

A226570


a(n) = Sum_{k=1..n} (k+1)! mod n.


1



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
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OFFSET

1,3


COMMENTS

Motivated by sequence A100083, numbers such that a(n) = 0.  Hasler
Note that this is a different sequence from A086330: in this one, the factorials are added up and then the remainder of the total divided by n is taken, whereas in A086330 each factorial is computed modulo n prior to being added up.  Alonso del Arte, Jun 11 2013


LINKS



FORMULA



EXAMPLE

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.


MATHEMATICA

Table[Mod[Sum[(k + 1)!, {k, n}], n], {n, 75}] (* Alonso del Arte, Jun 11 2013 *)


PROG

(PARI) a(n)=lift(sum(m=2, n1, Mod(m!, n)))
(Python)
a, c = 0, 1
for m in range(2, n):
c = c*m%n
if c==0:
break
a = (a+c)%n


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



