A279082 Smallest tetrahedral number with exactly n divisors, or 0 if no such number exists.

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, 349504, 0, 2362041, 0, 7770, 0, 0, 0, 3276, 0, 0, 0, 82160, 0, 0, 0, 4980598206895701865797099599990718771851543594988165352649, 19600, 0, 0, 7140, 0, 0, 0, 91625967616, 0, 155965244802415621, 0, 41664, 0, 0, 0, 2332880, 0, 0, 0, 59640, 0

Offset

1,3

Comments

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^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

Links

Table of n, a(n) for n=1..65.

Crossrefs

Cf. A000292, A279081.

Cf. A081978 (analogous sequence for triangular numbers).

Sequence in context: A051499 A333654 A008345 * A016488 A087212 A048870

Adjacent sequences: A279079 A279080 A279081 * A279083 A279084 A279085

Keyword

nonn

Author

Jon E. Schoenfield, Jan 06 2017, Jan 31 2021

Extensions

a(42)=a(50)=a(70)=a(78)=a(98)=0 from Max Alekseyev, Feb 03 2024

Status

approved