largest prime remainder from three consecutive primes

Question

From three consecutive primes 89 results from $223*211mod199$=89. Is this the largest prime remainder? For p(n)=then nth prime, the form is p(n+1)*p(n+2)modp(n)=prime.

Answer 1

THIS IS NOT A COMPREHENSIVE ANSWER, rather an extended comment that is too long for the 'comment' section.

Consider three consecutive prime numbers $p_n,\ p_{n+1},\ p_{n+2}$ with gaps $g_1=p_{n+1}-p_n$ and $g_2=p_{n+2}-p_{n+1}$. For all $p_n>2$, $g_1$ and $g_2$ are even.

$p_{n+1}\equiv g_1\mod{p_n}$ and $p_{n+2}\equiv g_1+g_2\mod{p_n}$ so $(p_{n+1})(p_{n+2})\equiv (g_1)(g_1+g_2)\mod{p_n}$.

Unless $(g_1)(g_1+g_2)>p_n$, $(g_1)(g_1+g_2)\mod{p_n}$ will be even and hence composite (see comment by lulu). Thus as larger candidate $p_n$ are examined, it would be necessary to focus on primes separated by larger gaps.

The size of gaps is subject to limits. In general $g_1\le \alpha p_n$, or $p_{n+1}\le (1+\alpha )p_n$ and $p_{n+2}\le (1+\alpha )p_{n+1}=(1+\alpha )^2p_n\approx (1+2\alpha )p_n$. So $\max(g_1)=(1+\alpha )p_n-p_n=\alpha p_n$ and $\max((g_1+g_2))\approx(1+2\alpha )p_n-p_n=2\alpha p_n$.

Thus $\max((g_1)(g_1+g_2))\approx 2{\alpha}^2{p_n}^2$. For $(g_1)(g_1+g_2)>p_n$ it must be the case that $p_n>\frac{1}{2{\alpha}^2}$.

For numbers greater than $2\cdot 10^7$, Schoenfeld cited here has given a numerical limit $\alpha \le \frac{1}{16597}$ requiring $p_n>\frac{16597^2}{2}\approx1.3\cdot 10^8$. Other non-numeric limits on $\alpha$ are also given in the Wikipedia article, which may be even smaller than the value of Schoenfeld. Note that even if candidates with $(g_1)(g_1+g_2)>p_n$ are found, there is no guarantee that the residue generated will be prime. This suggests that very large primes (much greater than those examined by gammatester) will have to be examined in the search for a prime remainder larger than $89$.