

A238002


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


2



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
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OFFSET

2,5


LINKS

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


FORMULA

a(p) = 0 for p prime.
a(2n) > a(2n + 1) for all n > 2.


EXAMPLE

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.


MAPLE

with(numtheory):
a:= n> add(add(`if`(i[1] in factorset(n), i[2], 0),
i=ifactors(j)[2]), j=1..n1):
seq(a(n), n=2..100); # Alois P. Heinz, Mar 17 2014


MATHEMATICA

cmpf[n_]:=Count[Flatten[Table[#[[1]], {#[[2]]}]&/@FactorInteger[ (n1)!]], _?( MemberQ[Transpose[FactorInteger[n]][[1]], #]&)]; Array[cmpf, 80] (* Harvey P. Dale, Jan 23 2016 *)


PROG

(Sage)
m=100 # change n for more terms
[sum(valuation(factorial(n1), p) for p in prime_divisors(n) if p in prime_divisors(factorial(n1))) for n in [2..m]] # Tom Edgar, Mar 14 2014
(PARI) a(n) = {nm = (n1)!; fn = factor(n); sum (i=1, #fn~, valuation(nm, fn[i, 1])); } \\ Michel Marcus, Mar 15 2014


CROSSREFS

Cf. A061006, A181569, A226198, A022559.
Sequence in context: A160214 A081087 A135031 * A016680 A062524 A152856
Adjacent sequences: A237999 A238000 A238001 * A238003 A238004 A238005


KEYWORD

nonn,easy


AUTHOR

Alonso del Arte, Feb 16 2014


STATUS

approved



