OFFSET
3,1
COMMENTS
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.
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
For all n < 171, a(n) > a(n-1) with the exception of a(11) < a(10). This occurs whenever m(n-1) > prime(n) * m(n).
The principle can be extended to r(n,i) = {-1,1,-p,p,-q_i,q_i} where q_i = prime(n+i). Such a sequence b for i = 1 would have b(3) = 12, as 5, 7, 11, 13, 17, 19 are all prime. This is the only number k for which all three of k+-1, k+-5, and k+-7 are prime. To satisfy the first requirement for k > 6, we need k == {0, 2, 8} (mod 10). Under this condition, one of k-5 or k+-7 will be divisible by 5. Since 12 - 7 = 5 is the only prime that is divisible by 5, k = 12 is the only k satisfying the condition.
If q, with a(n) + 1 < q < a(n) + prime(n)^2, is prime, the difference r(n) = q - a(n) is also prime. Proof: Per definition a(n) is divisible by all d < prime(n). It follows that, if r is divisible by any d, then so is q = a(n) + r, whence q is not prime. Thus, if q is prime, then r is either also prime or only has prime factors f >= prime(n), i.e., r >= prime(n)^2. See "Fortunate Numbers" (A005235).
There is no a(1) and a(2). Since prime(1) = 2, both k+1 and k+2 need to be prime. This is only true for k = 1, but 1 - 1 = 0 is not prime. For a(2) we have prime(2) = 3 and one of k+1, k-1, k+3 is divisible by 3.
LINKS
Florian Baur, Table of n, a(n) for n = 3..200
Charles Kusniec, An idea for a new class of some numbers, mersenneforum.org
FORMULA
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
EXAMPLE
a(3): The 3rd prime is 5. The smallest number divisible by all d < 5 is 12. Since 12 - 1 = 11, 12 + 1 = 13, 12 + 5 = 17, 12 - 5 = 7 are all prime, a(3) = 12.
a(5): The 5th prime is 11. The smallest number divisible by all d < 11 is 2520. However, 2520 - 1 = 2519 is not prime. The smallest number satisfying all conditions is 93240, since 93240 - 1, 93240 + 1, 93240 + 11, 93240 - 11 are all prime and 93240 is divisible by all d < 11. Thus, a(5) = 93240.
PROG
(PARI) A354882(n) = { my(s = 1, p = prime(n), c = lcm([1..p-1])); while(!(isprime(s*c+1) & isprime(s*c-1) & isprime(s*c+p) & isprime(s*c-p)), s++); return(s*c)} \\ Florian Baur, Jul 17 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Florian Baur, Jun 10 2022
STATUS
approved