A340865 Primes p such that (p^2 + 1)/2 and 2*p^2 - 1 are also prime.

3, 11, 59, 181, 199, 379, 409, 571, 739, 1039, 1439, 2239, 2269, 2351, 2381, 2671, 2719, 2789, 3049, 3529, 4021, 4201, 4721, 4999, 5431, 5531, 5839, 6329, 6619, 8329, 9241, 9419, 9631, 9689, 10151, 11329, 11551, 12071, 12421, 13339, 14489, 15091, 17419, 18301

Offset: 1

Author

Jon E. Schoenfield, Jan 24 2021

Keywords

• nonn

Comments

Intersection of A048161 and A106483.

How many triangular numbers with 6 divisors (A292989) can be divisible by the same squared prime p^2?

The k-th triangular number T(k) = A000217(k) = k*(k+1)/2 can be written as the product of two coprime factors A and B where A=k and B=(k+1)/2 for odd k, A=k/2 and B=k+1 for even k. If a triangular number has 6 divisors, then it is of the form p^2*q where p and q are distinct primes. We can identify four cases:

Case 1: A = k = p^2 and B = (k+1)/2 = q, so q = (p^2 + 1)/2; solutions occur at primes p in A048161.

Case 2: A = k = q and B = (k+1)/2 = p^2, so 2*p^2 - 1 = q; solutions occur at primes p in A106483.

Case 3: A = k/2 = p^2 and B = k+1 = q. In this case, 2*p^2 + 1 = q. For p = 2, we would get q = 9 (nonprime), so p must be odd. If prime p > 3 (so q > 19), we have p^2 == 1 (mod 3), so q == 0 (mod 3), hence nonprime. So the only solution for this case occurs at p=3, q=19, t = 3^2*19 = 171.

Case 4: A = k/2 = q and B = k+1 = p^2. In this case, 2*q + 1 = p^2, so p is odd, but then p^2 == 1 (mod 8), so q == 0 (mod 4), hence q is not prime: no solutions exist.

Since Case 4 has no solutions, at most three triangular numbers with 6 divisors can be divisible by the same squared prime p^2; Case 3 has a solution only at p=3 and, in fact, there are three triangular numbers with 6 divisors that are divisible by 3^2: t = 3^2*5 = 45 = T(9), t = 3^2*17 = 153 = T(17), and 3^2*19 = 171 = T(18).

For all primes p > 3, then, at most two triangular numbers with 6 divisors are divisible by p^2; this sequence (after the initial term, 3) lists the primes p such that p^2 divides exactly two triangular numbers that have 6 divisors.

Examples

			Both (3^2 + 1)/2 = 5 and 2*3^2 - 1 = 17 are prime, so 3 is in the sequence.
(5^2 + 1)/2 = 13 is prime, but 2*5^2 - 1 = 49 = 7^2 is not prime, so 5 is not in the sequence.
(7^2 + 1)/2 = 25 is not prime, so even though 2*7^2 - 1 = 97 is prime, 7 is not in the sequence.
Neither (23^2 + 1)/2 = 265 = 5*53 nor 2*23^2 - 1 = 1057 = 7*151 is prime, so 23 is not in the sequence.
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A048161, A106483, A292989.

Programs

PARI

isok(p) = (p>2) && isprime(p) && isprime((p^2+1)/2) && isprime(2*p^2-1); \\ Michel Marcus, Jan 25 2021