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a(1) = 0, a(n) = gcd(A051904(n), A051903(n)) for n >= 2.
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%I #16 Sep 12 2024 01:43:28

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

%C a(n) for n >= 2 equals GCD of minimum and maximum exponents in the prime factorization of n.

%C 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 Antti Karttunen, <a href="/A158378/b158378.txt">Table of n, a(n) for n = 1..65536</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.

%F For n >= 2 holds: a(n)*A157754(n) = A051904(n)*A051903(n).

%F 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 Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - _Amiram Eldar_, Sep 11 2024

%e For n = 12 = 2^2 * 3^1 we have a(12) = gcd(2,1) = 1.

%t Table[GCD @@ {Min@ #, Max@ #} - Boole[n == 1] &@ FactorInteger[n][[All, -1]], {n, 100}] (* _Michael De Vlieger_, Jul 12 2017 *)

%o (PARI)

%o A051903(n) = if((1==n),0,vecmax(factor(n)[, 2]));

%o A051904(n) = if((1==n),0,vecmin(factor(n)[, 2]));

%o A158378(n) = gcd(A051903(n),A051904(n)); \\ _Antti Karttunen_, Jul 12 2017

%o (PARI) a(n) = if(n == 1, 0, my(e = factor(n)[,2]); gcd(vecmin(e), vecmax(e))); \\ _Amiram Eldar_, Sep 11 2024

%Y Cf. A157754, A051904, A051903, A052409.

%K nonn

%O 1,4

%A _Jaroslav Krizek_, Mar 17 2009