cp's OEIS Frontend

A248229 Numbers k such that A248227(k+1) = A248227(k) + 1.

%I A248229 #4 Oct 08 2014 16:47:12
%S A248229 2,3,5,6,7,9,10,12,13,15,16,18,19,20,22,23,25,26,28,29,31,32,33,35,36,
%T A248229 38,39,41,42,43,45,46,48,49,51,52,54,55,56,58,59,61,62,64,65,67,68,69,
%U A248229 71,72,74,75,77,78,80,81,82,84,85,87,88,90,91,93,94,95
%N A248229 Numbers k such that A248227(k+1) = A248227(k) + 1.
%C A248229 Since A248227(k+1) - A248227(k) is in {0,1} for k >= 1, A248228 and A248229 are complementary.
%H A248229 Clark Kimberling, <a href="/A248229/b248229.txt">Table of n, a(n) for n = 1..500</a>
%e A248229 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 A248229 $MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];
%t A248229 N[Table[Zeta[4] - p[n], {n, 1, z/10}]]
%t A248229 f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];
%t A248229 u = Flatten[Table[f[n], {n, 1, z}]]   (* A248227 *)
%t A248229 Flatten[Position[Differences[u], 0]]  (* A248228 *)
%t A248229 Flatten[Position[Differences[u], 1]]  (* A248229 *)
%t A248229 f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}]  (* A248230 *)
%Y A248229 Cf. A248227, A248228, A248230.
%K A248229 nonn,easy
%O A248229 1,1
%A A248229 _Clark Kimberling_, Oct 05 2014