

A346094


a(n) = n / A275823(n), where A275823(n) is the least k such that n divides phi(k^2).


1



1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 4, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 2, 3, 4, 1, 6, 1, 2, 1, 2, 1, 4, 1, 2, 1, 4, 1, 6, 5, 2, 3, 2, 1, 4, 1, 2, 3, 4, 1, 2, 1, 4, 1, 2, 1, 2, 1, 2, 1, 2, 1, 6, 1, 4, 3, 2, 1, 6, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,6


COMMENTS

a(n) = n divided by the least k such that A002618(k) [= phi(k^2) = k*phi(k)] is a multiple of n.
It is easy to see that such k is always a divisor of n since k contains only some of prime factors of n and there cannot be other prime factor that does not divide n. In order to see this, let us assume p divides k (where p is prime that does not divide n) and (p1) contribute the division in A275823. At this case there is definitely smaller option to do this instead of p1 since it is always possible that k could contain necessary prime powers from factorization of p1 instead of p. At the same time, obviously A275823(n) <= n. So terms of this sequence are always integers.


LINKS

Table of n, a(n) for n=1..105.


MATHEMATICA

Array[#/Block[{k = 1}, While[! Mod[EulerPhi[k^2], #] == 0, k++]; k] &, 105] (* Michael De Vlieger, Jul 22 2021 *)


PROG

(PARI) A346094(n) = { my(k=1); while((k*eulerphi(k)) % n, k++); (n/k); };


CROSSREFS

Cf. A002618, A275823.
Sequence in context: A161305 A161280 A160984 * A082898 A095999 A334948
Adjacent sequences: A346091 A346092 A346093 * A346095 A346096 A346097


KEYWORD

nonn


AUTHOR

Antti Karttunen and Altug Alkan, Jul 21 2021


STATUS

approved



