A158378 a(1) = 0, a(n) = gcd(A051904(n), A051903(n)) for n >= 2.
0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2
Offset: 1
Keywords
Examples
For n = 12 = 2^2 * 3^1 we have a(12) = gcd(2,1) = 1.
Links
Programs
-
Mathematica
Table[GCD @@ {Min@ #, Max@ #} - Boole[n == 1] &@ FactorInteger[n][[All, -1]], {n, 100}] (* Michael De Vlieger, Jul 12 2017 *)
-
PARI
A051903(n) = if((1==n),0,vecmax(factor(n)[, 2])); A051904(n) = if((1==n),0,vecmin(factor(n)[, 2])); A158378(n) = gcd(A051903(n),A051904(n)); \\ Antti Karttunen, Jul 12 2017
-
PARI
a(n) = if(n == 1, 0, my(e = factor(n)[,2]); gcd(vecmin(e), vecmax(e))); \\ Amiram Eldar, Sep 11 2024
Formula
a(1) = 0, a(p) = 1, a(pq) = 1, a(pq...z) = 1, a(p^k) = k, for p = primes (A000040), pq = product of two distinct primes (A006881), pq...z = product of k (k > 2) distinct primes p, q, ..., z (A120944), p^k = prime powers (A000961(n) for n > 1), k = natural numbers (A000027).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Sep 11 2024
Comments