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Let $S_n=\Sigma^n_{k=1}p_k$, where $p_k$ is the $k$-th prime number.

Conjecture:

$$\forall p\in\mathbb P\exists n\in\mathbb N: p|S_n$$

Verified for the $1000$ first primes. Is there a proof for this result in general?

In the diagram primes are projected on the x-axis and $n$ (as in $S_n$) on the y-axis.

enter image description here

As H.H.Rugh commented there is a stronger conjecture for all positive integers m. Below a table of $n$-records for different $m\in\mathbb Z^+$ (some of them primes):

    m                          n factorization of Sn 
    1 (1)                      1 (2)
    3 (3)                     10 (3,43)
    6 (2,3)                   57 (2,3,5,229)
   12 (2,2,3)                 97 (2,2,3,1879)
   18 (2,3,3)                113 (2,3,3,41,43)
   35 (5,7)                  180 (5,7,2531)
   42 (2,3,7)                305 (2,2,2,3,7,7,239)
   90 (2,3,3,5)              357 (2,2,2,3,3,3,5,367)
  101 (101)                  422 (5,101,1129)
  137 (137)                  861 (2,137,9739)
  163 (163)                  902 (5,7,11,47,163)
  195 (3,5,13)               907 (2,3,3,5,13,2551)
  202 (2,101)               1207 (2,2,19,101,719)
  222 (2,3,37)              1359 (2,2,2,3,3,37,2671)
  252 (2,2,3,3,7)           1683 (2,2,3,3,7,17,37,71)
  326 (2,163)               1765 (2,2,3,23,163,277)
  474 (2,3,79)              2077 (2,2,2,3,5,79,1861)
  504 (2,2,2,3,3,7)         2133 (2,2,2,3,3,3,7,53,233)
  522 (2,3,3,29)            2379 (2,3,3,3,3,3,7,29,239)
  643 (643)                 2529 (2,3,3,11,211,643)
  647 (647)                 3092 (11,647,5791)
  658 (2,7,47)              3353 (2,3,7,43,47,577)
  700 (2,2,5,5,7)           3593 (2,2,5,5,7,103,787)
  817 (19,43)               4683 (2,3,11,19,43,1847)
  995 (5,199)               5329 (2,3,5,17,199,1291)
 1004 (2,2,251)             6415 (2,2,2,251,96643)
 1204 (2,2,7,43)            6533 (2,2,2,2,7,7,31,43,193)
 1459 (1459)                7241 (2,2,3,3,5,5,191,1459)
 1488 (2,2,2,2,3,31)        7307 (2,2,2,2,2,3,31,85909)
 1610 (2,5,7,23)            8079 (2,5,7,23,43,4567)
 1677 (3,13,43)            10171 (2,2,2,3,3,3,13,43,4259)
 1870 (2,5,11,17)          10331 (2,5,11,17,71,4003)
 2035 (5,11,37)            11459 (2,2,3,3,5,11,37,9029)
 2616 (2,2,2,3,109)        11753 (2,2,2,3,3,3,3,3,37,89,109)
 2672 (2,2,2,2,167)        18137 (2,2,2,2,7,167,93047)
 3420 (2,2,3,3,5,19)       21709 (2,2,3,3,3,5,13,19,137,139)
 3830 (2,5,383)            27617 (2,5,53,383,20749)
 4232 (2,2,2,23,23)        38861 (2,2,2,2,3,3,23,23,113189)
 7394 (2,3697)             45381 (2,107,3697,15083)
 7450 (2,5,5,149)          47323 (2,3,3,5,5,7,41,149,677)
Lehs
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  • Given a fixed $p$, all other primes will be asymptotically equidistributed (to a first order approximation) among units mod $p$, so statistically we should expect one of the partial sums to be zero mod $p$ almost surely. Primes are defined multiplicatively rather than additively, so many if not most existential conjectures about their additive nature only have known probabilistic answers. – anon Aug 04 '16 at 19:06
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    Seems also to work (numerically) for any integer. From probabilistic reason not surprising but right in an algebraist world that's not an argument. – H. H. Rugh Aug 04 '16 at 19:09
  • @H.H.Rugh: yes but I guess it's even more difficult to prove. Or? – Lehs Aug 04 '16 at 20:18
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    @arctictern, could this possibly be an proof idea? – Lehs Aug 04 '16 at 20:20
  • Could Dirichlet's theorem about primes in an arithmetic progression be of use? How about prime gaps? – marty cohen Aug 05 '16 at 03:20
  • Talking out of my ass here but it wouldn't surprise me if this weren't true but the first counter example would be very large and ... well it'd be impossible to verify there isn't a sum, wouldn't it? I mean it's true for small primes because small primes are close together their sums will have any variety of divisors. But for any of the monster huge primes there being a multiple is highly unlikely and as primes get far apart the sums need not mesh. – fleablood Aug 05 '16 at 21:54
  • take a prime, P, with say 20 digits. for a sum divisiblity is normally distributed we'd expect $P|S_{n\approx P}$ to be the first divisble S_n. But we don't expect sum divisibility to be normally distributed, as the difference between S_k become huge ... well, I don't see the sums becoming well distributed. – fleablood Aug 05 '16 at 22:06
  • We'd have to look at the distribution of $S_k$ mod $p$ for a fixed prime $p$. But, I have no idea how we proceed. – Sungjin Kim Aug 05 '16 at 22:13
  • Could someone explain the conjecture in plain english? I’m getting lost with all the symbols. – Stephen Aug 09 '16 at 21:06
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    @Lucas: Every prime divide some sum $2+3+5+...+p_k$. – Lehs Aug 10 '16 at 04:17

1 Answers1

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$a(n)$, the smallest $k$ such that the $n$-th prime divides the sum of the first $k$ primes, is tabulated at http://oeis.org/A111287. In the comments there, it says:

It follows from a theorem of Daniel Shiu that $k$ always exists. Shiu has proved that if $\gcd(a,b) = 1$ then the arithmetic progression $a, a + b, ..., a + kb, \dots$ contains arbitrarily long sequences of consecutive primes. Since, for any positive integer $b$, there are thus arbitrarily long sequences of consecutive primes congruent to 1 mod $b$, there must be infinitely many $a(n)$ that are divisible by $b$.

To clarify the previous comment: If the sum of the primes up to some point is $s \bmod b$, then we need exactly $b-s$ consecutive primes equal to $1 \bmod b$ to produce a sum divisible by $b$. Hence when there are $b-1$ consecutive primes congruent to $1 \bmod b$, then the sum of primes up to one of those primes will be divisible by $b$. [From T. D. Noe, Dec 02 2009]

Community
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Gerry Myerson
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  • The consecutive primes are not necessariy consecutive in the sequence, correct? – Jacob Wakem Aug 07 '16 at 16:54
  • Can you point us to the paper by Shiu? Is the proof difficult? – Jacob Wakem Aug 07 '16 at 17:01
  • Have you gone to the oeis link? There's a link there to the Shiu paper. – Gerry Myerson Aug 07 '16 at 22:50
  • Can you breach the paywall? My acceptance to Princeton doesn't seem to have been processed yet ;) – Jacob Wakem Aug 08 '16 at 03:34
  • If you have access to a library, you can ask for it on interlibrary loan. – Gerry Myerson Aug 08 '16 at 06:45
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    @Alephnull Yes, the breakthrough of Shiu’s theorem is that it says something useful about many consecutive primes. If they were also consecutive elements of the arithmetic progression, that would be 1) false without constraints on how large $k$ is relative to $b$, and 2) of comparable strength to the prime $k$-tuples conjecture. – Erick Wong Nov 20 '18 at 17:42