A238002 Count with multiplicity of prime factors of n in (n - 1)!.

0, 0, 1, 0, 4, 0, 4, 2, 8, 0, 12, 0, 11, 7, 11, 0, 21, 0, 19, 10, 19, 0, 28, 4, 23, 10, 26, 0, 44, 0, 26, 16, 32, 11, 47, 0, 35, 19, 43, 0, 61, 0, 42, 28, 42, 0, 63, 6, 56, 24, 50, 0, 72, 16, 58, 28, 54, 0, 94, 0, 57, 37, 57, 18, 98, 0, 67, 33, 91, 0, 99, 0, 71, 50, 74, 17, 113, 0, 92

Offset: 2

Author

Alonso del Arte, Feb 16 2014

Keywords

• nonn
• easy

Examples

			a(4) = 1 because 3! = 6 = 2 * 3, which has one prime factor (2) in common with 4.
a(5) = 0 because gcd(4!, 5) = 1.
a(6) = 4 because 5! = 120 = 2^3 * 3 * 5, which has four factors (2 thrice and 3 once) in common with 6.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A061006, A181569, A226198, A022559.

Programs

Maple

with(numtheory):
a:= n-> add(add(`if`(i[1] in factorset(n), i[2], 0),
        i=ifactors(j)[2]), j=1..n-1):
seq(a(n), n=2..100);  # Alois P. Heinz, Mar 17 2014

Mathematica

cmpf[n_]:=Count[Flatten[Table[#[[1]],{#[[2]]}]&/@FactorInteger[ (n-1)!]], ?( MemberQ[Transpose[FactorInteger[n]][[1]],#]&)]; Array[cmpf,80] (* _Harvey P. Dale, Jan 23 2016 *)

PARI

a(n) = {nm = (n-1)!; fn = factor(n); sum (i=1, #fn~, valuation(nm, fn[i,1]));} \\ Michel Marcus, Mar 15 2014

Sage

m=100 # change n for more terms
[sum(valuation(factorial(n-1),p) for p in prime_divisors(n) if p in prime_divisors(factorial(n-1))) for n in [2..m]] # Tom Edgar, Mar 14 2014

Formula

a(p) = 0 for p prime.

a(2n) > a(2n + 1) for all n > 2.