cp's OEIS Frontend

A190797 For primes p and q=p+6 create primitive Pythagorean triangles with sides (q^2 - p^2)/2, (p^2 + q^2)/2, and p*q. If the two remainders of the middle and longest side modulo the shortest side are both prime, then p is in the sequence.

%I A190797 #12 May 25 2025 20:43:10
%S A190797 11,23,41,83,107,167,191,263,307,347,367,461,641,653,877,881,1103,
%T A190797 1187,1367,2081,2393,2677,3607,4283,4357,4967,5081,5231,5387,5471,
%U A190797 5651,6037,6197,6311,6353,6857,7823,8117,8693,8747,9221,9743,9851,9923
%N A190797 For primes p and q=p+6 create primitive Pythagorean triangles with sides (q^2 - p^2)/2, (p^2 + q^2)/2, and p*q. If the two remainders of the middle and longest side modulo the shortest side are both prime, then p is in the sequence.
%C A190797 The short side is 6p+18, the middle side p^2+6p, the long side 6p+18+p^2.
%C A190797 The first few values have more terms == 3 (mod 4) than 1 (mod 4), but this does not appear to be the case for later terms. - _Franklin T. Adams-Watters_, May 22 2011
%F A190797 If p=6k+5, then the remainders are 7 + 12*k and 25 + 12*k.
%F A190797 If p=6k+1, then the remainders are 7 + 24*k and 25 + 24*k.
%e A190797 For p=41 and q=47, the sides are (47^2 - 41^20)/2=264, 41*47=1927 and (41^2 + 43^2)/2=1945; divide 1927 and 1945 through 264 to get remainders 79 and 97.  Since both are primes, p=41 is in the sequence.
%o A190797 (PARI) forprime(p=5,10000,if(isprime(q=p+6),x=(q^2-p^2)/2;if(isprime(((q^2+p^2)/2)%x)&isprime(p*q%x),print1(p", "))))
%Y A190797 Cf. A023201.
%K A190797 nonn
%O A190797 1,1
%A A190797 _J. M. Bergot_, May 20 2011
%E A190797 Corrected and extended by _Franklin T. Adams-Watters_, May 22 2011