P(2) = 6, and s(1) = 4 is the only term in s <= 6.
a(3) = 6 since P(3) = 30, and the set s(1..6) = {4, 8, 9, 16, 25, 27} contains k <= 30.
a(4) = 14 since P(4) = 210, and the set s(1..14) = {4, 8, 9, 16, 25, 27, 32, 49, 64, 81, 121, 125, 128, 169} contains k <= 210, etc.
		

Crossrefs

Cf. A036386, A246547.

Programs

Mathematica

Table[Sum[PrimePi@ Floor[#^(1/k)], {k, 2, Floor@ Log2[#]}] &[Product[Prime[i], {i, n}]], {n, 0, nn}]

Python

from sympy import primorial, primepi, integer_nthroot
def A380402(n):
    if n == 0: return 0
    m = primorial(n)
    return int(sum(primepi(integer_nthroot(m,k)[0]) for k in range(2,m.bit_length()))) # Chai Wah Wu, Jan 24 2025

Formula

a(n) = Sum_{k=2..floor(log_2(P(n)))} pi(floor(P(n)^(1/k))), where P(n) = A002110(n).

Extensions

a(24) corrected by Chai Wah Wu, Jan 25 2025

a(26) from Jinyuan Wang, Feb 25 2025