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.
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
Examples
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.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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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]
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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);}
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) = n / A375932(n).
a(n) = n if and only if n is a power of a squarefree number (A072774).
omega(a(n)) = A362611(n).
omega(a(n)) = 1 if and only if n is in A356862.
a(n!) = A060818(n) for n != 3.
Sum_{k=1..n} a(k) ~ c * n^2, where c = 3/Pi^2 = 0.303963... (A104141).
Comments