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
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).
Programs
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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)
Comments