A340865 - OEIS

%I #10 Jan 25 2021 01:10:15

%S 3,11,59,181,199,379,409,571,739,1039,1439,2239,2269,2351,2381,2671,

%T 2719,2789,3049,3529,4021,4201,4721,4999,5431,5531,5839,6329,6619,

%U 8329,9241,9419,9631,9689,10151,11329,11551,12071,12421,13339,14489,15091,17419,18301

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

%C Intersection of A048161 and A106483.

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

%C 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:

%C 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.

%C 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.

%C 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.

%C 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.

%C 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).

%C 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.

%H Jon E. Schoenfield, <a href="/A340865/b340865.txt">Table of n, a(n) for n = 1..10000</a>

%e Both (3^2 + 1)/2 = 5 and 2*3^2 - 1 = 17 are prime, so 3 is in the sequence.

%e (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.

%e (7^2 + 1)/2 = 25 is not prime, so even though 2*7^2 - 1 = 97 is prime, 7 is not in the sequence.

%e 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.

%o (PARI) isok(p) = (p>2) && isprime(p) && isprime((p^2+1)/2) && isprime(2*p^2-1); \\ _Michel Marcus_, Jan 25 2021

%Y Cf. A000217, A048161, A106483, A292989.

%K nonn

%O 1,1

%A _Jon E. Schoenfield_, Jan 24 2021