A057245 -id:A057245 - cp's OEIS frontend

A049782 a(n) = (0! + 1! + ... + (n-1)!) mod n.

0, 0, 1, 2, 4, 4, 6, 2, 1, 4, 1, 10, 10, 6, 4, 10, 13, 10, 9, 14, 13, 12, 21, 10, 14, 10, 10, 6, 17, 4, 2, 26, 1, 30, 34, 10, 5, 28, 10, 34, 4, 34, 16, 34, 19, 44, 18, 10, 48, 14, 13, 10, 13, 10, 34, 34, 28, 46, 28, 34, 22, 2, 55, 26, 49, 34, 65, 30, 67, 34, 68, 10, 55, 42, 64, 66, 34

Offset: 1

Clark Kimberling

Kurepa's conjecture is that gcd(!n,n!) = 2, n > 1. It is easy to prove that this is equivalent to showing that gcd(p,!p) = 1 for all odd primes p. In Guy, 2nd edition, it is stated that Mijajlovic has tested up to p = 10^6. Subsequently Gallot tested up to 2^26. I have continued up to just above p = 2^27, in fact to p < 144000000. There were no examples found where gcd(p,!p) > 1. - Paul Jobling, Dec 02 2004
According to Kellner, the conjecture has been proved by Barsky and Benzaghou. - T. D. Noe, Dec 02 2004
Barsky and Benzaghou withdrew their proof in 2011. I've extended the search up to 10^9; no counterexample was found. - Milos Tatarevic, Feb 01 2013

R. K. Guy, Unsolved Problems in Number Theory, B44: is a(n) > 0 for n > 2?

T. D. Noe, Table of n, a(n) for n = 1..10000
Romeo Mestrovic, The Kurepa-Vandermonde matrices arising from Kurepa's left factorial hypothesis, Filomat 29:10 (2015), 2207-2215. DOI:10.2298/FIL1510207M
Vladica Andrejic, Milos Tatarevic, Searching for a counterexample of Kurepa's Conjecture, arXiv:1409.0800 [math.NT], 2014.
D. Barsky and B. Benzaghou, Nombres de Bell et somme de factorielles, Journal de Théorie des Nombres de Bordeaux, 16:1, No. 17, 2004.
D. Barsky and B. Benzaghou, Erratum à l'article "Nombres de Bell et somme de factorielles", Journal de Théorie des Nombres de Bordeaux, 23:2 (2011), p. 527.
Y. Gallot, More information.
Bernd C. Kellner, Some remarks on Kurepa's left factorial, arXiv:math/0410477 [math.NT], 2004.
Romeo Mestrovic, Variations of Kurepa's left factorial hypothesis, arXiv preprint arXiv:1312.7037 [math.NT], 2013-2014.
Stephen A. Silver, C program to generate this sequence.
M. Tatarevic, Searching for a counterexample to the Kurepa's left factorial hypothesis (p < 10^9).

Cf. A000142, A057245.

Note that in the context of this sequence, !n is the left factorial A003422 not the subfactorial A000166.

List([1..80], n-> Sum([0..n-1], k-> Factorial(k)) mod n ); # G. C. Greubel, Dec 11 2019

a049782 :: Int -> Integer
a049782 n = (sum $ take n a000142_list) `mod` (fromIntegral n)
-- Reinhard Zumkeller, Nov 02 2011

[&+[Factorial(k-1): k in [1..n]] mod (n): n in [1..80]]; // Vincenzo Librandi, May 31 2019

a:= proc(n) local c, i, t; c, t:=1, 1;
      for i to n-1 do t:= (t*i) mod n; c:= c+t od; c mod n
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 16 2013

Table[Mod[Sum[ i!, {i, 0, n-1}], n], {n, 80}]
nn=80; With[{fcts=Accumulate[Range[0,nn]!]},Flatten[Table[Mod[Take[fcts,{n}], n], {n,nn}]]] (* Harvey P. Dale, Sep 22 2011 *)

a(n)=my(s=1,f=1); for(k=1,n, f=f*k%n; s+=f); s%n \\ Charles R Greathouse IV, Feb 07 2017

[mod(sum(factorial(k) for k in (0..n-1)), n) for n in (1..80)] # G. C. Greubel, Dec 11 2019

a(n) = A003422(n) mod n = !n mod n. - G. C. Greubel, Dec 11 2019

More terms from Erich Friedman, who observes that the first 500 terms are nonzero. Independently extended by Stephen A. Silver.

A064383 Integers n >= 1 such that n divides 0!-1!+2!-3!+4!-...+(-1)^{n-1}(n-1)!.

Original entry on oeis.org

1, 2, 4, 5, 10, 13, 20, 26, 37, 52, 65, 74, 130, 148, 185, 260, 370, 463, 481, 740, 926, 962, 1852, 1924, 2315, 2405, 4630, 4810, 6019, 9260, 9620, 12038, 17131, 24076, 30095, 34262, 60190, 68524, 85655, 120380, 171310, 222703, 342620, 445406, 890812, 1113515

Offset: 1

Kevin Buzzard (buzzard(AT)ic.ac.uk), Sep 28 2001

If a is in the sequence, then so are all its positive divisors. If a and b are coprime and in the sequence, then so is their product. Hence in extending the sequence, one may as well just look for primes in the sequence (and then check powers of these primes). Heuristically one might expect a very sparse but infinite set of primes in the sequence, but the largest one I know is p=463 and I've searched up to 600000. This sequence was brought to my attention by David Loeffler.
Also, n such that A000522(n)==1 (mod n^2). - Benoit Cloitre, Apr 15 2003
The primes in this sequence are the same as the terms > 1 in A124779. - Jonathan Sondow, Nov 09 2006
Also, n such that n|A(n-1), where A(0) = 1 and A(k) = k*A(k-1)+1 = A000522(k) for k > 0. - Jonathan Sondow, Dec 22 2006
Michael Mossinghoff has calculated that 2, 5, 13, 37, 463 are the only primes in the sequence up to 150 million. - Jonathan Sondow, Jun 12 2007

			4 is in the sequence because 4 divides 0!-1!+2!-3!=1-1+2-6=-4.

R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Springer-Verlag, 2004, B43.

J. Sondow, The Taylor series for e and the primes 2, 5, 13, 37, 463: a surprising connection
J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
Index entries for sequences related to factorial numbers

Cf. A000522, A057245, A064384, A124779, A129924.

s = 0; Do[ s = s + (-1)^(n)(n)!; If[ Mod[ s, n + 1 ] == 0, Print[ n + 1 ] ], {n, 0, 600000} ]
Divisors[4454060] (* From Formula above *) (* Harvey P. Dale, Aug 09 2012 *)

Up to n=600000, these are just the divisors of 4*5*13*37*463.

More terms from Sean A. Irvine, Jul 02 2023

A188775 Numbers k such that Sum_{j=1..k} j^j == -1 (mod k).

Original entry on oeis.org

1, 2, 3, 6, 14, 42, 46, 1806, 2185, 4758, 5266, 10895, 24342, 26495, 44063, 52793, 381826, 543026, 547311, 805002

Offset: 1

José María Grau Ribas, Apr 10 2011

Numbers k such that A001923(k) == -1 (mod k).
a(21) > 10^7. - Hiroaki Yamanouchi, Aug 25 2015
Numbers k such that k divides A062970(k). - Jianing Song, Feb 03 2019

			6 is a term because 1^1 + 2^2 + 3^3 + 4^4 + 5^5 + 6^6 = 50069 and 50069 + 1 = 6 * 8345. - _Bernard Schott_, Feb 03 2019

Cf. A128981 (sum == 0 (mod n)), A188776 (sum == 1 (mod n)).
Cf. A057245.
Cf. A001923, A062970.

isA188775 := proc(n) add( modp(k &^ k,n),k=1..n) ; if modp(%,n) = n-1 then true; else false; end if; end proc:
for n from 1 do if isA188775(n) then printf("%d\n",n) ; end if; end do: # R. J. Mathar, Apr 10 2011

Union@Table[If[Mod[Sum[PowerMod[i,i,n],{i,1,n}],n]==n-1,Print[n];n],{n,1,10000}]

f(n)=lift(sum(k=1,n,Mod(k,n)^k));
for(n=1,10^6,if(f(n)==n-1,print1(n,", "))) \\ Joerg Arndt, Apr 10 2011

m=0;for(n=1,1000,m=m+n^n;if((m+1)%n==0,print1(n,", "))) \\ Jinyuan Wang, Feb 04 2019

sum = 0
for n in range(10000):
    sum += n**n
    if sum % (n+1) == 0:
        print(n+1, end=',')
# Alex Ratushnyak, May 13 2013

a(12)-a(16) from Joerg Arndt, Apr 10 2011

a(17)-a(20) from Lars Blomberg, May 10 2011

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

Original entry on oeis.org

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

Offset: 1

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

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

			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.

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

Cf. A057245, A064384

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 *)

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

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

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

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

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

A216056 Primes p such that p divides the numerator of Sum( k!/2^k, k=1..p-1 ).

Original entry on oeis.org

234781, 115480283

Offset: 1

N. J. A. Sloane, Aug 30 2012

B. M. M. de Weger, Sums with factorials, NMBRTHRY list, Aug 28 2012

Cf. A215972, A215974, A057245

A225727 Numbers k such that sum of first k primorials (A143293) is divisible by k.

Original entry on oeis.org

1, 3, 17, 51, 967, 2901, 16439, 49317, 147951, 1331559

Offset: 1

Alex Ratushnyak, May 13 2013

a(5) = 967 is a prime,
a(6) = a(5) * 3,
a(7) = a(5) * 17,
a(8) = a(5) * 51,
a(9) = a(5) * 51 * 3,
a(10) = a(5) * 51 * 27.
The next term, if it exists, is greater than 15600000. - Alex Ratushnyak, Jun 16 2013

			Sum of first 3 primorials is 1+2+6=9, because 9 is divisible by 3, the latter is in the sequence.
Sum of first 17 primorials is A143293(17) = 1955977793053588026279. Because A143293(17) is divisible by 17, the latter is in the sequence.

Cf. A143293, A002110, A057245, A128981.

primes = [2,3]
def addPrime(k):
  for p in primes:
    if k%p==0:  return
    if p*p > k:  break
  primes.append(k)
for n in range(5,1000000,6):
  addPrime(n)
  addPrime(n+2)
sum_ = 0
primorial = n = 1
for p in primes:
  sum_ += primorial
  primorial *= p
  if sum_ % n == 0:  print(n, end=', ')
  n += 1

from itertools import chain, accumulate, count, islice
from operator import mul
from sympy import prime
def A225727_gen(): return (i+1 for i, m in enumerate(accumulate(accumulate(chain((1,),(prime(n) for n in count(1))), mul))) if m % (i+1) == 0)
A225727_list = list(islice(A225727_gen(),6)) # Chai Wah Wu, Feb 23 2022

cp's OEIS Frontend

A049782 a(n) = (0! + 1! + ... + (n-1)!) mod n.

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A064383 Integers n >= 1 such that n divides 0!-1!+2!-3!+4!-...+(-1)^{n-1}(n-1)!.

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A188775 Numbers k such that Sum_{j=1..k} j^j == -1 (mod k).

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

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A216056 Primes p such that p divides the numerator of Sum( k!/2^k, k=1..p-1 ).

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A225727 Numbers k such that sum of first k primorials (A143293) is divisible by k.

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