A279081 - OEIS

A279081

Number of divisors of the n-th tetrahedral number.

1, 3, 4, 6, 4, 8, 12, 16, 8, 12, 8, 12, 8, 20, 16, 20, 8, 24, 16, 24, 8, 16, 18, 24, 18, 36, 24, 24, 8, 24, 20, 24, 16, 48, 32, 24, 8, 32, 24, 32, 8, 24, 32, 48, 16, 20, 24, 45, 18, 36, 16, 36, 24, 96, 48, 32, 8, 24, 16, 24, 16, 56, 96, 56, 16, 24, 16, 48, 16

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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:

n mod 12 A B C factor that can be even

======== === ======= ======= =======================

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).

LINKS

Michel Marcus, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = A000005(A000292(n)) = A000005(n*(n+1)*(n+2)/6).

From Ridouane Oudra, Jan 25 2024: (Start)

a(6*n) = tau(2*n)*tau(6*n+1)*tau(6*n+2)/2;

a(6*n+1) = tau(6*n+1)*tau(3*n+1)*tau(2*n+1);

a(6*n+2) = tau(6*n+2)*tau(2*n+1)*tau(6*n+4)/2;

a(6*n+3) = tau(2*n+1)*tau(3*n+2)*tau(6*n+5);

a(6*n+4) = tau(6*n+4)*tau(6*n+5)*tau(2*n+2)/2;

a(6*n+5) = tau(6*n+5)*tau(n+1)*tau(6*n+7). (End)

EXAMPLE