A248233 - OEIS

%I #7 Oct 11 2014 14:16:06

%S 2,3,5,6,7,9,10,12,13,14,16,17,19,20,21,23,24,26,27,29,30,31,33,34,36,

%T 37,38,40,41,43,44,45,47,48,50,51,53,54,55,57,58,60,61,62,64,65,67,68,

%U 70,71,72,74,75,77,78,79,81,82,84,85,86,88,89,91,92,94

%N Numbers k such that A248231(k+1) = A248231(k) + 1.

%C Since A248231(k+1) - A248232(k) is in {0,1} for k >= 1, A248232 and A248233 are complementary.

%C This appears to be a duplicate of A097432. - _R. J. Mathar_, Oct 10 2014

%H Clark Kimberling, <a href="/A248233/b248233.txt">Table of n, a(n) for n = 1..1000</a>

%e The difference sequence of A248231 is (0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, ...), so that A248232 = (1, 4, 8, 11, 15, 18, 22, 25, 28,...) and A248233 = (2, 3, 5, 6, 7, 9, 10, 12, 13, 14, 16, 17,...), the complement of A248232.

%t z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]

%t f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]

%t u = Flatten[Table[f[n], {n, 1, z}]]   (* A248231 *)

%t Flatten[Position[Differences[u], 0]]  (* A248232 *)

%t Flatten[Position[Differences[u], 1]]  (* A248233 *)

%t Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}]  (* A248234 *)

%Y Cf. A248231, A248232, A248234, A248227.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 05 2014