A279082 - OEIS

%I #27 Feb 04 2024 03:29:08

%S 1,0,4,10,0,20,0,56,0,0,0,84,0,0,0,120,0,2300,0,560,0,0,0,1140,0,0,0,

%T 349504,0,2362041,0,7770,0,0,0,3276,0,0,0,82160,0,0,0,

%U 4980598206895701865797099599990718771851543594988165352649,19600,0,0,7140,0,0,0,91625967616,0,155965244802415621,0,41664,0,0,0,2332880,0,0,0,59640,0

%N Smallest tetrahedral number with exactly n divisors, or 0 if no such number exists.

%C 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:

%C .

%C   p^6      q^2       r           resulting equations

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

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

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

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

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

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

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

%C .

%C 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).

%C 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.

%C 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]

%C Solutions to 2*p^6 = q^2 + 1 correspond to (some) integral points (X,Y) = (p^3, 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

%Y Cf. A000292, A279081.

%Y Cf. A081978 (analogous sequence for triangular numbers).

%K nonn

%O 1,3

%A _Jon E. Schoenfield_, Jan 06 2017, Jan 31 2021

%E a(42)=a(50)=a(70)=a(78)=a(98)=0 from _Max Alekseyev_, Feb 03 2024