A343007 Relative position of the average value between two consecutive partial sums of the Leibniz formula for Pi.

6, 13, 26, 41, 62, 85, 114, 145, 182, 221, 266, 313, 366, 421, 482, 545, 614, 685, 762, 841, 926, 1013, 1106, 1201, 1302, 1405, 1514, 1625, 1742, 1861, 1986, 2113, 2246, 2381, 2522, 2665, 2814, 2965, 3122, 3281, 3446, 3613, 3786, 3961, 4142, 4325, 4514, 4705

Offset: 1

Author

Raphael Ranna, Apr 02 2021

Keywords

• nonn
• easy

Comments

Define L(n) to be the n-th partial sum of the Leibniz formula Pi = 4 - 4/3 + 4/5 - 4/7 + ..., i.e., L(n) = Sum_{j=1..n} 4*(-1)^(j+1)/(2*j-1). For every positive integer n, L(n+1) is closer to Pi than L(n) is. If we let V be the average of the two consecutive partial sums L(n) and L(n+1), then the partial sums that lie closest to V are L(a(n)-1) and L(a(n)+1) (one of which is above V, the other below).

Examples

			The first several partial sums are as follows:
  n      L(n)
  -  ------------
  1  4.0000000000
  2  2.6666666...
  3  3.4666666...
  4  2.8952380...
  5  3.3396825...
  6  2.9760461...
  7  3.2837384...
  8  3.0170718...
.
For n=1, the average of the partial sums L(1) and L(2) is V = (L(1) + L(2))/2 = (4 + 2.6666666...)/2 = 3.3333333...; the two partial sums closest to V are L(5)=3.3396825... and L(7)=3.2837384..., and V lies in the interval between them, so a(1)=6.
The formula as it is written works for all data in the sequence, but it needs to be proven that it works for all possible integer values of n.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Cf. A047550, A102083, A254527.

Programs

Mathematica

Rest@ CoefficientList[Series[x (6 + x + x^3)/((1 + x) (1 - x)^3), {x, 0, 48}], x] (* Michael De Vlieger, Apr 05 2021 *)

Formula

a(1) = 6; a(n) = a(n-1) + r(n), where r(n) = A047550(n) = 4*n - (-1)^n.

G.f.: x*(6 + x + x^3)/((1 + x)*(1 - x)^3). - Jinyuan Wang, Apr 03 2021

From Stefano Spezia, Apr 03 2021: (Start)

a(n) = (3 + (-1)^(n+1) + 4*n + 4*n^2)/2.

a(2*n) = A102083(n).

a(2*n-1) = A254527(n). (End)