A283680 Odd numbers of the form (p^2 - q^2) / 24 with primes p > q > 3.

1, 3, 5, 7, 11, 13, 15, 17, 21, 23, 25, 27, 33, 35, 37, 39, 43, 45, 47, 53, 55, 57, 63, 65, 67, 69, 73, 75, 77, 85, 87, 91, 93, 95, 97, 103, 105, 107, 113, 115, 117, 123, 127, 133, 135, 137, 143, 145, 147, 153, 155, 157, 163, 165, 167, 173, 175, 177, 183, 185, 187, 193, 195, 203, 205, 207, 209, 213, 215, 217

Offset: 1

Author

Thomas Ordowski and Altug Alkan, Mar 14 2017

Keywords

• nonn

Comments

If p == 3(mod 8) and p == 1(mod 3), and if q == 7(mod 8) and q == 1(mod 3); then (p^2 - q^2)/24 is odd. Thus, this sequence is infinite.

Note: if p - q > 12, then (p^2 - q^2)/24 is composite.

Theorem: (p^2 - q^2)/24 is an odd integer if and only if pq == +-3(mod 8). - Carl Pomerance, Mar 14 2017

The complement is: 9, 19, 29, 31, 41, 49, 51, 59, 61, 71, 79, 81, 83, 89, 99, 101, 109, 111, 119, 121, 125, 129, 131, 139, 141, 149, 151, 159, 161, 169, 171, ... - Robert G. Wilson v, Mar 14 2017

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A283562.

Programs

Mathematica

p = 7; lst = {}; While[p < 501, q = 5; While[q < p, If[Mod[p^2 - q^2, 24] == 0 && Mod[(p^2 - q^2)/24, 2] == 1, AppendTo[lst, (p^2 - q^2)/24]; lst = Union@ lst]; q = NextPrime@ q];  p = NextPrime@ p]; Take[lst, 75] (* Robert G. Wilson v, Mar 14 2017 *)

Formula

a(n) ~ cn(loglog n)^(3/2)/(log n)^(1+eta), where eta = 1-(1+loglog 2)/log 2 = 0.086..., perhaps with c about 4/3. - Conjectured by Carl Pomerance, Mar 15 2017