A276935 Number of distinct prime factors prime(k) of n such that prime(k)^k, but not prime(k)^(k+1) is a divisor of n.
0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0
Offset: 1
Examples
For n = 12 = 2*2*3 = prime(1)^2 * prime(2)^1, neither of the prime factors satisfies the condition, thus a(12) = 0. For n = 18 = 2*3*3 = prime(1)^1 * prime(2)^2, both prime factors satisfy the condition, thus a(18) = 1+1 = 2. For n = 750 = 2*3*5*5*5 = prime(1)^1 * prime(2)^1 * prime(3)^3, only the prime factors 2 and 5 satisfy the condition, thus a(750) = 1+1 = 2.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
f[p_, e_] := If[PrimePi[p] == e, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 30 2023 *)
-
PARI
a(n) = {my(f = factor(n)); sum(i = 1, #f~, primepi(f[i,1]) == f[i,2]);} \\ Amiram Eldar, Sep 30 2023
Formula
a(1) = 0, for n > 1, a(n) = a(A028234(n)) + [A067029(n) = A055396(n)], where [] is Iverson bracket, giving 1 as its result when the stated equivalence is true and 0 otherwise.
From Amiram Eldar, Sep 30 2023: (Start)
Additive with a(p^e) = 1 if e = primepi(p), and 0 otherwise.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} (1/prime(k)^k - 1/prime(k)^(k+1)) = 0.33083690651252383414... . (End)