1,2

n | sum_{m=1..n} (m+1)! => n | sum_{m=2..n+1} m! => n | sum_{m=2..n-1} m! for n>2 => sum_{m=2..n-1+k} (m! (mod n)) ==0 for all k>=0. If n is present and even then n/2 is present. - Robert G. Wilson v Nov 11 2004.

Numbers n such that n | (A007489(n+1)-1), also n | (A003422(n+2)-2).

The first few partial sums of (m+1)!, starting with m=1 are 2,8,32,152,872,5912,46232,409112. Of these, 2 is divisible by 1, 8 is divisible by 2, 152 is divisible by 4, but 32 is not divisible by 3. Therefore the first few terms of this sequence are 1,2,4.

s = -1; Do[s = s + n!; If[ Mod[s, n] == 0, Print[n]], {n, 50000}] (from Robert G. Wilson v Nov 15 2004)

(PARI) s=0:for(n=1, 5000, s=s+(n+1)!: if(s%n==0, print(n)))

Cf. A057245.

Sequence in context: A006398 A053364 A018355 this_sequence A053147 A128055 A061285

Adjacent sequences: A100080 A100081 A100082 this_sequence A100084 A100085 A100086

nonn

Mark Hudson (mrmarkhudson(AT)hotmail.com), Nov 08 2004

a(13) & a(14) from Robert G. Wilson v (rgwv(AT)rgwv.com), Nov 15 2004