%I #14 Jul 20 2023 12:12:17

%S 2,3,7,71,179,547,983,1283,1289,2909,3709,20269,40829,256579,772573

%N Prime numbers p such that A022559(p) is a multiple of A000720(p).

%C Also primes p such that the number of prime factors (with repetition) of p! is a multiple of the number of different prime factors of p! (Prime numbers in A088533).

%e A022559(7) = 8 is a multiple of A000720(7) = 4.

%t a = {2}; b = {1}; For[n = 3, n < 1000, n++, If[PrimeQ[n], AppendTo[b, 1], c = FactorInteger[n]; For[j = 1, j < Length[c] + 1, j++, b[[PrimePi[c[[j, 1]]]]] = b[[PrimePi[c[[j, 1]]]]] + c[[j, 2]]]]; If[Mod[Plus @@ b, Length[b]] == 0, If[PrimeQ[n], AppendTo[a, n]]]]; a

%t Select[Prime[Range[530]],Divisible[PrimeOmega[#!],PrimeNu[#!]]&] (* The program generates the first 11 terms of the sequence. To generate more, increase the Range constant, but the program will then take a long time to run. *) (* _Harvey P. Dale_, Jan 01 2020 *)

%o (PARI) for(x=2,10000,x1=x!;y=bigomega(x1)/omega(x1); if(y==floor(y),if(isprime(x), print1((x)","))))

%Y Cf. A000720, A022559, A088533.

%K nonn,hard,more

%O 1,1

%A _Cino Hilliard_, Nov 16 2003

%E Edited and extended by _Stefan Steinerberger_, Dec 11 2007

%E Offset corrected by _Mohammed Yaseen_, Jul 20 2023

%E a(14)-a(15) from _Alois P. Heinz_, Jul 20 2023