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A248228 Numbers k such that A248227(k+1) = A248227(k).

%I A248228 #6 Oct 08 2014 16:47:02
%S A248228 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 A248228 79,83,86,89,92,96,99,102,105,109,112,115,119,122,125,128,132,135,138,
%U A248228 141,145,148,151,154,158,161,164,167,171,174,177,180,184,187
%N A248228 Numbers k such that A248227(k+1) = A248227(k).
%C A248228 Since A248227(k+1) - A248227(k) is in {0,1} for k >= 1, A248228 and A248229 are complementary.
%H A248228 Clark Kimberling, <a href="/A248228/b248228.txt">Table of n, a(n) for n = 1..300</a>
%e A248228 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 A248228 $MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];
%t A248228 N[Table[Zeta[4] - p[n], {n, 1, z/10}]]
%t A248228 f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];
%t A248228 u = Flatten[Table[f[n], {n, 1, z}]]   (* A248227 *)
%t A248228 Flatten[Position[Differences[u], 0]]  (* A248228 *)
%t A248228 Flatten[Position[Differences[u], 1]]  (* A248229 *)
%t A248228 f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}]  (* A248230 *)
%Y A248228 Cf. A248227, A248229, A248230.
%K A248228 nonn,easy
%O A248228 1,2
%A A248228 _Clark Kimberling_, Oct 05 2014