A301317 a(n) = (n-1)! + 1 mod n^3.

0, 2, 3, 7, 25, 121, 35, 433, 226, 881, 495, 1, 676, 1233, 2701, 2049, 4420, 1, 4009, 1, 2647, 6425, 4945, 1, 626, 15393, 1, 1, 13137, 1, 21731, 1, 13069, 2041, 1, 1, 23532, 19153, 50194, 1, 14104, 1, 41237, 1, 1, 76729, 86433, 1, 1, 1, 78031, 1, 77645

Offset

1,2

Comments

There is no known number n > 1 for which a(n)=0.

For a(n) to equal 1, (n-1)! must be divisible by n^3 which tends to be the most frequent case for large n. For example, all n which are a product of three or more distinct primes belong to this category. So do all proper powers of primes except 2^2, 2^3, 2^4, 3^2, and 5^2.

Obviously, when a(n) = 1, then also A055976(n) = 1 and A301316(n) = 1.

If n is prime, a(n) is divisible by n. - Robert Israel, Mar 20 2018

Links

Stanislav Sykora, Table of n, a(n) for n = 1..10000

Wikipedia, Wilson prime

Formula

a(n) = ((n-1)! + 1) mod n^3. - Jon E. Schoenfield, Mar 18 2018

a(n) = A038507(n-1) mod A000578(n). - Michel Marcus, Mar 20 2018

Example

From Muniru A Asiru, Mar 20 2018: (Start)
((1-1)! + 1) mod 1^3 = (0! +1) mod 1 = 2 mod 1 = 0.
((2-1)! + 1) mod 2^3 = (1! +1) mod 8 = 2 mod 8 = 2.
((3-1)! + 1) mod 3^3 = (2! +1) mod 27 = 3 mod 27 = 3.
((4-1)! + 1) mod 4^3 = (3! +1) mod 64 = 7 mod 64 = 7.
((5-1)! + 1) mod 5^3 = (4! +1) mod 125 = 25 mod 125 = 25.
... (End)

Maple

seq((factorial(n-1)+1) mod n^3, n=1..60); # Muniru A Asiru, Mar 20 2018

Mathematica

Array[Mod[(# - 1)! + 1, #^3] &, 53] (* Michael De Vlieger, Mar 19 2018 *)

Prog

(PARI) a(n) = ((n-1)! + 1) % n^3; \\ Michel Marcus, Mar 18 2018
(GAP) List([1..60], n->(Factorial(n-1)+1) mod n^3); # Muniru A Asiru, Mar 20 2018

Crossrefs

Cf. A000578, A038507, A055976, A301316.

Sequence in context: A094697 A095910 A074189 * A325125 A091230 A063852

Adjacent sequences: A301314 A301315 A301316 * A301318 A301319 A301320

Keyword

nonn

Author

Stanislav Sykora, Mar 18 2018

Status

approved