cp's OEIS Frontend

The n-th tetrahedral number is A000292(n) = n*(n+1)*(n+2)/6. The only odd-valued terms are a(1)=1, a(2)=3, and a(48)=45, corresponding to the only nonzero tetrahedral numbers that are also square, i.e., A000292(1)=1, A000292(2)=4, and A000292(48)=19600.

We can write n*(n+1)*(n+2)/6 as the product of three pairwise coprime integers A, B, and C as follows, depending on the value of n mod 12:

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n mod 12 A B C factor that can be even

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0 n/3 n+1 (n+2)/2 A

1 n (n+1)/2 (n+2)/3 B

2 n/2 (n+1)/3 n+2 C

3 n/3 (n+1)/2 n+2 B

4 n n+1 (n+2)/6 A

5 n (n+1)/6 n+2 B

6 n/6 n+1 n+2 C

7 n (n+1)/2 (n+2)/3 B

8 n (n+1)/3 (n+2)/2 A

9 n/3 (n+1)/2 n+2 B

10 n/2 n+1 (n+2)/3 C

11 n (n+1)/6 n+2 B

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For all n > 6, A, B, and C are all greater than 1 and share no prime factors, so their product must contain at least three distinct prime factors; consequently, its number of divisors cannot be prime or semiprime. The only semiprimes in this sequence are a(3), a(4), and a(5), and the only prime is a(2).

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