%I #20 Apr 24 2020 22:49:01
%S 1,3,6,6,10,10,14,14,14,14,22,22,26,26,26,26,34,34,38,38,38,38,46,46,
%T 46,46,46,46,58,58,62,62,62,62,62,62,74,74,74,74,82,82,86,86,86,86,94,
%U 94,94,94,94,94,106,106,106,106,106,106,118,118,122,122,122,122
%N Least k such that n! divides phi(k!) (k > 0).
%H Robert Israel, <a href="/A276000/b276000.txt">Table of n, a(n) for n = 1..5000</a>
%F a(n) = 2*A007917(n) for n>2. - _Andrey Zabolotskiy_, Aug 16 2016
%e a(2) = 3 because phi(3!) is divisible by 2!.
%p N:= 100: # for a(1)..a(N)
%p V:= Vector(N): n:= 0:
%p for k from 1 while n < N do
%p r:= numtheory:-phi(k!);
%p for i from n+1 to N while r mod (i!) = 0 do
%p V[i]:= k; n:= i;
%p od;
%p od:
%p convert(V,list);# _Robert Israel_, Apr 24 2020
%t Array[Block[{k = 1}, While[Mod[EulerPhi[k!], #!] != 0, k++]; k] &, 64] (* _Michael De Vlieger_, Apr 24 2020 *)
%o (PARI) a(n) = {my(k = 1); while(eulerphi(k!) % n!, k++); k; }
%Y Cf. A048855, A007917.
%K nonn
%O 1,2
%A _Altug Alkan_, Aug 16 2016
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