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A343007
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Relative position of the average value between two consecutive partial sums of the Leibniz formula for Pi.
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0
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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
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1,1
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COMMENTS
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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).
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LINKS
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FORMULA
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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
a(n) = (3 + (-1)^(n+1) + 4*n + 4*n^2)/2.
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EXAMPLE
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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...
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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.
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MATHEMATICA
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Rest@ CoefficientList[Series[x (6 + x + x^3)/((1 + x) (1 - x)^3), {x, 0, 48}], x] (* Michael De Vlieger, Apr 05 2021 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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