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If the k-th tetrahedral number (i.e., A000292(k) = k*(k+1)*(k+2)/6) has exactly 42 divisors, it must be of the form p^6 * q^2 * r where p, q, and r are distinct primes, and it can be shown that p^6, q^2, and r must be k, (k+1)/2, and (k+2)/3, in some order. This results in six cases:

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p^6 q^2 r resulting equations

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k (k+1)/2 (k+2)/3 p^6 = 2*q^2 - 1 = 3*r - 2

k (k+2)/3 (k+1)/2 p^6 = 3*q^2 - 2 = 2*r - 1

(k+1)/2 k (k+2)/3 2*p^6 = q^2 + 1 = 3*r - 1

(k+1)/2 (k+2)/3 k 2*p^6 = 3*q^2 - 1 = r + 1

(k+2)/3 k (k+1)/2 3*p^6 = q^2 + 2 = 2*r + 1

(k+2)/3 (k+1)/2 k 3*p^6 = 2*q^2 + 1 = r + 2

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Cases 1 and 2 require p^6 + 1 = 2*q^2 and p^6 + 1 = 2*r, respectively, but both can be ruled out by factoring p^6 + 1 = (p^2 + 1)*(p^4 - p^2 + 1).

The four remaining cases require, respectively, the existence of a square of the form 2*p^6 - 1, (2*p^6 + 1)/3, 3*p^6 - 2, or (3*p^6 - 1)/2. An exhaustive search of the primes p up to 10^10 finds none that yields a square for any of those four forms, so if a(42) is not zero, then a(42) > (10^10)^18 / 6 = 1.666...*10^179.

Terms from a(66) through a(100): ?, 0, 375299968925696, 0, 0, 0, 215820, 0, 0, 0, 24019198012555264, 0, 0, 0, 88560, 0, 0, 0, 32016576, 0, 0, 0, 1431655424, 0, 2391444, 0, a(92), 0, 0, 0, 27720, 0, 0, 0, 40690000. (a(92), too large to show here, is p^22*((3*p^22 - 1)/2)*(3*p^22 - 2) where p = 10711.) [Updated by Max Alekseyev, Feb 03 2024]

Solutions to 2*p^6 = q^2 + 1 correspond to (some) integral points (X,Y) = (p^2, q) on the elliptic curve 2*X^3 = Y^2 + 1. All other cases for n = 42, as well as all the cases for n in {50, 70, 78, 98}, also correspond to elliptic curves, and I have computationally verified that all their integral points do not have the required form. Hence, a(42) = a(50) = a(70) = a(78) = a(98) = 0. - Max Alekseyev, Feb 03 2024

The only odd terms are 1, 3, and 45 (which correspond to the three positive tetrahedral numbers that are square, i.e., 1, 4, and 19600). It is easy to show that no term larger than 6 is semiprime.

A tetrahedral number with exactly 42 divisors would have to be of the form p^6 * q^2 * r, with p, q, and r distinct primes; does such a tetrahedral number exist?

A tetrahedral number with exactly 50 divisors would have to be of the form p^4 * q^4 * r, with p, q, and r distinct primes; does such a tetrahedral number exist?

Additional terms < 200 include (but may not be limited to) 44, 45, 48, 52, 54, 56, 60, 64, 68, 72, 76, 80, 84, 88, 90, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 150, 152, 156, 160, 162, 164, 168, 172, 176, 180, 184, 188, 192, 196