A226570 a(n) = Sum_{k=1..n} (k+1)! mod n.
0, 0, 2, 0, 2, 2, 4, 0, 8, 2, 10, 8, 8, 4, 2, 8, 11, 8, 7, 12, 11, 10, 19, 8, 12, 8, 8, 4, 15, 2, 0, 24, 32, 28, 32, 8, 3, 26, 8, 32, 2, 32, 14, 32, 17, 42, 16, 8, 46, 12, 11, 8, 11, 8, 32, 32, 26, 44, 26, 32, 20, 0, 53, 24, 47, 32, 63, 28, 65, 32, 66, 8, 53, 40, 62, 64, 32, 8, 18, 72, 62, 2, 25, 32, 62, 14, 44, 32, 74, 62, 60, 88, 62, 16, 7, 56, 78, 46, 98
Offset: 1
Keywords
Examples
a(3) = 2 because 2!, 3! and 4! are 2, 6 and 24 respectively, which add up to 32, and modulo 3 that is 2. a(4) = 0 because 2!, 3!, 4! and 5! add up to 152, and modulo 4 that is 0 (note that this is different from A086330(4) = 4).
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
Table[Mod[Sum[(k + 1)!, {k, n}], n], {n, 75}] (* Alonso del Arte, Jun 11 2013 *) -
PARI
a(n)=lift(sum(m=2,n-1,Mod(m!,n)))
-
PARI
a(n)=my(t=Mod(1,n)); lift(sum(m=2,n+1,t*=m)) \\ Charles R Greathouse IV, Jun 11 2013
-
Python
def A226570(n): a, c = 0, 1 for m in range(2,n): c = c*m%n if c==0: break a = (a+c)%n return a # Chai Wah Wu, Apr 16 2024
Formula
a(n) = A086330(n) mod n. - Chai Wah Wu, Apr 16 2024
a(n) = Sum_{k=1..A002034(n)-2} (k+1)! mod n. - David A. Corneth, Apr 16 2024
Comments