A279081 - OEIS

%I #18 Mar 12 2024 15:40:48

%S 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,

%T 24,24,8,24,20,24,16,48,32,24,8,32,24,32,8,24,32,48,16,20,24,45,18,36,

%U 16,36,24,96,48,32,8,24,16,24,16,56,96,56,16,24,16,48,16

%N Number of divisors of the n-th tetrahedral number.

%C 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.

%C 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:

%C .

%C n mod 12   A      B        C     factor that can be even

%C ========  ===  =======  =======  =======================

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

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

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

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

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

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

%C     6     n/6    n+1      n+2                 C

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

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

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

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

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

%C .

%C 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).

%H Michel Marcus, <a href="/A279081/b279081.txt">Table of n, a(n) for n = 1..10000</a>

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

%F From _Ridouane Oudra_, Jan 25 2024: (Start)

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

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

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

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

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

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

%e a(48) = tau(48*59*50/6) = tau(19600) = tau(2^4 * 5^2 * 7^2) = (4+1)*(2+1)*(2+1) = 5*3*3 = 45.

%p with(numtheory): seq(tau(n*(n+1)*(n+2)/6), n=1..70) ; # _Ridouane Oudra_, Jan 25 2024

%t DivisorSigma[0, Binomial[Range[100]+2, 3]] (* _Paolo Xausa_, Feb 19 2024 *)

%o (PARI) a(n) = numdiv(n*(n+1)*(n+2)/6); \\ _Michel Marcus_, Jan 07 2017

%Y Cf. A000292, A279082, A279083.

%K nonn

%O 1,2

%A _Jon E. Schoenfield_, Jan 06 2017