cp's OEIS Frontend

A346094 a(n) = n / A275823(n), where A275823(n) is the least k such that n divides phi(k^2).

%I A346094 #23 Aug 16 2021 22:47:35
%S A346094 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,
%T A346094 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,
%U A346094 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
%N A346094 a(n) = n / A275823(n), where A275823(n) is the least k such that n divides phi(k^2).
%C A346094 a(n) = n divided by the least k such that A002618(k) [= phi(k^2) = k*phi(k)] is a multiple of n.
%C A346094 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 (p-1) contribute the division in A275823. At this case there is definitely smaller option to do this instead of p-1 since it is always possible that k could contain necessary prime powers from factorization of p-1 instead of p. At the same time, obviously A275823(n) <= n. So terms of this sequence are always integers.
%t A346094 Array[#/Block[{k = 1}, While[! Mod[EulerPhi[k^2], #] == 0, k++]; k] &, 105] (* _Michael De Vlieger_, Jul 22 2021 *)
%o A346094 (PARI) A346094(n) = { my(k=1); while((k*eulerphi(k)) % n, k++); (n/k); };
%Y A346094 Cf. A002618, A275823.
%K A346094 nonn
%O A346094 1,6
%A A346094 _Antti Karttunen_ and _Altug Alkan_, Jul 21 2021