A343783 - OEIS

%I #23 May 05 2021 21:03:32

%S 1,1,2,2,2,2,6,2,6,2,2,2,6,6,2,2,2,6,6,2,6,2,2,2,2,6,6,6,2,2,30,2,2,2,

%T 6,6,6,6,6,2,2,6,6,2,6,2,2,2,6,2,2,6,2,6,2,6,6,2,2,2,30,30,6,2,6,2,6,

%U 2,2,6,2,6,6,6,2,6,30,6,6,2,6,2,2,6,2,6

%N a(n) is the largest primorial number (A002110) which divides phi(n).

%H Amiram Eldar, <a href="/A343783/b343783.txt">Table of n, a(n) for n = 1..10000</a>

%H Paul Pollack and Carl Pomerance, <a href="https://doi.org/10.1215/00192082-8591576">Phi, primorials, and Poisson</a>, Illinois J. Math., Vol. 64, No. 3 (2020), pp. 319-330; <a href="http://pollack.uga.edu/IJM-ppp.pdf">alternative link</a>.

%F a(n) = A053589(A000010(n)).

%F Let pr(n) be the largest prime divisor of a(n) (i.e., a(n) = pr(n)# = A034386(pr(n))). Then pr(n) ~ log(log(n))/log(log(log(n))) on a set of integers of asymptotic density 1 (Pollack and Pomerance, 2020).

%F From _Bernard Schott_, May 05 2021: (Start)

%F a(2n) = a(n) for n>=1.

%F a(n) = 1 iff n = 1 or n = 2.

%F a(n) = 2 iff 3 does not divide phi(n) (A088232)

%F a(n) >= 6 iff 3 divides phi(n) (A066498). (End)

%e a(3) = 2 since phi(3) = 2 and 2 = A002110(1).

%e a(5) = 2 since phi(5) = 4 and 2 = A002110(1) is the largest primorial dividing 4.

%e a(7) = 6 since phi(7) = 6 and 6 = A002110(2).

%t prim[n_] := Times @@ Prime[Range[n]]; gp[n_] := Module[{k = 1}, While[Divisible[n, prim[k]], k++]; prim[k - 1]]; a[n_] := gp[EulerPhi[n]]; Array[a, 100]

%o (PARI) f(n) = my(s=1); forprime(p=2, , if(n%p, return(s), s *= p)); \\ A053589

%o a(n) = f(eulerphi(n)); \\ _Michel Marcus_, May 01 2021

%Y Cf. A000010, A002110, A034386, A053589.

%K nonn

%O 1,3

%A _Amiram Eldar_, Apr 29 2021