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%I #4 Oct 15 2014 21:02:13
%S 2,4,6,7,9,10,12,13,15,16,17,19,20,21,23,24,25,27,28,29,31,32,33,35,
%T 36,37,39,40,41,43,44,45,46,48,49,50,52,53,54,56,57,58,59,61,62,63,65,
%U 66,67,68,70,71,72,74,75,76,77,79,80,81,83,84,85,86,88,89
%N Numbers k such that A248610(k+1) = A248610(k) + 1.
%H Clark Kimberling, <a href="/A248612/b248612.txt">Table of n, a(n) for n = 1..1500</a>
%e (A248610(k+1) - A248610(k)) = (0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, ...), so that A248611 = (1, 3, 5, 8, 11, 14, 18, 22, 26, ..) and A248612 = (2, 4, 6, 7, 9, 10, 12, 13, ...).
%t z = 300; p[k_] := p[k] = Sum[1/((h^2)*Binomial[2 h, h]), {h, 1, k}]
%t d = N[Table[Pi^2/18 - p[k], {k, 1, z/5}], 12]
%t f[n_] := f[n] = Select[Range[z], Pi^2/18 - p[#] < 1/3^n &, 1]
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248610 *)
%t d = Differences[u]
%t v = Flatten[Position[d, 0]] (* A248611 *)
%t w = Flatten[Position[d, 1]] (* A248612 *)
%Y Cf. A248610, A248611, A248608.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Oct 10 2014
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