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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

Table of n, a(n) for n=0..24.

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: A165501 A274614 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 May 25 15:22 EDT 2019. Contains 323572 sequences. (Running on oeis4.)