A077659 a(n) = smallest k>1 such that the sum n^k + (n+1)^k is prime, or -1 if no such k exists.

2, 2, 4, 2, 2, 4, 2, 4, 2, 32

Offset

1,1

Comments

Checking k up through 1024 suggests that the sequence may continue -1, 2, 4, 2, -1, 4, 2, -1, 2, -1, 16, 2, 8, 2, 2, 4, 4, -1, 2, 2, 4, 2, 4, 2, 2, 4, 4, 4, 2, ...

For any a>1 and b>1, a^k + b^k is composite for all odd k>1. Hence if n^k + (n+1)^k is prime then k must be a power of 2.

It is known that a(11) > 2^22. Is it possible that 11^2^m + 12^2^m is composite for all m > 0?

Links

Table of n, a(n) for n=1..10.

T. D. Noe, Factorizations of Generalized Fermat Numbers 12^2^k + 11^2^k

Example

a(3)=4 because 3^2 + 4^2 = 25 is not prime, but 3^4 + 4^4 = 337 is prime.

Mathematica

lst={}; For[n=1, n<=100, n++, k=2; While[k<=2^10 && !PrimeQ[n^k+(n+1)^k], k=2*k]; If[k<=2^10, AppendTo[lst, k], AppendTo[lst, -1]]]; lst

Crossrefs

Cf. A078902.

Cf. A080121.

Sequence in context: A073103 A247257 A069177 * A367953 A212595 A087692

Adjacent sequences: A077656 A077657 A077658 * A077660 A077661 A077662

Keyword

hard,more,nonn

Author

T. D. Noe, Nov 14 2002

Status

approved