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A375931
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
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, 4, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 8, 41, 42, 43, 4, 9, 46, 47, 16, 49, 25, 51, 4, 53, 27, 55, 8, 57, 58, 59, 4, 61, 62, 9, 64, 65, 66, 67, 4, 69, 70, 71
OFFSET
1,2
COMMENTS
Differs from A327526 at n = 12, 20, 28, 40, 44, 45, ... .
Each positive number appears in this sequence either once or infinitely many times:
1. If m is squarefree then the only solution to a(x) = m is x = m.
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.
LINKS
FORMULA
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.
a(n) = A261969(n)^A051903(n).
a(n) = n / A375932(n).
a(n) = n if and only if n is a power of a squarefree number (A072774).
A051903(a(n)) = A051903(n).
omega(a(n)) = A362611(n).
omega(a(n)) = 1 if and only if n is in A356862.
Omega(a(n)) = A051903(n) * A362611(n).
a(n!) = A060818(n) for n != 3.
Sum_{k=1..n} a(k) ~ c * n^2, where c = 3/Pi^2 = 0.303963... (A104141).
EXAMPLE
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.
MATHEMATICA
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]
PROG
(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); }
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Sep 03 2024
STATUS
approved