A276000 Least k such that n! divides phi(k!) (k > 0).

1, 3, 6, 6, 10, 10, 14, 14, 14, 14, 22, 22, 26, 26, 26, 26, 34, 34, 38, 38, 38, 38, 46, 46, 46, 46, 46, 46, 58, 58, 62, 62, 62, 62, 62, 62, 74, 74, 74, 74, 82, 82, 86, 86, 86, 86, 94, 94, 94, 94, 94, 94, 106, 106, 106, 106, 106, 106, 118, 118, 122, 122, 122, 122

Offset: 1

Author

Altug Alkan, Aug 16 2016

Keywords

• nonn

Examples

			a(2) = 3 because phi(3!) is divisible by 2!.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A048855, A007917.

Programs

Maple

N:= 100: # for a(1)..a(N)
V:= Vector(N): n:= 0:
for k from 1 while n < N do
  r:= numtheory:-phi(k!);
  for i from n+1 to N while r mod (i!) = 0 do
    V[i]:= k; n:= i;
  od;
od:
convert(V,list);# Robert Israel, Apr 24 2020

Mathematica

Array[Block[{k = 1}, While[Mod[EulerPhi[k!], #!] != 0, k++]; k] &, 64] (* Michael De Vlieger, Apr 24 2020 *)

PARI

a(n) = {my(k = 1); while(eulerphi(k!) % n!, k++); k; }

Formula

a(n) = 2*A007917(n) for n>2. - Andrey Zabolotskiy, Aug 16 2016