A248609 -id:A248609 - cp's OEIS frontend

A248607 Least k such that Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..k} < 1/3^n.

1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 37, 39, 40, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 58, 59, 61, 62, 64, 65, 67, 69, 70, 72, 73, 75, 76, 78, 80, 81, 83, 84, 86, 87, 89, 91, 92, 94, 95, 97, 98, 100, 102

Offset: 1

Clark Kimberling, Oct 10 2014

This sequence provides insight into the manner of convergence of sum{2^h/((2h+1)*C(2h,h)), h = 1..k} to Pi/2. Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248608 and A248609 partition the positive integers.

			Let s(n) = Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..n}.  Approximations follow:
n ... s(n) ...... 1/3^n
1 ... 0.23746 ... 0.333333
2 ... 0.10413 ... 0.111111
3 ... 0.04698 ... 0.037037
4 ... 0.02159 ... 0.012345
5 ... 0.01004 ... 0.004115
6 ... 0.00471 ... 0.001371
7 ... 0.00223 ... 0.000472
a(5) = 7 because s(7) < 1/3^5 < s(6).

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 20.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A248608, A248609, A248610.

z = 300; p[k_] := p[k] = Sum[2^h/((2 h + 1) Binomial[2 h, h]), {h, 0, k}]
d = N[Table[Pi/2 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], Pi/2 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248607 *)
d = Differences[u]
v = Flatten[Position[d, 1]] (* A248608 *)
w = Flatten[Position[d, 2]] (* A248609 *)

A248608 Numbers k such that A248607(k+1) = A248607(k) + 1.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 20, 22, 24, 26, 29, 31, 33, 35, 38, 40, 42, 45, 47, 49, 52, 54, 56, 59, 61, 63, 66, 68, 70, 73, 75, 77, 80, 82, 85, 87, 89, 92, 94, 96, 99, 101, 103, 106, 108, 111, 113, 115, 118, 120, 122, 125, 127, 130, 132, 134, 137, 139

Offset: 1

Clark Kimberling, Oct 10 2014

			(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, ...).

Clark Kimberling, Table of n, a(n) for n = 1..400

Cf. A248607, A248609, A248611.

z = 300; p[k_] := p[k] = Sum[2^h/((2 h + 1) Binomial[2 h, h]), {h, 0, k}]
d = N[Table[Pi/2 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], Pi/2 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248607 *)
d = Differences[u]
v = Flatten[Position[d, 1]] (* A248608 *)
w = Flatten[Position[d, 2]] (* A248609 *)

cp's OEIS Frontend

A248607 Least k such that Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..k} < 1/3^n.

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