A100083 Numbers n such that n divides Sum_{m=1..n} (m+1)!.

1, 2, 4, 8, 31, 62, 124, 248, 373, 746, 1492, 2984, 11563, 23126, 46252, 92504

Offset: 1

Author

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

Keywords

• nonn
• more

Comments

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! == 0 (mod n) for all k>=0. If n is present and even, then n/2 is present. - Robert G. Wilson v, Nov 11 2004

The terms occur in groups of 4, in a series of n, 2n, 4n, and 8n. Is there any way of calculating the next term in the reduced series: 1, 31, 373, 11563? - Harvey P. Dale, Jun 11 2013

If n is in the sequence and k|n, then k is also in the sequence. In the other direction, if s and t is in the sequence and gcd(s,t)=1, then n=s*t is also in the sequence. Therefore, we need to check only the prime powers, after which we can easily build the rest of the sequence. The prime powers in the sequence begin with 2, 4, 8, 31, 373, ... - Robert Gerbicz, Jun 11 2013

For any terms in this sequence, their LCM also belongs to this sequence. If a(17) exists, it is prime. - Max Alekseyev, Jun 11 2013

a(17) > 74*10^7. - Lars Blomberg, Jun 15 2013

Integers n such that n | (A003422(n) - 2). - David W. Wilson, Jul 20 2013

Examples

			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.
		

Links

• R. Gerbicz (and others), Re: A100083, SeqFan list, Jun 11 2013

Crossrefs

Cf. A057245, A064384

Programs

Mathematica

s = -1; Do[s = s + n!; If[ Mod[s, n] == 0, Print[n]], {n, 50000}] (* Robert G. Wilson v, Nov 15 2004 *)
Take[Flatten[Select[MapIndexed[List,Accumulate[Range[2,24000]!]], Divisible[#[[1]],#[[2,1]]]&]],{2,-1,2}] (* Harvey P. Dale, Jun 11 2013 *)

PARI

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

PARI

is(n)=my(t=Mod(1,n));sum(m=2,n+1,t*=m)==0 \\ Charles R Greathouse IV, Jun 11 2013

Python

from itertools import count, islice
def A100083_gen(startvalue=1): # generator of terms >= startvalue
    for n in count(max(startvalue,1)):
        a, c = 0, 1
        for m in range(2,n):
            c = c*m%n
            if c==0:
                break
            a = (a+c)%n
        if not a:
            yield n
A100083_list = list(islice(A100083_gen(),20)) # Chai Wah Wu, Apr 16 2024

Formula

Numbers n such that n | (A007489(n+1)-1), also n | (A003422(n+2)-2), n | A054116(n+1).

Extensions

a(13)-a(14) from Robert G. Wilson v, Nov 15 2004

a(15) from Harvey P. Dale, Jun 11 2013

a(16) from Giovanni Resta, confirmed by Charles R Greathouse IV and Robert G. Wilson v, Jun 11 2013

Edited by Max Alekseyev, Mar 27 2015