OFFSET
1,2
COMMENTS
There is no square > 1 in this sequence, because if f(n) = n^4 + n^3 + n^2 + n + 1, then f(n^2) = f(n)*f(-n). Actually, f(x) divides f(x^m) for all m not in 5Z. So the only perfect powers in this sequence can be 5th, 25th, 125th... powers. The least perfect power > 1 in this sequence is 22^5. - M. F. Hasler, Feb 09 2012
The corresponding prime numbers n^4 + n^3 + n^2 + n + 1 are in A088548. - Bernard Schott, Dec 19 2012
This is also the list of bases where 11111 is a prime number. - Christian N. K. Anderson, Mar 28 2013
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
MAPLE
A049409:=n->if isprime(n^4+n^3+n^2+n+1) then n fi; seq(A049409(n), n=1..300); # Wesley Ivan Hurt, Dec 28 2013
MATHEMATICA
lst={}; Do[p=n^0+n^1+n^2+n^3+n^4; If[PrimeQ[p], AppendTo[lst, n]], {n, 300}]; lst (* Vladimir Joseph Stephan Orlovsky, Jun 10 2009 *)
Select[Range[300], PrimeQ[1+Total[#^Range[4]]]&] (* Harvey P. Dale, Mar 12 2018 *)
PROG
(PARI) for(n=1, 1000, ispseudoprime(n^4+n^3+n^2+n+1) & print1(n", ")) \\ M. F. Hasler, Feb 09 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved