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A124598
Primes p of the form k^2+s where k > 1 and 1 <= s < (k+1)^2, such that q = k^4+s is prime and larger than p.
3
5, 7, 11, 17, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 89, 97, 101, 107, 109, 127, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 307, 311, 331, 337, 347
OFFSET
1,1
COMMENTS
The terms of this sequence illustrate a special case of the conjecture from A126769.
EXAMPLE
5 = 2^2+1 is prime, 17 = 2^4+1 is a larger prime and 1 < 3^2, hence 5 is a term.
29 = 4^2+13 is prime, 269 = 4^4+13 is a larger prime and 13 < 5^2, hence 29 is a term.
805499 = 897^2+890 is prime, 647395643771 = 897^4+890 is a larger prime and 890 < 898^2, hence 805499 is a term.
Prime number 19 has the form k^2+s with s < (k+1)^2 in two ways, as 3^2+10 and 4^2+3. Neither 3^4+10 = 91 nor 4^4+3 = 259 is prime, hence 19 is not in the sequence.
PROG
(PARI) m=19; v=[]; for(k=2, m, for(s=1, (k+1)^2-1, if((p=k^2+s)<m^2&&isprime(p)&&(q=k^4+s)>p&&isprime(q), v=concat(v, p)))); print(Set(v)) \\
(PARI) upto(n)=my(res = List()); forprime(p = 5, n, for(k = ceil(sqrt(p / 2 + 1/4) - 0.5), sqrtint(p-1), if(isprime(k^4 + p - k^2), listput(res, p); next(2)))); res \\ David A. Corneth, Apr 08 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Tomas Xordan, Mar 02 2007
EXTENSIONS
Edited, corrected and extended by Klaus Brockhaus, Mar 05 2007
STATUS
approved