OFFSET
1,2
COMMENTS
a(n) = 0 only for n = 1. a(n) = k occurs infinitely many times for k >= 1. The numbers n = 2^e * m = 2^A007814(n) * A000265(n) for which a(n) = k and their asymptotic density are as follows:
1. k = 1: n is an odd squarefree number (A056911) and the density is d(1) = 2/(3*zeta(2)) = 0.405284... (A185199).
2. k >= 3 is odd: e < (k+1)/2 and m is a (k+1)-free number that is not a k-free number: d(k) = (1 - 1/2^((k+1)/2)) * (f(k+1)/zeta(k+1) - f(k)/zeta(k)), where f(k) = 1 - 1/2^k.
3. k >= 2 is even: e = k/2 and m is a (k+1)-free number, or e < k/2 and m is a (k+1)-free number that is not a k-free number: d(k) = (1/2^(k/2+1)) * f(k+1)/zeta(k+1) + (1-1/2^(k/2)) * (f(k+1)/zeta(k+1) - f(k)/zeta(k)), where f(k) is defined above.
The asymptotic mean of this sequence is Sum_{k>=1} k * d(k) = 2.64836785173193409440576... .
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Gaussian Prime.
Wikipedia, Gaussian integer: Unique factorization.
EXAMPLE
a(2) = 2 because 2 = -i * (1+i)^2.
MATHEMATICA
a[n_] := Max[FactorInteger[n, GaussianIntegers -> True][[;; , 2]]]; a[1] = 0; Array[a, 100]
(* or *)
a[n_] := Module[{e = IntegerExponent[n, 2], od, em}, odd = n / 2^e; Max[2*e, If[odd == 1, 0, Max[FactorInteger[odd][[;; , 2]]]]]]; Array[a, 100]
PROG
(PARI) a(n) = if(n == 1, 0, vecmax(factor(n*I)[, 2]));
(PARI) a(n) = my(e = valuation(n, 2), es = factor(n >> e)[, 2]); max(2*e, if(#es, vecmax(es), 0));
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Oct 01 2024
STATUS
approved