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a(n) = floor((1287/545)^n).
5

%I #17 Feb 20 2020 01:29:17

%S 2,5,13,31,73,173,409,967,2283,5392,12735,30073,71017,167706,396032,

%T 935217,2208486,5215270,12315692,29083113,68678837,162182870,

%U 382989640,904417737,2135753445,5043513182,11910094433,28125305569,66417005997

%N a(n) = floor((1287/545)^n).

%C The first eight terms are primes. Does there exist a number theta such that the floor of theta^n is always prime?

%D Richard Crandall and Carl Pomerance, Prime Numbers - a Computational Perspective, Springer, 2001, page 69, exercise 1.75.

%H Harry J. Smith, <a href="/A063636/b063636.txt">Table of n, a(n) for n = 1..300</a>

%e (1287/545)^3 = 13.16879..., so a(3)=13.

%o (PARI) { for (n=1, 300, write("b063636.txt", n, " ", 1287^n \ 545^n); ) } \\ _Harry J. Smith_, Aug 26 2009

%Y Cf. A000040, A051254, A060449, A060699.

%K nonn

%O 1,1

%A _Jud McCranie_, Aug 10 2001