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The product of the prime powers in the prime factorization of n that have an exponent that is equal to the maximum exponent in this factorization.
3

%I #15 Sep 08 2024 08:07:11

%S 1,2,3,4,5,6,7,8,9,10,11,4,13,14,15,16,17,9,19,4,21,22,23,8,25,26,27,

%T 4,29,30,31,32,33,34,35,36,37,38,39,8,41,42,43,4,9,46,47,16,49,25,51,

%U 4,53,27,55,8,57,58,59,4,61,62,9,64,65,66,67,4,69,70,71

%N The product of the prime powers in the prime factorization of n that have an exponent that is equal to the maximum exponent in this factorization.

%C Differs from A327526 at n = 12, 20, 28, 40, 44, 45, ... .

%C Each positive number appears in this sequence either once or infinitely many times:

%C 1. If m is squarefree then the only solution to a(x) = m is x = m.

%C 2. If m = s^k is a power of a squarefree number s with k >= 2, then x = m * i is a solution to a(x) = m for all numbers i that are k-free numbers (i.e., having exponents in their prime factorizations that are all less than k) that are coprime to m.

%H Amiram Eldar, <a href="/A375931/b375931.txt">Table of n, a(n) for n = 1..10000</a>

%F If n = Product_{i} p_i^e_i (where p_i are distinct primes) then a(n) = Product_{i} p_i^(e_i * [e_i = max_{j} e_j]), where [] is the Iverson bracket.

%F a(n) = A261969(n)^A051903(n).

%F a(n) = n / A375932(n).

%F a(n) = n if and only if n is a power of a squarefree number (A072774).

%F A051903(a(n)) = A051903(n).

%F omega(a(n)) = A362611(n).

%F omega(a(n)) = 1 if and only if n is in A356862.

%F Omega(a(n)) = A051903(n) * A362611(n).

%F a(n!) = A060818(n) for n != 3.

%F Sum_{k=1..n} a(k) ~ c * n^2, where c = 3/Pi^2 = 0.303963... (A104141).

%e 180 = 2^2 * 3^2 * 5, and the maximum exponent in the prime factorization of 180 is 2, which is the exponent of its prime factors 2 and 3. Therefore a(180) = 2^2 * 3^2 = (2*3)^2 = 36.

%t a[n_] := Module[{f = FactorInteger[n], p, e, i, m}, p = f[[;; , 1]]; e = f[[;; , 2]]; m = Max[e]; i = Position[e, m] // Flatten; (Times @@ p[[i]])^m]; Array[a, 100]

%o (PARI) a(n) = {my(f = factor(n), p = f[,1], e = f[,2], m); if(n == 1, 1, m = vecmax(e); prod(i = 1, #p, if(e[i] == m, p[i], 1))^m);}

%Y Cf. A001221 (omega), A001222 (Omega), A005117, A051903, A060818, A072774, A104141, A261969, A327526, A356862, A362611, A375932.

%K nonn,easy

%O 1,2

%A _Amiram Eldar_, Sep 03 2024