A226198 -id:A226198 - cp's OEIS frontend

A056653 Composite numbers together with 1 but excluding 4.

1, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95

Offset: 1

Robert G. Wilson v, Aug 30 2000

These are also the numbers k such that k divides (k-1)!.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A018252, A002808, A226198, A014076 (odd terms).

for n from 1 to 100 do
  if irem(factorial(n-1),n) = 0 then print(n) end if;
end do: # Peter Bala, Jan 24 2017

Select[ Range[ 1, 100 ], Mod[ (# - 1)!, # ] == 0 & ]
Join[{1},Select[Range[5,100],CompositeQ]] (* Harvey P. Dale, Jun 14 2024 *)

from sympy import composite
def A056653(n): return composite(n) if n>1 else 1 # Chai Wah Wu, Jul 31 2024

Edited by Vladeta Jovovic, Apr 30 2003

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

Original entry on oeis.org

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

Alonso del Arte, Feb 16 2014

			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.

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

Cf. A061006, A181569, A226198, A022559.

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

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

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

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

a(p) = 0 for p prime.

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

A242426 a(n) = floor(n! / n^3).

Original entry on oeis.org

1, 0, 0, 0, 0, 3, 14, 78, 497, 3628, 29990, 277200, 2834328, 31770514, 387459072, 5108103000, 72397196844, 1097800704000, 17735107218083, 304112751022080, 5516784599040000, 105559797875432727, 2124765080865042873, 44881973505008640000, 992717442773183102976

Offset: 1

Alex Ratushnyak, May 14 2014

nonn

Cf. A000142, A000290, A000578, A072230, A242427.

Cf. A226198 (floor(n!/n^2)).

import math
for i in range(1,32): print(math.factorial(i)//(i**3), end=', ')

a(n) = floor(A000142(n-1) / A000290(n)).

Formula corrected by David Radcliffe, Aug 07 2025

cp's OEIS Frontend

A056653 Composite numbers together with 1 but excluding 4.

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

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A242426 a(n) = floor(n! / n^3).

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