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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

.

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).

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