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I am currently working on a sequence, $a_n$, that is defined as follows:

$$a_n\text{ is the smallest product of }n\text{ distinct odd primes, }m=p_1p_2\dots p_n\text{, such that }\frac{d+\frac{m}{d}}{2}\text{ is prime for all }d\text{ dividing }m.$$

This sequence is currently known to five terms: $3, 21, 105, 1365, 884037$ (OEIS A128281). I have coded a program in an attempt to find a sixth term. During my search, I was able to determine that all members of the sequence except the first are congruent to $1{\pmod{4}}$. My best method currently is to test every number congruent to $1{\pmod{4}}$ starting at $255257 = 3\cdot5\cdot7\cdot11\cdot13\cdot17+2$. My program tests first if each number is a product of six distinct odd primes, then tests if it fits the divisor requirements for the sequence. With my program I have determined that $a_6>10^9$. Are there any optimizations I could make so that my program takes less time to search? Is there any reason to suspect that $a_6$ does not exist? If so, would this be the case for further terms as well?

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  • $\begingroup$ An acceleration is surely to permute over the six-tuples containing distinct odd primes. But probably the user posting this OEIS sequence has done long calculations already. Maybe, you have to be very very patient. $\endgroup$ – Peter Aug 27 at 11:56
  • $\begingroup$ This sequence specifically hadn't been touched since its creation in 2007, so I thought I might have a decent chance at finding the next term. $\endgroup$ – Kirk Fox Aug 28 at 4:50
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    $\begingroup$ As of Aug 27 2020, it appears it was checked up to $a_6\gt 2\cdot 10^{11}$. $\endgroup$ – Vepir Aug 30 at 18:47
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    $\begingroup$ I can approve that $a_6>10^{10}$ and probably even larger than $10^{12}$. I could find an example such that none of the $64$ numbers has a factor less than $293$ , so I think $a_6$ exists. But since we have several relatively large numbers that have to be prime and in total $32$ (the others are equal to those), I think, $a_6$ is very large, maybe $20$ digits or more. $\endgroup$ – Peter Aug 31 at 11:56

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