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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)
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  • $\begingroup$ 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. $\endgroup$
    – arctic tern
    Aug 4, 2016 at 19:06
  • 1
    $\begingroup$ Seems also to work (numerically) for any integer. From probabilistic reason not surprising but right in an algebraist world that's not an argument. $\endgroup$
    – H. H. Rugh
    Aug 4, 2016 at 19:09
  • $\begingroup$ @H.H.Rugh: yes but I guess it's even more difficult to prove. Or? $\endgroup$
    – Lehs
    Aug 4, 2016 at 20:18
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    $\begingroup$ @arctictern, could this possibly be an proof idea? $\endgroup$
    – Lehs
    Aug 4, 2016 at 20:20
  • 1
    $\begingroup$ @Lucas: Every prime divide some sum $2+3+5+...+p_k$. $\endgroup$
    – Lehs
    Aug 10, 2016 at 4:17

1 Answer 1

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+50
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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]

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  • $\begingroup$ The consecutive primes are not necessariy consecutive in the sequence, correct? $\endgroup$
    – Jacob Wakem
    Aug 7, 2016 at 16:54
  • $\begingroup$ Can you point us to the paper by Shiu? Is the proof difficult? $\endgroup$
    – Jacob Wakem
    Aug 7, 2016 at 17:01
  • $\begingroup$ Have you gone to the oeis link? There's a link there to the Shiu paper. $\endgroup$
    – Gerry Myerson
    Aug 7, 2016 at 22:50
  • $\begingroup$ Can you breach the paywall? My acceptance to Princeton doesn't seem to have been processed yet ;) $\endgroup$
    – Jacob Wakem
    Aug 8, 2016 at 3:34
  • 1
    $\begingroup$ @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. $\endgroup$
    – Erick Wong
    Nov 20, 2018 at 17:42

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