A278924 A sequence showing numerators in ratios tending to the constant Pi/4 = 0.785398163397448... .

4, 296, 36772, 1288688, 96641548, 26576092808, 8637277012172, 1079658805128928, 91770997994914276, 43591225139846360008

Offset: 1

Author

Sanjar Abrarov, Dec 01 2016

Keywords

• nonn
• frac

Comments

The ratios c(n)/d(n) rapidly tend to the constant Pi/4 = 0.785398163397448... with increasing index n: abs(Pi/4 - c(1)/d(1)) > abs(Pi/4 - c(2)/d(2)) > abs(Pi/4 - c(3)/d(3)) > abs(Pi/4 - c(4)/d(4)) ..., where c(n) = A278924(n) and d(n) = A278364(n) are even and odd positive integers, respectively.

Examples

			------------------------------------------------
n    c(n)                   d(n)
------------------------------------------------
1    4                      5
2    296                    375
3    36772                  46875
4    1288688                1640625
5    96641548               123046875
6    26576092808            33837890625
7    8637277012172          10997314453125
8    1079658805128928       1374664306640625
9    91770997994914276      116846466064453125
10   43591225139846360008   55502071380615234375
------------------------------------------------
At n = 6 the ratio c(6)/d(6) = 26576092808/33837890625 is close to Pi/4. However, at n = 10 the ratio c(10)/d(10) = 43591225139846360008/55502071380615234375 becomes more closer to Pi/4.
		

Links

• Sanjar Abrarov, Table of n, a(n) for n = 1..49
• S. M. Abrarov and B. M. Quine, A generalized Viéte's-like formula for pi with rapid convergence, arXiv:1610.07713 [math.GM], (2016).

Crossrefs

Cf. A278364, A003881, A096954, A096955.

Programs

Mathematica

x := 1;(* argument x *)
M := 1;(* initial value for the integer M *)
n := 1; (* index *)
(* Note that arctan(1) = Pi/4 *)
atan := I*Sum[(1/(2*m - 1))*(1/(1 + 2*(I/x))^(2*m - 1) - 1/(1 - 2*(I/x))^(2*m - 1)),{m, 1, Floor[M/2] + 1}];
sqn := {};(* initiate the sequence *)
AppendTo[sqn,{"Index n", "Numerators", "Denominators"}];
While[M <= 20, AppendTo[sqn,{n, Numerator[atan], Denominator[atan]}];
{M = M + 2, n++}];
Print[MatrixForm[sqn]]

Formula

arctan(1) = I*lim_{M -> inf}Sum_{m = 1..floor(M/2) + 1}(1/(2*m - 1))*(1/(1 + 2*I)^(2*m - 1) - 1/(1 - 2*I)^(2*m - 1))