A283562 -id:A283562 - cp's OEIS frontend

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

Thomas Ordowski and Altug Alkan, Mar 14 2017

nonn

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

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

Cf. A283562.

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 *)

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

A283792 Primes of the form (p^2 + q^2) / 2 such that (p^2 - q^2) / 24 is prime, where primes p > q > 3.

Original entry on oeis.org

109, 157, 229

Offset: 1

Thomas Ordowski and Altug Alkan, Mar 16 2017

Union of primes of the form:
t^2 + 6^2 such that t and p = t+6 and q = t-6 are primes,
(2t)^2 + 3^2 such that t and p = 2t+3 and q = 2t-3 are primes,
(3t)^2 + 2^2 such that t and p = 3t+2 and q = 3t-2 are primes,
(6t)^2 + 1^2 such that t and p = 6t+1 and q = 6t-1 are primes.
Note: this last subset is empty.
We have p*q*(p^2-q^2)*(p^2+q^2) = p^5*q - p*q^5 == 0 (mod 5), so at least one of p, q, p^2-q^2, or p^2+q^2 must be divisible by 5. Thus, this sequence is finite and 229 is the last term. - Robert Israel, Mar 16 2017

			Prime 109 = (13^2 + 7^2)/2 is a term since (13^2 - 7^2)/24 = 5 is prime.
Note: 109 = (2*5)^2 + 3^2, 157 = 11^2 + 6^2, and 229 = (3*5)^2 + 2^2.

Cf. A103739, A283562.

list(lim)=my(v=List(), p2, q2, t); lim\=1; lim=min(max(lim,9),229); forprime(p=3, sqrtint(2*lim-9), p2=p^2; forprime(q=3, min(sqrtint(2*lim-p2), p-2), q2=q^2; if((p2-q2)%24==0 && isprime(t=(p2+q2)/2) && isprime((p2-q2)/24), listput(v, t)))); Set(v) \\ Charles R Greathouse IV, Mar 17 2017

cp's OEIS Frontend

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

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