Florian Baur, <a href="/A354882/b354882_1.txt">Table of n, a(n) for n = 3..200</a>

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a(n) is the smallest number k that is divisible by all numbers d with d < p = prime(n), and such that all of k+1, k-1, k+p, k-p are prime.

Suggested by Charles Kusniec on the mersenneforum.org message board (see "Links" section): a number c(n) = k(n) + r(n), where k(n) = m * A099795(n), r(n) = {-1, 1, p, -p} and p = prime(n), is not divisible by any d with d < p and the density of primes among c(n) is expected to be higher than for random numbers of that magnitude.

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The probability that an arbitrary c(n) is prime is higher than that of an arbitrary number of the same magnitude. Let t(n) denote that probability, then t(n) = 1/(((1/2)*(2/3)*(4/5)*(6/7)*...*(prime(n-1)-1)/prime(n-1)) * log(c(n))). According to Mertens's third theorem the value asymptotically approaches 1/((e^gamma/log(prime(n-1))) * log(c(n))). log(c(n)) can be approximated as prime(n-1). This yields t(n) ~ log(prime(n-1))/(e^gamma*prime(n-1)). The probability that c(n) is prime for a random value of m is t(n)^4. Thus, the sequence is expected to grow with (prime(n-1)/log(prime(n-1))^4*A099795(n). For n < 201 the arithmetic mean of m(n)*t(n)^4 is 1.1. - Florian Baur, Jul 12 2023

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a(n) = m(n) * A099795(n). Specifically, m(3) = m(4) = 1. For all other n < 201: , 25 < m(n) < 333054037. and m(n) cannot have prime(n) as a factor. - Florian Baur, Jul 12 2023

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The probability that an arbitrary c(n) is prime is higher than that of an arbitrary number of the same magnitude. Let t(n) denote that probability, then t(n) = 1/(((1/2)*(2/3)*(4/5)*(6/7)*...*(prime(n-1)-1)/prime(n-1)) * log(c(n))). According to Mertens's third theorem the value asymptotically approaches 1/((e^gamma/log(prime(n-1))) * log(c(n))). log(c(n)) can be approximated as prime(n-1). This yields t(n) ~= log(prime(n-1))/(e^gamma*prime(n-1). The probability that c(n) is prime for a random value of m is t(n)^4. Thus, the sequence is expected to grow with (prime(n-1)/log(prime(n-1))^4*A099795(n). For n < 201 the arithmetic mean of m(n)*t(n)^4 is 1.1. - Florian Baur, Jul 12 2023

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