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The terms of this sequence are integer values of consecutive denominators (with signs) from the fractional expansion (using only fractions with numerators to be positive 1's) of the BBP polynomial ( 4/(8*k+1) - 2/(8*k+4) - 1/(8*k+5) - 1/(8*k+6) ) for all k (starting from 0 to infinity).
3

%I #15 Oct 09 2023 04:46:45

%S 1,1,1,1,-2,-5,-6,3,9,-5,-13,-14,5,30,510,-10,-21,-22,7,59,5163,

%T 53307975,-14,-29,-30

%N The terms of this sequence are integer values of consecutive denominators (with signs) from the fractional expansion (using only fractions with numerators to be positive 1's) of the BBP polynomial ( 4/(8*k+1) - 2/(8*k+4) - 1/(8*k+5) - 1/(8*k+6) ) for all k (starting from 0 to infinity).

%C The Egyptian fraction expansion is applied to the first fraction (that is, 4/(8*k+1) ) of the BBP polynomial ( 4/(8*k+1) - 2/(8*k+4) - 1/(8*k+5) - 1/(8*k+6) ) for k >= 1. R. Knott's converter calculator #1 (http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#calc1) is used for such conversion. Note that in the case of k=0, 4/(8*k+1) = 4 and could be trivially expressed as 1/1 + 1/1 + 1/1 + 1/1. It remains to be seen how the above described Pi presentation relates to Engel's presentation of Pi, which also consists of an infinite sum of fractions whose numerators are all 1's.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula">Bailey-Borwein-Plouffe formula</a>.

%e For k=1, 4/(8*k+1) = 4/9 = 1/3 + 1/9, thus the first (smallest) denominator is 3 so a(7)=3.

%e For k=1, 4/(8*k+1) = 4/9 = 1/3 + 1/9 and the second (next to smallest) denominator is 9 so a(8)=9.

%Y Cf. A154429.

%K sign,uned

%O 0,5

%A _Alexander R. Povolotsky_, Jan 17 2009, Jan 18 2009