# Primes p(n) such that p(n) + p(n+3) = p(n+1) + p(n+2) and p(n) + p(n+4) = p(n+2) + p(n+3) - Conjecture [closed]

I have been studying this sequence (A266882 in the OEIS) and found the following pattern:

$13 + 17 + 19 + 23 + 29 = 101$ (101 is prime)

37 does not hold.

$223 + 227 + 229 + 233 + 239 = 1151$ (1151 is prime)

The same is true for 1087, 1423, 1483, and 2683.

So, is 37 the only prime of this sequence (A266882) for whom the sum is not also a prime?

No. A simple check shows that it is already false for the next few primes in the sequence, namely for 4783, 6079, 7331. In fact, one could even ask whether these first numbers are the only ones for which this sum is prime, but it turns out that it later holds also for 11057 and 12269.

• Indeed, and among the first 100 admissible primes, the conjecture holds for 27 of those and is false for the remaining ones. Jul 5, 2016 at 22:31

Here's a Haskell program to display the values for which the conjecture is false.

Within a minute, the largest value it found was the following, rather nice quintuple: $$[468883,468887,468889,468893,468899]$$

import Math.NumberTheory.Primes

specialPrime n = ((nthPrime n) + (nthPrime (n + 3)) == (nthPrime (n + 1)) + (nthPrime (n + 2))) && ((nthPrime n) + (nthPrime (n + 4)) == (nthPrime (n + 2)) + (nthPrime (n + 3)))

specialPrimes = filter (specialPrime) [1..]

specialTuples n = let v = specialPrimes !! (n-1)
in  fmap nthPrime [v..(v + 4)]

checkConjecture n = isPrime (foldr (+) 0 $specialTuples n) falseValues = fmap specialTuples$ filter (not . checkConjecture) [1..]

main = do
sequence_ \$ fmap print falseValues

• Passing the 1 million cap [1008407,1008409,1008419,1008421,1008433] Jul 5, 2016 at 22:57