# Any relation between primorial numbers and oblong (n(n+1)) numbers?

Just noticed that some primorial numbers are oblong:

$$\prod\limits_{i=1}^{3}p_i = 5 \cdot 6$$

$$\prod\limits_{i=1}^{4}p_i = 14 \cdot 15$$

$$\prod\limits_{i=1}^{7}p_i = 714 \cdot 715$$

Does anyone know if there are infinite cases of numbers that are at the same time oblong and primorial?

• Do you mean with "oblong" twice a triangle number ? Then, a number is "oblong" if and only if $4k+1$ is a perfect square. – Peter Jun 2 at 19:15
• A related question would suggest that such numbers are quite rare. – Keith Backman Jun 2 at 19:31
• Does this answer your question? Products involving the first $n$ primes, wherein two products differ by 1 (You are looking at the A161620 sequence. There are no known examples larger than $714\cdot 715$) – Vepir Jun 2 at 20:08
• It seems that $510510$ is the largest primorial of the form $n(n+1)$. If another one exists, ist must be very large. – Peter Jun 2 at 20:10
• Note that 714 and 715 are Babe Ruth's and Hank Aaron's career home run records. Pal Erdos had a baseball with those numbers on it, signed by Hank Aaron. That means Aaron's Erdos number is 1. I think Ron Graham has the baseball now. The problem of finding the next pair of such numbers is in Guy's UPINT. It's a fairly famous unsolved problem. – B. Goddard Jun 2 at 21:19

This PARI/GP program

(22:05) gp > n=1;s=1;while(n<10^6,n=nextprime(n+1);s=s*n;if(issquare(4*s+1)==1,print(n," ",s)))
2 2
3 6
5 30
7 210
17 510510
(22:11) gp >


shows that upt $$10^6$$# , there is no larger primorial than $$510510$$, which has the form $$n(n+1)$$. This leads to the strong conjecture that there are no more. To prove it, will be very difficult, I guess.

• The OEIS entry apparently claims that the list is complete upto $10^{11}$#, my search limit however lets already almost no hope for another solution. – Peter Jun 2 at 20:18