A248360 -id:A248360 - cp's OEIS frontend

A248630 Numbers k such that A248630(k+1) = A248360(k) + 1.

7, 11, 15, 18, 22, 25, 28, 31, 34, 37, 39, 42, 45, 48, 51, 53, 56, 59, 61, 64, 67, 69, 72, 75, 77, 80, 82, 85, 88, 90, 93, 95, 98, 101, 103, 106, 108, 111, 113, 116, 118, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 154, 156, 159, 161

Offset: 1

Clark Kimberling, Oct 10 2014

			(A248630(k+1) = A248360(k)) = (2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, ...), so that A248630 = (7, 11, 15, 18, ...).

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

Cf. A248630, A248629, A248632.

z = 300; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}]
d = N[Table[6 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248629 *)
d = Differences[u]
v = Flatten[Position[d, 1]]  (* A248630 *)

A248347 a(n) = floor(1/(Pi - 2^(n+1)*sin(Pi/2^(n+1)))).

Original entry on oeis.org

3, 12, 49, 198, 792, 3170, 12681, 50727, 202909, 811636, 3246545, 12986183, 51944732, 207778928, 831115713, 3324462855, 13297851421, 53191405684, 212765622737, 851062490950, 3404249963800, 13616999855201, 54467999420806, 217871997683226, 871487990732903

Offset: 1

Clark Kimberling, Oct 05 2014

Let Arch(n) = 2^(n+1)*sin(Pi/2^(n+1)) be the Archimedean approximation to Pi (Finch, pp. 17 and 23) given by a regular polygon of 2^(n+1) sides. A248347 provides insight into the manner of convergence of Arch(n) to Pi. Another provider is the fact that the least k for which Arch(k) < 1/4^n is A000027(n) = n. (For the closely related function arch, see A248355.)

			n    Pi - Arch(n)      1/(Pi - Arch(n))
1    0.313166...         3.1932...
2    0.0801252...       12.4805...
3    0.0201475...       49.6339...
4    0.00504416...     198.249...
5    0.0012615...      792.709...

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

Cf. A000027, A248355, A248357, A248355, A248360.

z = 200; p[k_] := p[k] = (2^(k + 1))*Sin[Pi/2^(k + 1)]
Table[Floor[1/(Pi - p[n])], {n, 1, z}]  (* A248347  *)

a(n) ~ 6 * 4^(n+1) / Pi^3. - Vaclav Kotesovec, Oct 09 2014

A248359 Least number k such that cos(Pi/k) + 1/(k*n) > 1.

Original entry on oeis.org

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 79, 84, 89, 94, 99, 104, 109, 114, 119, 124, 129, 134, 139, 144, 149, 153, 158, 163, 168, 173, 178, 183, 188, 193, 198, 203, 208, 213, 218, 223, 227, 232, 237, 242, 247, 252, 257, 262, 267, 272, 277

Offset: 1

Clark Kimberling, Oct 07 2014

It appears that a(n+1) - a(n) is in {4,5} for n >= 1.

Lim_{n->infinity} a(n)/n = Pi^2/2 = 4.9348022..., but lim_{n->infinity} (a(n+1) - a(n)) does not exist; Pi^2/2 is only a mean value of these differences. - Vaclav Kotesovec, Oct 09 2014

			Taking n = 2, we have cos(Pi/9) + 1/(18) = 0.99524... < 1 < 1.0010565... = cos(Pi/10) + 1/(20), so that a(2) = 10, as corroborated for n = 2 in the following list of approximations:
n ... cos(Pi/a(n)) + 1/(n*a(n))
1 ... 1.009016994
2 ... 1.001056516
3 ... 1.000369823
4 ... 1.000188341
5 ... 1.000114701
6 ... 1.000077451

Clark Kimberling and Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (first 500 terms from Clark Kimberling)

Cf. A102753, A248360.

z = 800; f[n_] := f[n] = Select[Range[z], Cos[Pi/#] + 1/(#*n) > 1 &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248359 *)
Table[Floor[1/(1 - Cos[Pi/n])], {n, 1, z/10}]  (* A248360 *)
Table[k=1; While[Cos[Pi/k]+1/(k*n)<=1,k++]; k,{n,1,100}] (* Vaclav Kotesovec, Oct 09 2014 *)

a(n) ~ n*Pi^2/2 = n*A102753. - Vaclav Kotesovec, Oct 09 2014

cp's OEIS Frontend

A248630 Numbers k such that A248630(k+1) = A248360(k) + 1.

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A248347 a(n) = floor(1/(Pi - 2^(n+1)*sin(Pi/2^(n+1)))).

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A248359 Least number k such that cos(Pi/k) + 1/(k*n) > 1.

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