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a(n) = 4 iff either n is in A005383 or n/2 is in A005384.

a(n) is odd iff n is in A001108.

a(n) = 6 if either n = 18 or n = q^2 where q is in A048161 or n = 2 q^2 - 1 where q is in A106483. - Robert Israel, Oct 26 2015

From Bernard Schott, Aug 29 2020: (Start)

a(n-1) is the number of solutions in positive integers (x, y, z) to the simultaneous equations (x + y - z = n, x^2 + y^2 - z^2 = n) for n > 1. See the British Mathematical Olympiad link. In this case, one always has z > x and z > y.

For n = 12 as in the Olympiad problem, the a(11) = 8 solutions are (13,78,79), (14,45,47), (15,34,37), (18,23,29), (23,18,29), (34,15,37), (45,14,47), (78,13,79). (End)

Not every term T(k) has the same prime signature as its successor triangular number T(k+1); the first counterexample is the pair (T(52), T(53)) = (1378, 1431) = (2 * 13 * 53, 3^3 * 53), each of which has 8 divisors. The first counterexample in which the two triangular numbers have the same number of distinct prime factors is (T(45630), T(45631)) = (1041071265, 1041116896) = (3^3 * 5 * 13^2 * 45631, 2^5 * 23 * 31 * 45631), each of which has 48 divisors.

Number of divisors of A001110(n).

Since each term in A001110 is a square, each term in this sequence is an odd number.

Adding terms to this sequence requires obtaining the prime factorization of terms from A001110. This is facilitated by the observation that A001110(n) = (A000129(n)*A001333(n))^2, but after a few hundred terms, it becomes difficult to obtain the prime factorization of some of the values of A000129(n) and A001333(n).

Obviously, a(1) is the only term whose value is 1. It can be shown (see A081978) that 15 appears only at a(7).

Conjectures:

(1) There exist only 5 triangular numbers with exactly 9 divisors: the 2nd, 3rd, 5th, 29th, and 59th terms of A001110.

(2) A001110(4)=41616 is the only triangular number with exactly 45 divisors.

(3) A001110(6)=48024900 is the only triangular number with exactly 405 divisors.

(4) A001110(8)=55420693056 is the only triangular number with exactly 189 divisors.

Every number with an odd number of divisors is a square, and no two consecutive positive triangular numbers are squares, so for all odd n, a(n) = 1 if a triangular number with n divisors exists, 0 otherwise.

T(190254647) = 18098415447674628 is the smallest triangular number that begins a run of 9 consecutive triangular numbers with 48 divisors. Does a longer run exist? - Jon E. Schoenfield, May 29 2022

From Jon E. Schoenfield, Mar 17 2023: (Start)

9 <= a(48) <= 16.

a(48) < 17 because a run of 17 consecutive triangular numbers with 48 divisors each would require (see the Example section at A319038) a run of 9 consecutive odd numbers 2k+1, 2k+3, ..., 2k+17 with tau2 divisors each and a run of 9 consecutive integers -- either k, k+1, ..., k+8 or k+1, k+2, ..., k+9 -- with tau1 divisors each, with tau1*tau2 = 48. There exists no run of 8 or more consecutive integers with tau1 < 12 divisors each, nor is there any run of 9 or more consecutive odd numbers with tau1 <= 4 divisors each (tau1 = 4 is impossible because among any 9 or more consecutive odd numbers there is at least one that is a multiple of 9, and the only such multiple with exactly 4 divisors is 3^3 = 27).

Similarly, it can be shown that if a(48) = 16, there must exist a run of 8 consecutive odd numbers 2k+1, 2k+3, ..., 2k+15 with tau2 = 4 divisors each and a run of 9 consecutive integers k, k+1, ..., k+8 with tau1 = 12 divisors each. It seems to me that such solutions should exist (although it could be very difficult to find any). E.g., the 9 consecutive integers could be numbers of the forms 2*13^2*p, 3*5^2*q, 2^5*r, s^2*t*u, 2*3^2*v, 7^2*w*x, 2^2*5*y, 3*z^2*a, and 2*b^2*c, respectively, and the 8 consecutive odd numbers could be of the forms 17*d, 7*e, 3*f, 5*g, 11*h, 3*i, 13*j, and 19*m, respectively, where each letter (other than k) represents a distinct prime. (End)

Further known terms are a(12)=A000045(91); a(16)=A000045(44); a(24)=A000045(50); a(32)=A000045(30); a(36)=A000045(24); a(48)=A000045(56); a(64)=A000045(54); a(80)=A000045(36); a(96)=A000045(182); a(128)=A000045(128); a(168)=A000045(48); a(192)=A000045(110); a(256)=A000045(80), ..., a(688128)=A000045(240) from the Kelly factorizations. - R. J. Mathar, Apr 05 2007

For n prime, a(n) = q*p^(n-1) or p^(2n-1) for some primes p and q since those are the only numbers with 2*n divisors. a(8) = 2584. - Chai Wah Wu, Dec 08 2014

The sequence is restricted to even numbers of divisors since 1 and 144 are the only Fibonacci numbers with an odd number of divisors (because they are the only positive Fibonacci numbers that are squares, see A227875). - Amiram Eldar, Jul 02 2023

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:

.

p^6 q^2 r resulting equations

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

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

.

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^2, 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

The only primes p for which a(p) > 0 are those for which both 2*3^(p-1) - 1 and 2*3^(p-1) + 1 are prime: 2, 3, and any other primes p such that p-1 appears both in A003307 and A003306. (If such a prime p > 3 exists, then p exceeds 1360105.)

Conjecture: The only primes p for which a(p) > 0 are 2 and 3.

The number of terms in the n-th row is A319037(n). Row n has no terms here iff A081978(n) = 0 (i.e., there is no triangular number with n divisors; this is the case for n = 3, 5, 7, 11, 13, 17, 19, 21, 23, 25, 29, 31, 33, 35, 37, 39, 41, 43, 47, 49, 51, 53, 55, 57, 59, 61, ...).

Additional known terms include a(15)=270480, a(16)=77309214720, a(19)=117261433825538425475625, a(20)=7874496, a(22)=0, a(23)=316659361382400, a(24)=100472400, a(25)=0, a(27)=18951806016, a(28)=35184372088827805696000000, a(31)=20752587086144471040, a(32)=3877678080.

It is known (see the comments and links at A081978) that a(n)=0 for every n such that n*(n+1)/2 is an odd composite not divisible by 3; this includes n = 10, 13, 22, 25, ..., i.e., all n such that n mod 12 is 1 or 10.

Conjectures:

1. a(n) > 0 for every n such that n*(n+1)/2 is even.

2. a(n) = 0 for every n such that n*(n+1)/2 is odd except n = 1, 5, and 9 (whose corresponding values of n*(n+1)/2 are 1, 15, and 45, respectively). Can this be proved for any of the values of n in {14, 17, 18, 21, 26, 29, 30}?