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A154925 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).

%I A154925 #15 Oct 09 2023 04:46:45
%S A154925 1,1,1,1,-2,-5,-6,3,9,-5,-13,-14,5,30,510,-10,-21,-22,7,59,5163,
%T A154925 53307975,-14,-29,-30
%N A154925 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 A154925 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 A154925 Wikipedia, <a href="https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula">Bailey-Borwein-Plouffe formula</a>.
%e A154925 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 A154925 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 A154925 Cf. A154429.
%K A154925 sign,uned
%O A154925 0,5
%A A154925 _Alexander R. Povolotsky_, Jan 17 2009, Jan 18 2009