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 A248232 Numbers k such that A248231(k+1) = A248231(k). 4

%I #4 Oct 08 2014 16:49:11

%S 1,4,8,11,15,18,22,25,28,32,35,39,42,46,49,52,56,59,63,66,69,73,76,80,

%T 83,87,90,93,97,100,104,107,110,114,117,121,124,128,131,134,138,141,

%U 145,148,151,155,158,162,165,168,172,175,179,182,186,189,192,196

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

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

%H Clark Kimberling, <a href="/A248232/b248232.txt">Table of n, a(n) for n = 1..500</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, A248233, A248234, A248227.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Oct 05 2014

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Last modified September 15 03:41 EDT 2024. Contains 375931 sequences. (Running on oeis4.)