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A190797
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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.
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0
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11, 23, 41, 83, 107, 167, 191, 263, 307, 347, 367, 461, 641, 653, 877, 881, 1103, 1187, 1367, 2081, 2393, 2677, 3607, 4283, 4357, 4967, 5081, 5231, 5387, 5471, 5651, 6037, 6197, 6311, 6353, 6857, 7823, 8117, 8693, 8747, 9221, 9743, 9851, 9923
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OFFSET
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1,1
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COMMENTS
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The short side is 6p+18, the middle side p^2+6p, the long side 6p+18+p^2.
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
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LINKS
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FORMULA
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If p=6k+5, then the remainders are 7 + 12*k and 25 + 12*k.
If p=6k+1, then the remainders are 7 + 24*k and 25 + 24*k.
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EXAMPLE
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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.
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PROG
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(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", "))))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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