Simon Plouffe was one of the discoverers of the nice pi formula
inf 4 2 1 1 -k pi = sum (---- - ---- - ---- - ----) 16 k=0 8k+1 8k+4 8k+5 8k+6
which allows computing individual bits of pi without computing the preceding part.
It seems to me that there should be another such formula, in which the mod 8 residues are rearranged a little. Perhaps with 8k+3 and 8k+7 replacing 1 & 5, or maybe just 1&3 instead of 1&5, or with 8k+2 in place of 8k+6.
Has anyone seen variations like this?
You don't need lattice numerology for these. Here are the closed forms-- rub them together symbolically: inf ==== \ 1 sqrt(2)
------------- = (sqrt(2) log(sqrt(2) + 1) + 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 1) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) + log(5) + 2 atan(2))/8 inf ==== \ 1 log(3) + 2 acot(2)
------------- = ------------------ / k 4 ==== (8 k + 2) 16 k = 0
inf ==== \ 1 sqrt(2)
------------- = (sqrt(2) log(sqrt(2) + 1) - 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 3) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) - log(5) + 2 atan(2))/4 inf ==== \ 1
------------- = acoth(4) / k ==== (8 k + 4) 16 k = 0 inf ==== \ 1 sqrt(2) ------------- = (sqrt(2) log(sqrt(2) + 1) + 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 5) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) - log(5) - 2 atan(2))/2 inf ==== \ 1
------------- = log(3) - 2 acot(2) / k ==== (8 k + 6) 16 k = 0
inf ==== \ 1 sqrt(2)
------------- = sqrt(2) log(sqrt(2) + 1) - 2 sqrt(2) atan(-------) / k 2 ==== (8 k + 7) 16 k = 0
- sqrt(2) log(sqrt(2) - 1) + log(5) - 2 atan(2)
The sums of 1/(2n+1)^k, for even k, give rational multiples of pi^k, and the alternating sums, for odd k, also give rational multiples of pi^k. So 1 - 1/27 + 1/125 - ... = pi^3 / 32. Perhaps there's some relationship between the leftovers: gamma, Catalan, zeta(odd), etc?
Rich
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