OFFSET
1,2
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
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
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Ordowski and Altug Alkan, Mar 14 2017
STATUS
approved