Dtyjlui

I have a function involving polynomials and the centre of the Binomial Triangle and I'd like to prove that the function produces a positive integer infinitely many times. I don't have any interest in what values the integers take so much, merely that they exist.

The function I have is:

$$f(n) = dfrac{15n^5+23n^4-20n^3-56n^2-48n-16}{n^3(n+1)(n+2)^4}begin{pmatrix}2n \ n end{pmatrix} + 4dfrac{3n^2+6n+4}{n^3 (n+2)^3}$$

The only method I have thought of is that I could try to prove that $sinleft(pi f(n)right)$ has an infinte number of roots. But that feels like kicking the can down the road as I don't know how to do that either!

Thank you for any and all help. Ben

Using partial fractions I get:

$$f(n) = frac{left( 4n^3+28n^2+84n+76 right) begin{pmatrix} 2n \ n end{pmatrix} - 2n-4}{(n+2)^4} - frac{begin{pmatrix} 2n \ n end{pmatrix} - 2}{n^3} - frac{4begin{pmatrix} 2n \ n end{pmatrix}}{n+1} $$

From here I can find conditions for each fraction separately.

$frac{begin{pmatrix} 2n \ n end{pmatrix} - 2}{n^3} in mathbb{N} implies n text{ is prime > 3 (I found this on A000984)}$

$frac{4begin{pmatrix} 2n \ n end{pmatrix}}{n+1} in mathbb{N} implies n in mathbb{N} text{ (Always true)}$

$$f(n) = frac{left( 4n^3+28n^2+84n+76 right) begin{pmatrix} 2n \ n end{pmatrix} - 2n-4}{(n+2)^4} - color{Blue}{frac{begin{pmatrix} 2n \ n end{pmatrix} - 2}{n^3} - frac{4begin{pmatrix} 2n \ n end{pmatrix}}{n+1}} $$

This leaves the final condition:

$frac{left( 4n^3+28n^2+84n+76 right) begin{pmatrix} 2n \ n end{pmatrix} - 2n-4}{(n+2)^4} in mathbb{N} \ implies left( 4n^3+28n^2+84n+76 right) begin{pmatrix} 2n \ n end{pmatrix} equiv 2n+4pmod{n^4+8n^3+24n^2+32n+16}$

This seems to hold true when $n+2$ is a prime.

So if this function outputs an infinite number of integers, the twin-prime conjecture should be true.
If $n$ is the lower of a pair of twin primes, then $f(n)$ is an integer.

Here is a proof that if both $n$ and $n+2$ are prime ($n>3$), then the output is integer:
clearly it suffices to show that if $n+2$ is prime then $$ left(4n^3+28n^2+84n+76right)binom{2n}{n} equiv 2(n+2) bmod{(n+2)^4}.$$
Let $p=n+2$ be prime, then this is equivalent to showing that:
$$tag1 2left(p^3+p^2+5p-3right)binom{2p-4}{p-2} equiv p bmod{p^4}.$$

Proof

Since $n>3$, $p>3$ and by Wolstenhome's theorem we have:
begin{align}
binom{2p-1}{p-1}&equiv 1 bmod p^3\
pbinom{2p-1}{p-1}&equiv p bmod p^4\
pfrac{2p-1}{p-1}frac{2p-2}{p-2}frac{2p-3}{p-3}binom{2p-4}{p-4}&equiv p bmod p^4.
end{align}

But $$ binom{2p-4}{p-2}=binom{2p-4}{p-4}frac{p-1}{p-3}frac{p}{p-2}.$$
Then
begin{align}
pfrac{2p-1}{p-1}frac{2p-2}{p-2}frac{2p-3}{p-3}frac{p-3}{p-1}frac{p-2}{p}binom{2p-4}{p-2}&equiv p bmod p^4
end{align}
That is
begin{align}
frac{(2p-1)(2p-2)(2p-3)}{(p-1)^2}binom{2p-4}{p-2}&equiv p bmod p^4
end{align}

But by doing the power expansion of $frac{1}{(p-1)^2}$ , we see that
$$ frac{1}{(p-1)^2} equiv 1+2p+3p^2+4p^3 bmod p^4$$
and we are done since it is easy to see by expansion that $$(2p-1)(2p-2)(2p-3)(1+2p+3p^2+4p^3) equiv 2(-3+5p+p^2+p^3) bmod p^4.$$

I does seem that for the OP function to bring an integer, the argument must be the smallest prime of a twin prime pair. But proving this seems very difficult. What about trying to disprove it by computer search of a counterexemple ?