A321616 Primes p = k^2 + (k-1)^2 such that k^p - (k-1)^p is prime.

5, 61, 113, 1741

Offset

1,1

Comments

Conjecture: generally, these are primes p = a^2 + b^2 with a > b > 0 such that (a^p - b^p)/(a-b) is prime, so must be a-b = 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)/(m-1) 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

a(5) > 109045. - J.W.L. (Jan) Eerland, Dec 11 2022

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 + (k-1)^2 ; seqQ[k_]:=Module[{p=f[k]}, PrimeQ[p] && PrimeQ[k^p - (k-1)^p ]]; f[Select[Range[30], seqQ]] (* Amiram Eldar, Nov 15 2018 *)
pQ[k_]:=Module[{c=k^2+(k-1)^2}, If[AllTrue[{c, k^c-(k-1)^c}, PrimeQ], c, Nothing]]; Array[pQ, 30] (* Harvey P. Dale, Aug 27 2023 *)

Prog

(PARI) lista(nn) = {for (k=1, nn, if (isprime(p=k^2 + (k-1)^2) && isprime(k^p - (k-1)^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