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Position of n^e in the joint ranking of {h^e} and {k^Pi}, h > 1, k > 1.
2

%I #5 Sep 12 2016 17:10:29

%S 1,3,5,7,8,10,12,13,15,16,18,20,21,23,25,26,28,29,31,32,34,36,37,39,

%T 40,42,43,45,46,48,50,51,53,54,56,57,59,60,62,63,65,66,68,69,71,72,74,

%U 76,77,79,80,82,83,85,86,88,89,91,92,94,95,97,98,100,101

%N Position of n^e in the joint ranking of {h^e} and {k^Pi}, h > 1, k > 1.

%H Clark Kimberling, <a href="/A276224/b276224.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = n + floor(n^(e/Pi)); the complement is given by n + floor(n^(Pi/e)).

%e The first numbers in the joint ranking are

%e 2^e < 2^Pi < 3^e < 3^Pi < 4^e < 4^Pi , so that a(n) = (1,3,5,...).

%t z = 150; r = N[E, 100]; s = N[Pi, 100];

%t u = Table[n + Floor[n^(s/r)], {n, 2, z}];

%t v = Table[n + Floor[n^(r/s)], {n, 2, z^(s/r)}];

%t w = Union[u, v];

%t Flatten[Table[Position[w, u[[n]]], {n, 1, z}]] (* A276223 *)

%t Flatten[Table[Position[w, v[[n]]], {n, 1, z}]] (* A276224 *)

%Y Cf. A276223 (complement).

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 06 2016