

A343007


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


0



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
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OFFSET

1,1


COMMENTS

Define L(n) to be the nth 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*j1). 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).


LINKS



FORMULA

a(1) = 6; a(n) = a(n1) + 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
a(n) = (3 + (1)^(n+1) + 4*n + 4*n^2)/2.


EXAMPLE

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.


MATHEMATICA

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


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



