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In a recent question, I defined a function $n!\#$ as follows:

$$n!\# = \prod_{i=1}^n (p_i + i) = (2 + 1)(3 + 2)(5 + 3)(7 + 4) \dots (p_n + n)$$

Modified Question

In light of such a trivial solution to my original question (see below), I am going to ask something a bit more interesting:

Question: Does it happen infinitely often that the primorial of $p_n$ divides the above function? More precisely, are there infinite integer solutions to:

$$\dfrac{n!\#}{p_n\#}$$

Aside: Any computations for $n$ showing either divisibility or indivisibility of $n!\#$ by $p_n\#$ are welcome.

Notes

I am quite skeptical as to the truth of this. Here is some initial progress:

  • $p_n$ divides $p_n\#$
  • Therefore we would have to have $p_n$ divides $n!\#$
  • So $p_n$ would have to divide $p_i + i$ for some $i$
  • But $p_i + i \leq p_n + n < 2p_n$
  • So we're forced to have $p_i + i = p_n$

An interesting related question is then, are there infinite solutions to $p_i + i = p_n$? If this is false, then the above is false, so it is a weaker statement.

Another interesting thing to try and prove is, does indivisibility happen infinitely often? This should be easier, but I am still stumped for ideas at the minute.

Original Question

Question: Does there exist any natural number $n$ such that the primorial of $p_n$ divides the above function? More precisely, is the following term ever an integer for some $n$:

$$\dfrac{n!\#}{p_n\#}$$

An equivalent statement: let $k_n$ be the largest prime number such that $k_n\#$ divides $n!\#$. Is it ever the case that $k_n \geq p_n$?

Solved: By the commenter Mark S., who shamed me into noticing that for the humble $n=3$ we have $n!\#$ divides $2, 3, 5$, so the statement is true.

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    $\begingroup$ Isn't it an integer for $n=3$? The numerator is $120$ and the denominator is $30$. $\endgroup$ – Mark S. Dec 18 '15 at 12:34
  • $\begingroup$ Actually @MarkS. I think my mathematical pride is telling me to update the question. I will add your case for $n=3$ and ask if there are any other cases. Thank you. $\endgroup$ – Colm Bhandal Dec 18 '15 at 12:42
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This is not a proof, but here is some relevant numerical evidence for the conjecture that $n=3$ is the only time (not counting $n=0$) this fraction is an integer.

Firstly, it's the only solution for $n$ up to $10000$. (All Mathematica code below.)

More importantly, the number of primes in the denominator that are not cancelled out appears to grow nearly linearly with $n$, yet it would have to drop to $0$ for the fraction to be an integer.

Here is a plot for $0\le n\le100$: plot of the number of prime factors in the denominator that are not cancelled out for n up to 100

Here is a plot for $0\le n\le2000$ with the line $y=\frac{7}{10}x$ drawn on top: plot of the number of prime factors in the denominator that are not cancelled out for n up to 2000

According to Mathematica, the best line to fit that up to $n=2000$ is actually closer to $y=0.699932x-26.1072$.

As an aside, the $p_n+n$ which appear in your mesh function are conjectured to be primes infinitely often, but it may not be known for sure.


(*Mathematica Code*)

f[0]=1;
f[n_]:=f[n]=f[n-1]*(Prime[n]+n);

p[0]=1;
p[n_]:=p[n]=p[n-1]*Prime[n];

Print[Table[ If[IntegerQ[f[n]/p[n]], n, Nothing], {n, 0, 10000}]];

countnegfactors[n_] := countnegfactors[n] = Length[Transpose[Select[FactorInteger[f[n]/p[n]],Function[x, x[[2]] == -1]]][[1]]];

Print[ListPlot[Table[{n, If[n == 0 || n == 3, 0, countnegfactors[n]]}, {n, 
   0, 100}]]];

m=Table[{n, If[n == 0 || n == 3, 0, countnegfactors[n]]}, {n, 0, 2000}];

Print[Show[ListPlot[m],Plot[(7/10)*(x),{x,0,2000},PlotStyle->Orange]]];

LinearModelFit[m, x, x]
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