1,1
For n = 2*k + 1, n^4 + 4^n = (n^2 + n*2^(k + 1) + 2^n) * (n^2 - n*2^(k + 1) + 2^n) The sequence gives those values of n for which both parentheses are primes. No further terms were found for k<=5000.
Table of n, a(n) for n=1..5.
Ignacio Larrosa Canestro et al., Find all primes of the form 4^n + n^4. Discussion in newsgroup sci.math.
a(1)=3 because 3^4+4^3=145=5*29, a(2)=5 because 5^4+4^5=1649=17*97.
Cf. A001589.
Sequence in context: A148500 A148501 A148502 * A146212 A018516 A138017
Adjacent sequences: A089482 A089483 A089484 * A089486 A089487 A089488
more,nonn
Hugo Pfoertner, Nov 11 2003
approved
