A248611 -id:A248611 - cp's OEIS frontend

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

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 *)

A248610 Least k such that (Pi^2)/18 - sum{1/(h^2*C(2h,h)), h = 1..k} < 1/3^n.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 30, 30, 31, 32, 33, 34, 34, 35, 36, 37, 37, 38, 39, 40, 40, 41, 42, 43, 44, 44, 45, 46, 47, 47, 48, 49, 50

Offset: 1

Clark Kimberling, Oct 10 2014

This sequence provides insight into the manner of convergence of sum{1/(h^2*C(2h,h)), h = 1..k} to (Pi^2)/18. Since a(n+1) - a(n) is in {0,1} for n >= 1, the sequences A248611 and A248612 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 ... 0.0483114 ... 0.333333
2 ... 0.0066446 ... 0.111111
3 ... 0.0010891 ... 0.037037
4 ... 0.0001962 ... 0.012345
5 ... 0.00003754 .. 0.004115
a(5) = 3 because s(3) < 1/3^5 < s(2).

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

Clark Kimberling, Table of n, a(n) for n = 1..2000

Cf. A248611, A248612, A248607.

z = 300; p[k_] := p[k] = Sum[1/((h^2)*Binomial[2 h, h]), {h, 1, k}]
d = N[Table[Pi^2/18 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], Pi^2/18 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248610 *)
d = Differences[u]
v = Flatten[Position[d, 0]]  (* A248611 *)
w = Flatten[Position[d, 1]]  (* A248612 *)

A248609 Numbers k such that A248607(k+1) = A248607(k) + 2.

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 14, 16, 18, 19, 21, 23, 25, 27, 28, 30, 32, 34, 36, 37, 39, 41, 43, 44, 46, 48, 50, 51, 53, 55, 57, 58, 60, 62, 64, 65, 67, 69, 71, 72, 74, 76, 78, 79, 81, 83, 84, 86, 88, 90, 91, 93, 95, 97, 98, 100, 102, 104, 105, 107, 109, 110, 112

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 A248567 = (2, 4, 6, 8, 10, 12, 14, 16, ...).

Clark Kimberling, Table of n, a(n) for n = 1..500

Cf. A248607, A248608, 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 *)

A248612 Numbers k such that A248610(k+1) = A248610(k) + 1.

Original entry on oeis.org

2, 4, 6, 7, 9, 10, 12, 13, 15, 16, 17, 19, 20, 21, 23, 24, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 45, 46, 48, 49, 50, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 65, 66, 67, 68, 70, 71, 72, 74, 75, 76, 77, 79, 80, 81, 83, 84, 85, 86, 88, 89

Offset: 1

Clark Kimberling, Oct 10 2014

			(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, ...).

Clark Kimberling, Table of n, a(n) for n = 1..1500

Cf. A248610, A248611, A248608.

z = 300; p[k_] := p[k] = Sum[1/((h^2)*Binomial[2 h, h]), {h, 1, k}]
d = N[Table[Pi^2/18 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], Pi^2/18 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248610 *)
d = Differences[u]
v = Flatten[Position[d, 0]]  (* A248611 *)
w = Flatten[Position[d, 1]]  (* A248612 *)

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

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A248610 Least k such that (Pi^2)/18 - sum{1/(h^2*C(2h,h)), h = 1..k} < 1/3^n.

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A248609 Numbers k such that A248607(k+1) = A248607(k) + 2.

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A248612 Numbers k such that A248610(k+1) = A248610(k) + 1.

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