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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). 3
 1, 1, 1, 1, -2, -5, -6, 3, 9, -5, -13, -14, 5, 30, 510, -10, -21, -22, 7, 59, 5163, 53307975, -14, -29, -30 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS 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) comes to 4 and could be trivially expressed as 1/1 + 1/1 + 1/1 + 1/1 . It remains to be seen how above described Pi presentation relates to the Engel's presentation of Pi, which also consists of the infinite sum of fractions, whose numerators are all 1's. LINKS EXAMPLE For n = 7 the a(7) = 3 because for the k=1 the 4/(8*k+1) comes to 4/9=1/3+1/9, thus the first (smallest) denominator is 3 so a(7)=3 For n = 8 the a(8) = 9 because for the k=1 the 4/(8*k+1) comes to 4/9=1/3+1/9 and the second (next to smallest) denominator is 9 so a(8)=9 . CROSSREFS Cf. A154429 Sequence in context: A340858 A309364 A062825 * A154962 A091655 A021979 Adjacent sequences:  A154922 A154923 A154924 * A154926 A154927 A154928 KEYWORD sign AUTHOR Alexander R. Povolotsky, Jan 17 2009, Jan 18 2009 STATUS approved

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Last modified June 19 20:12 EDT 2021. Contains 345144 sequences. (Running on oeis4.)