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%I #4 Oct 15 2014 21:01:23
%S 1,3,5,7,9,11,13,15,17,20,22,24,26,29,31,33,35,38,40,42,45,47,49,52,
%T 54,56,59,61,63,66,68,70,73,75,77,80,82,85,87,89,92,94,96,99,101,103,
%U 106,108,111,113,115,118,120,122,125,127,130,132,134,137,139
%N Numbers k such that A248607(k+1) = A248607(k) + 1.
%H Clark Kimberling, <a href="/A248608/b248608.txt">Table of n, a(n) for n = 1..400</a>
%e (A248607(k+1) - A248607(k)) = (1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2,...), so that A248608 = (1, 3, 5, 7, 9, 11, 13, 15, 17, ..) and A248609 = (2, 4, 6, 8, 10, 12, 14, 16, ...).
%t z = 300; p[k_] := p[k] = Sum[2^h/((2 h + 1) Binomial[2 h, h]), {h, 0, k}]
%t d = N[Table[Pi/2 - p[k], {k, 1, z/5}], 12]
%t f[n_] := f[n] = Select[Range[z], Pi/2 - p[#] < 1/3^n &, 1]
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248607 *)
%t d = Differences[u]
%t v = Flatten[Position[d, 1]] (* A248608 *)
%t w = Flatten[Position[d, 2]] (* A248609 *)
%Y Cf. A248607, A248609, A248611.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 10 2014
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