This site is supported by donations to The OEIS Foundation.

# User:Jaume Oliver Lafont/BBP

$log(2)={\frac {1}{64}}\sum _{k=0}^{\infty }\left({\frac {64}{6k+1}}-{\frac {32}{6k+2}}-{\frac {8}{6k+3}}-{\frac {8}{6k+4}}+{\frac {4}{6k+5}}+{\frac {1}{6k+6}}\right){\frac {1}{64^{k}}}=\sum _{n=1}^{\infty }{\frac {1}{cos({\frac {n\pi }{3}})n2^{n}}}$ (A176900, A002162)

$log(3)={\frac {1}{32}}\sum _{k=0}^{\infty }\left({\frac {16}{6k+1}}+{\frac {24}{6k+2}}+{\frac {16}{6k+3}}+{\frac {6}{6k+4}}+{\frac {1}{6k+5}}\right){\frac {1}{64^{k}}}=\sum _{n=1}^{\infty }{\frac {1-cos({\frac {n\pi }{3}})}{n2^{n-1}}}$ ## Two series related to A014480

${\sqrt {2}}\log {(1+{\sqrt {2}})}=\sum _{k=0}^{\infty }{\frac {1}{(2k+1)2^{k}}}$ ${\sqrt {2}}\arctan {\frac {1}{\sqrt {2}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)2^{k}}}={\frac {1}{2}}\sum _{k=0}^{\infty }\left({\frac {2}{4k+1}}-{\frac {1}{4k+3}}\right){\frac {1}{4^{k}}}$ ## A ternary zero relation

The sum of zero relations (75) and (76) (103) and (104) in Bailey can be written in six terms.

$0=\sum _{k=0}^{\infty }{\frac {1}{(-27)^{k}}}\left({{\frac {9}{6k+1}}-{\frac {9}{6k+2}}-{\frac {12}{6k+3}}-{\frac {3}{6k+4}}+{\frac {1}{6k+5}}}\right)$ (check)

This formula is similar to equation (17) in Broadhurst, 1998

$S_{2}={\frac {4}{81}}\sum _{k=0}^{\infty }\left(-{\frac {1}{27}}\right)^{k}\left({\frac {9}{(6k+1)^{2}}}-{\frac {9}{(6k+2)^{2}}}-{\frac {12}{(6k+3)^{2}}}-{\frac {3}{(6k+4)^{2}}}+{\frac {1}{(6k+5)^{2}}}\right)$ ## Obtention of a zero relation of the form 0=P(1,2^20,40,A)

From Bellard's formula 

${\frac {\pi }{4}}={\frac {1}{256}}\sum _{k=0}^{\infty }{\frac {1}{(-1024)^{k}}}\left({\frac {512}{20k+2}}-{\frac {160}{20k+5}}-{\frac {128}{20k+6}}-{\frac {8}{20k+10}}-{\frac {8}{20k+14}}-{\frac {5}{20k+15}}+{\frac {2}{20k+18}}\right)$ In P notation,

${\frac {\pi }{4}}={\frac {1}{2^{8}}}P(1,-2^{10},20,(0,512,0,0,-160,-128,0,0,0,-8,0,0,0,-8,-5,0,0,2,0,0))$ Equivalently,

${\frac {\pi }{4}}={\frac {1}{2^{18}}}P(1,2^{20},40,(0,2^{19},0,0,-5*2^{15},-2^{17},0,0,0,-2^{13},0,0,0,-2^{13},-5*2^{10},0,0,2^{11},0,0,0,-2^{9},0,0,5*2^{5},2^{7},0,0,0,2^{3},0,0,0,2^{3},5,0,0,-2,0,0))$ If Ferguson's formula is written with length 40 and both expressions for ${\frac {\pi }{4}}$ subtracted, the following zero relation is obtained.

$0=P(1,2^{20},40,(2^{19},-3*2^{19},2^{18},0,2^{19},3*2^{17},-2^{16},0,2^{15},2^{16},2^{14},0,-2^{13},3*2^{13},2^{14},0,2^{11},-3*2^{11},2^{10},0,-2^{9},3*2^{9},-2^{8},0,-2^{9},-3*2^{7},2^{6},0,-2^{5},-2^{6},-2^{4},0,2^{3},-3*2^{3},-2^{4},0,-2,6,-1,0))\,$ $a[4k]=0\,$ This result can be written as a simple linear combination of the three formulas with the same parameters in , namely (eq.67)-(eq.66)-(eq.68) -(eq.96)+(eq.97)-(eq.98).

## Obtention of a zero relation of the form 0=P(1,2^12,24,A)

Setting a=4 in equation (3) from ,

$atan(1/3)={\frac {1}{2^{11}}}P(1,2^{12},8,(2^{9},2^{8},2^{6},0,-8,-4,-1,0))$ $={\frac {1}{2^{11}}}P(1,2^{12},24,3*(0,0,2^{9},0,0,2^{8},0,0,2^{6},0,0,0,0,0,-8,0,0,-4,0,0,-1,0,0,0))$ Setting a=2 in equation (4) from ,

$atan(1/3)={\frac {1}{2^{4}}}P(1,2^{4},8,(2^{3},-2^{3},2^{2},0,-2,2,-1,0))$ $={\frac {1}{2^{12}}}P(1,2^{12},24,(2^{11},-2^{11},2^{10},0,-2^{9},2^{9},-2^{8},0,2^{7},-2^{7},2^{6},0,-2^{5},2^{5},-2^{4},0,2^{3},-2^{3},2^{2},0,-2,2,-1,0))$ Subtracting both results,

$0=P(1,2^{12},24,(2^{11},-2^{11},-2^{11},0,-2^{9},-2^{10},-2^{8},0,-2^{8},-2^{7},2^{6},0,-32,32,32,0,8,16,4,0,4,2,-1,0))\,$ This zero relation can also be written as a BBP-type formula of base -2^6 and is a linear combination of three formulas with the same parameters in , namely -(eq.61)-4*(eq.63)-4*(eq.65) -(eq.91)-4*(eq.93)-4*(eq.95)

## Binary and ternary formulas

$log(2^{b}-1)=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {b}{2^{k}}}-{\frac {1}{2^{bk}}}\right)\,$ $log(3^{b}-1)=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {3b}{3^{k}}}-{\frac {b}{9^{k}}}-{\frac {1}{3^{bk}}}\right)\,$ (s,p,q,t,r) in Mathar's table 1. $log(2)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {1}{2^{k}}}\right)$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {2}{3^{k}}}-{\frac {1}{9^{k}}}\right)$ $log(3)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {2}{2^{k}}}-{\frac {1}{4^{k}}}\right)$ (1,3,2,2,-1/4) $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {3}{3^{k}}}-{\frac {1}{9^{k}}}\right)$ $log(5)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {2}{2^{k}}}+{\frac {1}{4^{k}}}-{\frac {1}{16^{k}}}\right)$ (1,5,2,2,1/4) $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {4}{3^{k}}}-{\frac {1}{81^{k}}}\right)$ $log(7)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {3}{2^{k}}}-{\frac {1}{8^{k}}}\right)$ (1,7,3,2,-1/8) $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {5}{3^{k}}}-{\frac {1}{9^{k}}}+{\frac {1}{27^{k}}}-{\frac {1}{729^{k}}}\right)$ $log(11)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {3}{2^{k}}}+{\frac {1}{4^{k}}}+{\frac {1}{32^{k}}}-{\frac {1}{1024^{k}}}\right)$ $={\frac {1}{2}}\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {13}{3^{k}}}-{\frac {4}{9^{k}}}-{\frac {1}{243^{k}}}\right)\,$ $log(13)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {4}{2^{k}}}-{\frac {1}{4^{k}}}+{\frac {1}{16^{k}}}+{\frac {1}{64^{k}}}-{\frac {1}{4096^{k}}}\right)$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {3}{2^{k}}}+{\frac {1}{4^{k}}}+{\frac {1}{8^{k}}}+{\frac {1}{16^{k}}}-{\frac {1}{4096^{k}}}\right)$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {7}{3^{k}}}-{\frac {2}{9^{k}}}-{\frac {1}{27^{k}}}\right)\,$ $log(17)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {4}{2^{k}}}+{\frac {1}{16^{k}}}-{\frac {1}{256^{k}}}\right)$ (1,17,4,2,1/16) $log(19)\,$ $=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {3}{2^{k}}}+{\frac {3}{4^{k}}}+{\frac {1}{512^{k}}}-{\frac {1}{262144^{k}}}\right)$ ### log(23)

$log(23)=\sum _{k=1}^{\infty }\left({\frac {4}{2^{k}}}+{\frac {1}{3^{k}}}-{\frac {1}{24^{k}}}\right)$ 23 is the smallest prime whose logarithm is not known to have a binary BBP-type formula.

### log(1)

From the identity

$1={\frac {(2^{6}-1)}{(2^{3}-1)(2^{2}-1)^{2}}}$ the following zero relation is obtained:

$0=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {1}{2^{k}}}-{\frac {2}{2^{2k}}}-{\frac {1}{2^{3k}}}+{\frac {1}{2^{6k}}}\right)$ This can be shown to be formula (62) in the Compendium by Bailey. See also , page 186.

A064078(6)=1.

### 0=atan(1)-atan(1/2)-atan(1/3)

$0\,$ $={\frac {1}{16}}P(1,16,8,(8,8,4,0,-2,-2,-1,0))$ (15 in )
$-{\frac {1}{16}}P(1,16,8,(0,16,0,0,0,-4,0,0))\,$ $-{\frac {1}{16}}P(1,16,8,(8,-8,4,0,-2,2,-1,0))\,$ $={\frac {1}{16}}P(1,16,8,(0,0,0,0,0,0,0,0))\,$ (NOT 61 in )

### Primes as $\Pi _{i}(2^{s_{i}}-1)^{q_{i}}$ (A144755, A161509)

$3=(2^{2}-1)\,$ $5={\frac {(2^{4}-1)}{(2^{2}-1)}}$ $7=(2^{3}-1)\,$ $11={\frac {(2^{10}-1)}{(2^{5}-1)(2^{2}-1)}}$ $13={\frac {(2^{12}-1)(2^{2}-1)}{(2^{6}-1)(2^{4}-1)}}={\frac {(2^{12}-1)}{(2^{2}-1)(2^{3}-1)(2^{4}-1)}}$ $17={\frac {(2^{8}-1)}{(2^{4}-1)}}$ $19={\frac {(2^{18}-1)}{(2^{9}-1)(2^{2}-1)^{3}}}$ $31=(2^{5}-1)\,$ $41={\frac {(2^{20}-1)(2^{2}-1)^{2}}{(2^{10}-1)(2^{4}-1)^{2}}}\,$ (M0.xxx)

$43={\frac {(2^{14}-1)}{(2^{7}-1)(2^{2}-1)}}\,$ (M0.xxx)

$73={\frac {(2^{9}-1)}{(2^{3}-1)}}\,$ (M0.xxx)

$127=(2^{7}-1)\,$ $151={\frac {(2^{15}-1)}{(2^{5}-1)(2^{3}-1)}}\,$ (M0.xxx)

$241={\frac {(2^{24}-1)(2^{4}-1)}{(2^{12}-1)(2^{8}-1)}}$ $257={\frac {(2^{16}-1)}{(2^{8}-1)}}$ (M0.xxx)

$331={\frac {(2^{30}-1)(2^{5}-1)}{(2^{15}-1)(2^{10}-1)(2^{2}-1)}}$ $337={\frac {(2^{21}-1)}{(2^{7}-1)(2^{3}-1)^{2}}}$ $683={\frac {(2^{22}-1)}{(2^{11}-1)(2^{2}-1)}}$ (M0.xxx) first given by Richard J. Mathar in his table of integrals (page 27)