cp's OEIS Frontend

All terms are odd and squarefree.

Generalized Bunyakovsky conjecture implies for any odd prime p there are infinitely many terms of the form p*q where q is prime.

If a(n) = pq, where p > q are both prime, then p is the hypotenuse and q is a leg of a primitive Pythagorean triple. (x^4-y^4 = (x^2+y^2)(x+y)(x-y), hence x-y=1 and x^2+y^2 and x+y are both prime. Note that x^2+y^2 can never be (x+y)^2 so a(n) is never the cube of a prime.)

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.

Expressions of the form m^j + 1 can be factored (e.g., m^3 + 1 = (m + 1)*(m^2 - m + 1)) for any positive integer j except when j is a power of 2, so (p^j + 1)/2 for prime p cannot be prime unless j is a power of 2. A005383, A048161, A176116, A340480, A341210, A341224, A341229, A341230, A341234, and this sequence list primes of the form (p^j + 1)/2 for j=2^0=1, j=2^1=2, ..., j=2^9=512, respectively.

Except 2, all terms are divisible by 10, and p-1 and q-1 are divisible by 100.

Numbers m such that p = m^2+1 and p + m^4/2 are both prime. - Chai Wah Wu, May 01 2025

See comments in A174825

c is the prime hypotenuse c, i. e. c of a primitive Pythagorean triple: a^2 + b^2 = c^2

Semiprimes which are a leg of an integral right triangle whose hypotenuse is also semiprime. This is to A048161 as semiprimes A001358 are to primes A000040. All terms must be odd (else r is not an integer).

Expressions of the form m^j + 1 can be factored (e.g., m^3 + 1 = (m + 1)*(m^2 - m + 1)) for any positive integer j except when j is a power of 2, so (p^j + 1)/2 for prime p cannot be prime unless j is a power of 2. A005383, A048161, A176116, A340480, A341210, A341224, A341229, A341230, A341234, A341264, and this sequence list primes of the form (p^j + 1)/2 for j=2^0=1, j=2^1=2, ..., j=2^10=1024, respectively.