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%I #6 Oct 08 2014 16:47:02
%S 1,4,8,11,14,17,21,24,27,30,34,37,40,44,47,50,53,57,60,63,66,70,73,76,
%T 79,83,86,89,92,96,99,102,105,109,112,115,119,122,125,128,132,135,138,
%U 141,145,148,151,154,158,161,164,167,171,174,177,180,184,187
%N Numbers k such that A248227(k+1) = A248227(k).
%C Since A248227(k+1) - A248227(k) is in {0,1} for k >= 1, A248228 and A248229 are complementary.
%H Clark Kimberling, <a href="/A248228/b248228.txt">Table of n, a(n) for n = 1..300</a>
%e The difference sequence of A248227 is (0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, ...), so that A248228 = (1, 4, 8, 11, 14, 17, 2,...) and A248229 = (2, 3, 5, 6, 7, 9, 10, 12, 13, 15, 16, 18,...), the complement of A248228.
%t $MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];
%t N[Table[Zeta[4] - p[n], {n, 1, z/10}]]
%t f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248227 *)
%t Flatten[Position[Differences[u], 0]] (* A248228 *)
%t Flatten[Position[Differences[u], 1]] (* A248229 *)
%t f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}] (* A248230 *)
%Y Cf. A248227, A248229, A248230.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 05 2014
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