A385196 The number of integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is a prime number.

0, 1, 1, 0, 1, 3, 1, 0, 0, 5, 1, 3, 1, 7, 6, 0, 1, 8, 1, 3, 8, 11, 1, 7, 0, 13, 0, 3, 1, 14, 1, 0, 12, 17, 10, 0, 1, 19, 14, 7, 1, 20, 1, 3, 8, 23, 1, 15, 0, 24, 18, 3, 1, 26, 14, 7, 20, 29, 1, 18, 1, 31, 8, 0, 16, 32, 1, 3, 24, 34, 1, 0, 1, 37, 24, 3, 16, 38

Offset: 1

Author

Amiram Eldar, Jun 21 2025

Keywords

• nonn
• easy

Examples

			For n = 6, the greatest divisor of k that is a unitary divisor of 6 for k = 1 to 6 is 1, 2, 3, 2, 1 and 6, respectively. 3 of the values are primes, and therefore a(6) = 3.
		

Links

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

Crossrefs

The unitary analog of A117494.

Cf. A010051, A047994, A065463, A077610, A384047.

The number of integers k from 1 to n such that the greatest divisor of k that is a unitary divisor of n is: A047994 (1), A384048 (squarefree), A384049 (cubefree), A384050 (powerful), A384051 (cubefull), A384052 (square), A384053 (cube), A384054 (exponentially odd), A384055 (odd), A384056 (power of 2), A384057 (3-smooth), A384058 (5-rough), A385195 (1 or 2), this sequence (prime), A385197 (noncomposite), A385198 (prime power), A385199 (1 or prime power).

Programs

Mathematica

f[p_, e_] := p^e - 1; a[1] = 0; a[n_] := Module[{fct = FactorInteger[n]}, (Times @@ f @@@ fct)*(Total[Boole[# == 1] & /@ fct[[;; , 2]]/(fct[[;; , 1]] - 1)])]; Array[a, 100]

PARI

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

Python

from sympy.ntheory import factorint
from sympy import Rational
def a(n: int) -> int:
    if n == 1: return 0
    S, P, F = 0, 1, factorint(n)
    for p, e in F.items():
        P *= p**e - 1
        if e == 1: S += Rational(1, p - 1)
    return int(P * S)
print([a(n) for n in range(1, 79)])  # Peter Luschny, Jun 22 2025

Formula

The unitary convolution of A047994 (the unitary totient phi) with A010051 (the characteristic function of prime numbers): a(n) = Sum_{d | n, gcd(d, n/d) == 1} A047994(d) * A010051(n/d).

a(n) = uphi(n) * Sum_{p || n} (1/(p-1)), where uphi = A047994, and p || n denotes that p unitarily divides n (i.e., the p-adic valuation of n is 1).

a(n) = A385197(n) - A047994(n).

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = c1 * c2 = 0.21890744964919019488..., c1 = Product_{p prime}(1 - 1/(p*(p+1))) = A065463, and c2 = Sum_{p prime}((p^2-1)/(p^2*(p^2+p-1))) = 0.31075288978811405615... .