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There is an exact $n$th prime number formula from mathworld

A double sum for the nth prime pn is $$p_n=1+\sum_{k=1}^{2(\lfloor n\ln n\rfloor+1)}\Biggl[1-\Biggl\lfloor\frac{\sum_{j=2}^k 1+\lfloor s(j)\rfloor}n\Biggr\rfloor\Biggr],\tag{13}$$ where $$s(j)\equiv-\frac{\sum_{s=1}^j \bigl(\bigl\lfloor\frac js\bigr\rfloor-\bigl\lfloor\frac{j-1}s\bigr\rfloor\bigr)-2}j\tag{14}$$

I tried to check it numerically but there is no correct answer :

s[j_] := -(Sum[(Floor[j/s] - Floor[(j - 1)/s]) - 2, {s, 1, j}]/j)

p[n_] := 1 + Sum[1 - Floor[Sum[1 + Floor[s[j]], {j, 2, k}]/n], {k, 1, 
2*(Floor[n Log[n]] + 1)}]


p[30]

-1106

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  • $\begingroup$ Note that the website cites this: gallup.unm.edu/~smarandache/articulo/artic.htm $\endgroup$
    – Feyre
    Commented Dec 13, 2016 at 11:12
  • $\begingroup$ Have you seen this? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Dec 13, 2016 at 11:15
  • $\begingroup$ @J.M. Are you saying mathworld (and UNM) simply mistate the formula? $\endgroup$
    – Feyre
    Commented Dec 13, 2016 at 11:18
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    $\begingroup$ Looks like it. Thus: s[j_] := 2 (1 + Sum[Floor[(j - 1)/s] - Floor[j/s], {s, 1, Sqrt[j]}])/j $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Dec 13, 2016 at 11:56
  • $\begingroup$ @J.M. hence there is a mistake in mathword article $\endgroup$
    – vito
    Commented Dec 13, 2016 at 12:16

1 Answer 1

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J. M.'s correction:

s[j_] := 2 (1 + Sum[⌊(j - 1)/s⌋ - ⌊j/s⌋, {s, 1, Sqrt[j]}])/j

p[n_] := 1 + Sum[1 - ⌊Sum[1 + ⌊s[j]⌋, {j, 2, k}]/n⌋, {k, 1, 2*(⌊n Log[n]⌋ + 1)}]

p[30]       (* 113 *)

Prime[30]   (* 113 *)
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  • 1
    $\begingroup$ I'd use Quotient[] instead of Floor[] + division myself (just like in the article I linked to). Nevertheless, it should be emphasized that this is of course an expensive way to generate primes. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Dec 13, 2016 at 12:17

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