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A158378 a(1) = 0, a(n) = gcd(A051904(n), A051903(n)) for n >= 2.

%I A158378 #16 Sep 12 2024 01:43:28
%S A158378 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,
%T A158378 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,
%U A158378 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
%N A158378 a(1) = 0, a(n) = gcd(A051904(n), A051903(n)) for n >= 2.
%C A158378 a(n) for n >= 2 equals GCD of minimum and maximum exponents in the prime factorization of n.
%C A158378 a(n) for n >= 2 it deviates from A052409(n), first different term is a(10800) = a(2^4*3^3*5^2), a(10800) = gcd(2,4) = 2, A052409(10800) = gcd(2,3,4) = 1.
%H A158378 Antti Karttunen, <a href="/A158378/b158378.txt">Table of n, a(n) for n = 1..65536</a>
%H A158378 <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.
%F A158378 For n >= 2 holds: a(n)*A157754(n) = A051904(n)*A051903(n).
%F A158378 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).
%F A158378 Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - _Amiram Eldar_, Sep 11 2024
%e A158378 For n = 12 = 2^2 * 3^1 we have a(12) = gcd(2,1) = 1.
%t A158378 Table[GCD @@ {Min@ #, Max@ #} - Boole[n == 1] &@ FactorInteger[n][[All, -1]], {n, 100}] (* _Michael De Vlieger_, Jul 12 2017 *)
%o A158378 (PARI)
%o A158378 A051903(n) = if((1==n),0,vecmax(factor(n)[, 2]));
%o A158378 A051904(n) = if((1==n),0,vecmin(factor(n)[, 2]));
%o A158378 A158378(n) = gcd(A051903(n),A051904(n)); \\ _Antti Karttunen_, Jul 12 2017
%o A158378 (PARI) a(n) = if(n == 1, 0, my(e = factor(n)[,2]); gcd(vecmin(e), vecmax(e))); \\ _Amiram Eldar_, Sep 11 2024
%Y A158378 Cf. A157754, A051904, A051903, A052409.
%K A158378 nonn
%O A158378 1,4
%A A158378 _Jaroslav Krizek_, Mar 17 2009