Prime numbers as the product of an arithmetic sequence

Question

Are there infinitely many prime numbers that can be represented as the product of three rational numbers bigger than 0 that create an arithmetic sequence? For example, the product of the arithmetic sequence {1, 1.5, 2} is the prime 3.

Answer 1

If $p$ is your prime and your three rationals are $x-d$, $x$ and $x+d$, you want $p = x^3 - x d^2$ where $x$ and $d$ are rational. $ x^3 - x d^2 - p$ is an elliptic curve. Writing $x = p/s$ and $d = -t/s$ we get the Weierstrass form $s^3 + t^2 - p^2$. If the curve has a rational point that is not a torsion point, it will generate infinitely many rational points using the group operation.

For example, with $p=3$ a generator of the rational points is (according to Sage) $(s,t) = (2,1)$, corresponding to $(x,d) = (3/2, -1/2)$. Three times this point is $(x,d) = [1323/629, -2870/13209]$, which corresponds to the three rationals $1369/357, 1323/629, 289/777$.

Another prime that works is $p=11$, where a generator is $(s,t) = (-12,43)$, and one solution is the three rationals $$\frac{299255401}{253518426},\ \frac{ 1089903276}{440588231}, \ \frac{648669961}{172194246}$$

If my use of Sage is correct, the first few primes that work are $3,11,13,17,29,31,43,47,53$. Sage seemed unable to compute the rank for $p=61$. The sequence $3,11,13,17,29,31,43,47,53$ does not seem to be in OEIS.

EDIT: More generally, if we ignore the requirement for $p$ to be prime, we have solutions for $$p = 1, 3, 6, 8, 10, 11, 13, 14, 15, 17, 21, 24, 25, 27, 28, 29, 31, 34, 35, 36, 39, 42, 43, 45, 46, 47, 48, 49, 52, 53, 55, 57, ...$$ This is not in OEIS yet either.

EDIT: Now it is: A328505