A364554 a(n) = number of primes of the form T(k)/n, for some k, where T(k)=A000217(k) is a triangular number.

1, 2, 3, 1, 3, 2, 2, 1, 3, 1, 2, 2, 1, 2, 4, 1, 0, 2, 1, 1, 4, 2, 2, 2, 1, 2, 2, 0, 1, 3, 1, 0, 4, 1, 3, 2, 2, 1, 3, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 1, 4, 1, 1, 3, 1, 1, 4, 1, 1, 3, 1, 0, 2, 1, 2, 1, 0, 2, 2, 2, 0, 0, 1, 1, 4, 1, 1, 2, 1, 0, 2, 1, 2, 2, 2, 2, 4, 0, 1, 4

Offset: 1

Author

Lamine Ngom, Jul 28 2023

Keywords

• nonn

Comments

Implementing a suggestion in the comment section of sequences A154296, ..., A154304, this sequence computes the number of primes of the form T(k)/n.

Equivalently, number of primes p such that 8*n*p+1 is a perfect square.

Let's consider, for all primes p, the set of linear recurrences {b(m)} defined as follows:

If p = 2, then {b(m)} = A074378 (numbers of the form x*(4*x -+ 1)); otherwise, b(m) = b(m-1) + 2*b(m-2) - 2*b(m-3) - b(m-4) + b(m-5) with initial terms b(0) = 0, b(1) = (p-1)/2, b(2) = (p+1)/2, b(3) = 2*p-1 and b(4) = 2*p+1. Numbers of the form x*(p*x -+ 1)/2.

Then a(n) = number of sequences {b(m)} in which n is a term.

This implies that:

i) for any n, the largest prime of the form T(k)/n is at most 2*n+1;

ii) if n is prime, then a(n) < 4. (3 and 5 are the only primes p such that a(p) = 3; primes p such that a(p) = 0 are A109998.)

Deeper in the examination of these results, we notice that the set of primes p of the form T(k)/n arises from the factorization of n. This set is exactly all primes p of the form (2*r -+ 1)/d or (r -+ 1)/(2*d), where d is some divisor of n and r is the ratio n/d. (Proof is welcome.)

Indices k of corresponding triangular numbers T(k) such that T(k) = n*p are then:

2*r if p = (2*r + 1)/d,

2*r - 1 if p = (2*r - 1)/d,

r if p = (r + 1)/(2*d),

r - 1 if p = (r - 1)/(2*d).

And pluging the value of p in the equivalent definition, the expression 8*n*p+1 yields respectively to following perfect squares: (4*r+1)^2, (4*r-1)^2, (2*r+1)^2 and (2*r-1)^2.

Examples

			a(15) = 4 since there are exactly 4 triangular numbers T(k) such that T(k) = 15*p, with p prime.
T(9)/15 = 45/15 = 3, T(14)/15 = 105/15 = 7, T(29)/15 = 435/15 = 29 and T(30)/15 = 465/15 = 31.
a(17) = 0 since there is no triangular number T(k) such that T(k) = 17*p, with p prime.
		

Crossrefs

Cf. A000217, A154296, ..., A154304, A074378, A001318, A057569, A057570, A109998.

Cf. A364555 (indices of 0's).

Formula

Conjecture: a(n) = number of primes in the union of sets {(2*r -+ 1)/d; (r -+ 1)/(2*d)}, with d divisor of n and r = n/d.