A180492 -id:A180492 - cp's OEIS frontend

A180491 Product of remainders of n mod k, for k = 2,3,4,...,n-1.

1, 1, 1, 0, 2, 0, 6, 0, 0, 0, 720, 0, 2160, 0, 0, 0, 2419200, 0, 65318400, 0, 0, 0, 754427520000, 0, 0, 0, 0, 0, 32953394073600000, 0, 311409573995520000, 0, 0, 0, 0, 0, 37269497815783833600000, 0, 0, 0, 7890485108998805913600000000, 0

Offset: 1

Carl R. White, Sep 08 2010

nonn

a(n) is zero where n is composite and is trivially less than or equal to n! when n is prime or 1.

a(n)=0 iff n is composite. See A180492. - Robert G. Wilson v, Sep 09 2010

			a(7) = (7 mod 2) * (7 mod 3) * (7 mod 4) * (7 mod 5) * (7 mod 6) = 1 * 1 * 3 * 2 * 1 = 6.

Cf. A034386, A000142, A004125, A180492, A180493.

Cf. A080339, A173392.

a:=proc(n) if n=1 then 1; elif isprime(n)=true then mul(n mod i, i=2..n-1); else 0; fi: end: seq(a(n), n=1..60); # Ridouane Oudra, Nov 01 2024

f[n_] := Times @@ Mod[n, Range[2, n - 1]]; Array[f, 42] (* Robert G. Wilson v, Sep 09 2010 *)

a(n) = A080339(n)*A173392(n). - Ridouane Oudra, Nov 01 2024

A180493 Numbers n such that prime(n)! is divisible by A180491(prime(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 29, 30, 35, 38, 39, 40, 53

Offset: 1

Carl R. White, Sep 08 2010

Also numbers n such that prime(n)! is divisible by A180492(n).
Is this sequence finite?
From Robert G. Wilson v, Sep 09 2010: (Start)
Checked to 10000 for more terms.
Conjecture: This sequence is finite and all its terms are present. (End)

Cf. A034386, A000142, A004125, A180491, A180492.

f[n_] := Times @@ Mod[n, Range[2, n - 1]]; k = 1; lst = {}; While[k < 10001, If[ Divisible[ Prime@k!, f@Prime@k], AppendTo[lst, k]; Print@k]; k++ ]; lst (* Robert G. Wilson v, Sep 09 2010 *)

cp's OEIS Frontend

A180491 Product of remainders of n mod k, for k = 2,3,4,...,n-1.

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