cp's OEIS Frontend

Terms may be categorized as belonging to the following types:

type 1: products of 3 distinct primes p,q,r such that 2*p*q + 1 = r: 78, 406, 465, ... (27108 of the first 100000 terms);

type 2: products of 3 distinct primes p,q,r such that 2*p*q - 1 = r: 66, 190, 435, ... (26848 of the first 100000 terms);

type 3: products of 3 distinct primes p,q,r such that p*q + 1 = 2*r: 231, 561, 1653, ... (23050 of the first 100000 terms);

type 4: products of 3 distinct primes p,q,r such that p*q - 1 = 2*r: 105, 595, 741, ... (22983 of the first 100000 terms);

type 5: products of the cube of a prime p and a distinct prime q such that 2*p^3 + 1 = q: 136, 31375, 3544453, ... (6 of the first 100000 terms);

type 6: products of the cube of a prime p and a distinct prime q such that 2*p^3 - 1 = q: 1431, 1774977571, 12642646591, ... (4 of the first 100000 terms);

type 7: products of the cube of a prime p and a distinct prime q such that p^3 - 1 = 2*q: the only term of this type is 351 = 3^3 * 13.

(No term is a product of the cube of a prime p and a distinct prime q such that p^3 + 1 = 2*q.)

This sequence contains no primes (since any prime p has an absolute difference of p from the zeroth triangular number, A000217(0) = 0*(0+1)/2 = 0).

The smallest numbers in this sequence having fewer than 8 divisors are

a(82) = 65341 = A000217(361) = 19^2 * 181,

a(248) = 354061 = A000217(841) = 29^2 * 421,

a(1431) = 6924781 = A000217(3721) = 61^2 * 1861,

a(2021) = 12708361 = A000217(5041) = 71^2 * 2521, and

a(2589) = 19478161 = A000217(6241) = 79^2 * 3121, each of which is a triangular number with exactly 6 divisors (A292989).

Conjectures:

(1) This sequence is a subset of the triangular numbers (A000217).

(2) This sequence includes no semiprimes.

Any number having an odd number of divisors is a square, so each term in this sequence is a term of A001110 (numbers that are both triangular and square). Since A001110(k) = (A000129(k)*A001333(k))^2, A001110(k) will have exactly 9 divisors iff A000129(k) and A001333(k) are both prime (i.e., k is in both A096650 and A099088); the first 5 values of k at which this occurs are 2, 3, 5, 29, and 59.

Conjecture: a(5) is the final term of this sequence.

Intersection of A048161 and A106483.

How many triangular numbers with 6 divisors (A292989) can be divisible by the same squared prime p^2?

The k-th triangular number T(k) = A000217(k) = k*(k+1)/2 can be written as the product of two coprime factors A and B where A=k and B=(k+1)/2 for odd k, A=k/2 and B=k+1 for even k. If a triangular number has 6 divisors, then it is of the form p^2*q where p and q are distinct primes. We can identify four cases:

Case 1: A = k = p^2 and B = (k+1)/2 = q, so q = (p^2 + 1)/2; solutions occur at primes p in A048161.

Case 2: A = k = q and B = (k+1)/2 = p^2, so 2*p^2 - 1 = q; solutions occur at primes p in A106483.

Case 3: A = k/2 = p^2 and B = k+1 = q. In this case, 2*p^2 + 1 = q. For p = 2, we would get q = 9 (nonprime), so p must be odd. If prime p > 3 (so q > 19), we have p^2 == 1 (mod 3), so q == 0 (mod 3), hence nonprime. So the only solution for this case occurs at p=3, q=19, t = 3^2*19 = 171.

Case 4: A = k/2 = q and B = k+1 = p^2. In this case, 2*q + 1 = p^2, so p is odd, but then p^2 == 1 (mod 8), so q == 0 (mod 4), hence q is not prime: no solutions exist.

Since Case 4 has no solutions, at most three triangular numbers with 6 divisors can be divisible by the same squared prime p^2; Case 3 has a solution only at p=3 and, in fact, there are three triangular numbers with 6 divisors that are divisible by 3^2: t = 3^2*5 = 45 = T(9), t = 3^2*17 = 153 = T(17), and 3^2*19 = 171 = T(18).

For all primes p > 3, then, at most two triangular numbers with 6 divisors are divisible by p^2; this sequence (after the initial term, 3) lists the primes p such that p^2 divides exactly two triangular numbers that have 6 divisors.

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}?