A368460 a(n) = card(k: prime(n)^2 <= k < prime(n + 1)^2 and k term of A368458).

1, 1, 1, 1, 2, 2, 2, 3, 3, 2, 5, 3, 3, 4, 5, 4, 3, 9, 5, 4, 7, 5, 6, 14, 6, 4, 7, 3, 5, 22, 6, 6, 4, 16, 5, 12, 12, 8, 15, 13, 5, 19, 5, 10, 4, 30, 26, 10, 6, 12, 11, 5, 28, 19, 15, 20, 8, 19, 9, 7, 22

Offset: 1

Author

Peter Luschny, Dec 26 2023

Keywords

• nonn
• more

Comments

If A368458 is written as an irregular triangle for n >= 3, then a(n) is the length of row n.

Conjecture: For all n >= 5, there is at least one j such that b(j) = 2 * (Bacher(j) - sigma(j)) + j + 1 is > 0 and prime(n)^2 < b(j) < prime(n + 1)^2. In other words, a(n) > 1 for n >= 5.

Examples

			a(11) = 5 because 31^2 = 961, 1073, 1147, 1271, 1333, 1369 = 37^2 and all the terms are in that order in A368458.
		

Crossrefs

Cf. A000203, A001248, A050216 (Brocard's Conjecture), A368207 (Bacher), A368457, A368458.

Programs

SageMath

# using A368207
def A368460(n):
    pn = nth_prime(n); pn1 = nth_prime(n + 1)
    A368457 = lambda n: 2 * (A368207(n) - sigma(n)) + n + 1
    return sum(1 for n in range(pn ** 2, pn1 ** 2) if A368457(n) > 0)
print([A368460(n) for n in range(1, 25)])