cp's OEIS Frontend

a(p)=0 if p is an odd prime. If n is an odd composite number, then a(n) is a square; see A001110 for numbers that are both triangular and square. - Victoria A Sapko (vsapko(AT)frc.mass.edu), Sep 28 2007

From Jon E. Schoenfield, May 25 2014: (Start)

If n is an odd semiprime, then a triangular number t having exactly n divisors must be of the form t = p^(2r) * q^(2s) = (p^r * q^s)^2, where p and q are distinct primes (p < q) and r and s are positive integers such that (2r+1)*(2s+1) = n.

If t is the k-th triangular number k(k+1)/2, it can be factored as t = u * v where

u = k and v = (k+1)/2 if k is odd, or

u = k/2 and v = k+1 if k is even.

Since neither p nor q (each of which is greater than 1) can divide both k and (k+1)/2, or both k/2 and k+1, only four cases need to be considered:

Case 1: k is even, q^(2s) = k/2, p^(2r) = k+1

Case 2: k is even, p^(2r) = k/2, q^(2s) = k+1

Case 3: k is odd, q^(2s) = k, p^(2r) = (k+1)/2

Case 4: k is odd, p^(2r) = k, q^(2s) = (k+1)/2

These yield the following equations:

Case 1: 2 * q^(2s) + 1 = p^(2r)

Case 2: 2 * p^(2r) + 1 = q^(2s)

Case 3: 2 * p^(2r) - 1 = q^(2s)

Case 4: 2 * q^(2s) - 1 = p^(2r)

Case 1 can be ruled out: since q > p, q is odd, so 2 * q^(2s) + 1 == 3 (mod 16), but p^(2r) cannot be 3 (mod 16).

For a Case 2 solution, since q is odd, 2 * p^(2r) + 1 = q^(2s) == 1 (mod 8), so p^(2r) == 0 (mod 4), so p must be even. Therefore, p = 2, and we must satisfy the equation 2 * 2^(2r) + 1 = q^(2s), whose left-hand side, which is divisible by 3 for every nonnegative integer r, is thus the square of a prime iff it is 3^2. So r=1, q=3, and s=1, yielding t = 2^2 * 3^2 = 36, which is the smallest triangular number with exactly 9 divisors, so a(9)=36.

In Case 3, both p and q must be odd, and p^r must be a number w having the property that 2*w^2 - 1 is square (i.e., (q^s)^2); every such number w is in A001653 (1, 5, 29, 169, 985, ...), and the corresponding value of q^s = sqrt(2*w^2 - 1) is in A002315 (1, 7, 41, 239, 1393, ...). (Note that A001653 and A002315 are defined with offsets of 1 and 0, respectively, so A001653(j) corresponds to A002315(j-1).) However, for odd semiprime n > 9, we need r > 1 and/or s > 1. The only nontrivial power (i.e., number of the form x^m, where both x and m are integers greater than 2) in A001653 is A001653(4) = 169 = 13^2 [Pethö], so the only Case 3 solution with r > 1 is 2 * 13^4 - 1 = 239^2, which yields the 15-divisor triangular number 13^4 * 239^2 = 1631432881 = a(15). A Case 3 solution with r = 1 and s > 1 would require 2 * p^2 - 1 = q^(2s), which is impossible since p < q.

Finally, in Case 4, both p and q must be odd, q^s must be in A001653, and p^r must be the corresponding term in A002315. However, using the only nontrivial power in A001653 (i.e., 169 = 13^2) as q^s would not yield a valid solution here because it would mean p = 239 and q = 13 (contradicting p < q). Thus, if a Case 4 solution exists for odd semiprime n > 9, we must have s = 1 and r > 1, so n = (2r+1)*(2s+1) = (2r+1)*3, where 2r+1 is prime. Such a solution requires an index j satisfying two conditions: (1) A001653(j) = q^1 = q is prime, and (2) the corresponding term A002315(j-1) = p^r is a nontrivial prime power. There are no nontrivial powers (whether of primes or composites) among the terms in A002315 below 10^5000. Moreover, the terms in A001653 are the odd-indexed terms from A000129 (Pell numbers), so condition (1) requires that j satisfy A000129(2j-1) = q. A096650 lists the indices of all prime or probable prime Pell numbers up to 100000. A check of the value A002315(j-1) corresponding to each prime or probable prime among the odd-indexed Pell numbers A000129(2j-1) up to j=50000 determined that none were nontrivial powers, so if any Case 4 solution with n > 9 exists, it will yield a triangular number t = p^(2r) * q^2 = (2 * q^2 - 1) * q^2, where q >= A000129(100001) = 3.16...*10^38277, so t > 10^153110. Since there are no nontrivial powers at all in A002315 below 10^5000, and since prime Pell numbers above A000129(50000) seem so scarce, it seems extremely unlikely that any such solution exists.

Thus, a(n) = 0 for every odd semiprime n that is not divisible by 3, and assuming that no Case 4 solution for odd semiprime n > 9 exists, the only nonzero a(n) where n is an odd semiprime greater than 9 is a(15) = 13^4 * 239^2 = 1631432881.

If j is prime and n=2j, then a(n) (if nonzero) must be of the form p^r * q, where p and q are distinct primes, r = j-1, and q is one of 3 functions of p^r:

q = f1(p^r) = 2*p^r - 1

q = f2(p^r) = 2*p^r + 1

q = f3(p^r) = (p^r - 1)/2

Of these, q = f1(p^r) for all but two cases among n < 1000:

at n=362, q = f2(p^r), with p=3;

at n=514, q = f3(p^r), with p=331.

Conjecture: a(2j) > 0 for all j > 1. (This conjecture holds at least through n = 2j = 1000. The largest a(n) for even n <= 1000 is a(898) = 20599^448 * (2 * 20599^448 - 1) = 3.21...*10^3865.) (End)

For more known terms, and information about unknown terms, see Links. - Jon E. Schoenfield, May 26 2014

If d(k*(k+1)/2) = 21, note that 2*q^2 = p^6 + 1 = (p^2 + 1)*(p^4 - p^2 + 1) has no prime solutions, so then k = p^2 and k+1 = 2*q^6, where p and q are distinct primes. We can prove 2*x^3 - y^2 = 1 has only one positive solution (1, 1) which shows that p^2 + 1 = 2*q^6 has no prime solutions. In the ring of Gaussian integers, x^3 = (1+y*i)*(1-y*i)/((1+i)*(1-i)) and (1+y*i)/(1+i) is coprime to (1-y*i)/(1-i), thus (1+y*i)/(1+i) = (u+v*i)^3 and (1-y*i)/(1-i) = (u-v*i)^3 for some integers u and v. Note that 1+y*i = (1+i)*(u+v*i)^3 = (u+v)*(u^2+v^2-4*u*v) + (u-v)*(u^2+v^2+4*u*v)*i, we have (u+v)*(u^2+v^2-4*u*v) = 1. Therefore, u = 1 and v = 0 if u > v, which means y = (u-v)*(u^2+v^2+4*u*v) = 1. This implies that a(21) = 0. - Jinyuan Wang, Aug 22 2020

a(n) is a perfect number for all n such that n/2 is in A000043. - J. Lowell, Mar 16 2024

Previous Name: Primes found among the denominators of the continued fraction rational approximations to sqrt(2).

See A096650 for the indices. - Jon E. Schoenfield, Jan 25 2017

A056869 is essentially the same sequence. - Jianing Song, Jan 02 2019

Note that for A001333(n) to be prime, the index n must be prime or a power of 2. The indices greater than 421 yield probable primes.

Numbers n for which ((1+sqrt(2))^n + (1-sqrt(2))^n)/2 is prime. - Artur Jasinski, Dec 10 2006

First differs from A264137 (Largest prime factor of the n-th Pell number) at n=17; see Example section. - Jon E. Schoenfield, Dec 10 2016

Index must be prime. The indices greater than 293 yield probable primes.

Conjecture: for n>1 a(n) is a multiple of 3.

The indices greater than 363 yield probable primes. - Vaclav Kotesovec, Oct 28 2013

Next term > 100000. - Tyler Busby, Mar 29 2024

a(29) >= 709. - Hugo Pfoertner, Jul 29 2019

859, 937, 1087, 1151, and 1193 belong to the sequence. 709 and 787 have not yet been ruled out. The next candidate after these appears to be 1471. - Daniel M. Jensen, Oct 18 2019

Primes p such that ((1+sqrt(2))^p - (1-sqrt(2))^p) / (2*sqrt(2)) is a composite number.

For all corresponding values of k, A000129(k) is a composite number. In other words, k cannot be a term of A096650.

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.