

A321616


Primes p = k^2 + (k1)^2 such that k^p  (k1)^p is prime.


0




OFFSET

1,1


COMMENTS

Conjecture: generally, these are primes p = a^2 + b^2 with a > b > 0 such that (a^p  b^p)/(ab) is prime, so must be ab = 1. It seems that there are no primes (a^q + b^q)/(a+b) for primes q = a^2 + b^2 > 5. Especially, there are probably no primes q = m^2 + 1 > 5 such that (m^q  1)/(m1) is prime or (m^q + 1)/(m+1) is prime. How to prove it?
No more terms up to the prime 19801 = 100^2 + 99^2.  Amiram Eldar, Nov 15 2018


LINKS

Table of n, a(n) for n=1..4.


EXAMPLE

The prime 5 = 2^2 + 1^2 and 2^5  1^5 = 31 is prime.
We have 61 = 6^2 + 5^2, 113 = 8^2 + 7^2, 1741 = 30^2 + 29^2.


MATHEMATICA

f[k_]:=k^2 + (k1)^2 ; seqQ[k_]:=Module[{p=f[k]}, PrimeQ[p] && PrimeQ[k^p  (k1)^p ]]; f[Select[Range[30], seqQ]] (* Amiram Eldar, Nov 15 2018 *)


PROG

(PARI) lista(nn) = {for (k=1, nn, if (isprime(p=k^2 + (k1)^2) && isprime(k^p  (k1)^p), print1(p, ", ")); ); } \\ Michel Marcus, Nov 18 2018


CROSSREFS

Cf. A002144, A121091.
Subsequence of A027862.
Sequence in context: A139915 A174053 A107191 * A182352 A142643 A201848
Adjacent sequences: A321613 A321614 A321615 * A321617 A321618 A321619


KEYWORD

nonn,more


AUTHOR

Thomas Ordowski, Nov 15 2018


STATUS

approved



