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# User:Jaume Oliver Lafont/Pi

${\displaystyle {\frac {\pi }{3{\sqrt {3}}}}=\sum _{k=0}^{\infty }\left({\frac {1}{3k+1}}-{\frac {1}{3k+2}}\right)}$ check

may be added as a comment to sequences

0,1,-1 period 3

(9*n^2+9*n+2)/2

1,2,4,5,7,8,10,11, (not divisible by 3)

The same constant -not on the OEIS, but see A086089- appears in a faster series:

http://mathworld.wolfram.com/PiFormulas.html (27) (Gosper).

${\displaystyle {\frac {\pi }{2{\sqrt {2}}}}}$(A093954)${\displaystyle =\sum _{k=0}^{\infty }\left({\frac {1}{8k+1}}+{\frac {1}{8k+3}}-{\frac {1}{8k+5}}-{\frac {1}{8k+7}}\right)=\sum _{k=0}^{\infty }{\frac {{\sqrt {2}}\sin \left({\frac {(2k+1)\pi }{4}}\right)}{2k+1}}}$

${\displaystyle {\frac {\pi }{3}}=\sum _{k=0}^{\infty }\left({\frac {1}{12k+1}}+{\frac {1}{12k+5}}-{\frac {1}{12k+7}}-{\frac {1}{12k+11}}\right)}$ check

This formula appears as a comment to A007310 and is equivalent to (33) in http://mathworld.wolfram.com/PiFormulas.html.

A similar one can be written for ${\displaystyle {\frac {\pi }{2}}}$.

${\displaystyle {\frac {\pi }{2}}=\sum _{k=0}^{\infty }\left({\frac {1}{12k+1}}+{\frac {2}{12k+3}}+{\frac {1}{12k+5}}-{\frac {1}{12k+7}}-{\frac {2}{12k+9}}-{\frac {1}{12k+11}}\right)}$

${\displaystyle {\sqrt {3}}={\frac {\sum _{k=0}^{\infty }\left({\frac {1}{12k+1}}+{\frac {1}{12k+5}}-{\frac {1}{12k+7}}-{\frac {1}{12k+11}}\right)}{\sum _{k=0}^{\infty }\left({\frac {1}{3k+1}}-{\frac {1}{3k+2}}\right)}}}$

(how can this be checked with a link to WolframAlpha?)

Similarly,

${\displaystyle {\sqrt {2}}={\frac {\sum _{k=0}^{\infty }\left({\frac {1}{8k+1}}+{\frac {1}{8k+3}}-{\frac {1}{8k+5}}-{\frac {1}{8k+7}}\right)}{\sum _{k=0}^{\infty }\left({\frac {1}{4k+1}}-{\frac {1}{4k+3}}\right)}}}$

a(n)=((n+1)!*sum(k=0,n,[1,0,1,0,-1,0,-1,0][k%8+1]/(k+1)))

1, 2, 8, 32, 136, 816, 4992, 39936, 399744, 3997440, 47600640, 571207680,

and

b(n)=((n+1)!*sum(k=0,n,[1,0,-1,0][k%4+1]/(k+1)))

1, 2, 4, 16, 104, 624, 3648, 29184, 302976, 3029760, 29698560, 356382720,

with ${\displaystyle {\frac {a(n)}{b(n)}}}$ approaching ${\displaystyle {\sqrt {2}}}$ (slowly) as ${\displaystyle n}$ grows.

Other sequences with the same property can be defined, given the zeros at even positions in both sequences, as follows:

a(n)=((2*(n+1))!/(n+1)!/2^(n+1)*sum(k=0,n,[1,1,-1,-1][k%4+1]/(2*k+1)))

1, 4, 17, 104, 1041, 12396, 150753, 2126160,

(closely related to A127676 and first four terms equal)

and

b(n)=((2*(n+1))!/(n+1)!/2^(n+1)*sum(k=0,n,[1,-1][k%2+1]/(2*k+1)))

1, 2, 13, 76, 789, 7734, 110937, 1528920,

(a shifted version of A024199)

For faster convergence, try the same with pairs ((13),(15)) -for ${\displaystyle {\sqrt {2}}}$- and ((13),(16)) -for ${\displaystyle {\sqrt {3}}}$- taken from http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf.

From (15)

a(n)=(n+1)!*sum(k=0,n,[2,0,1,0,2,0,-2,0,-1,0,-2,0][k%12+1]/(k+1)/2^(k/2))

2, 4, 13, 52, 272, 1632, 11244, 89952, 807048, 8070480, 88548480, 1062581760,

with a(n)/(n+1)! -> Pi*sqrt(2)/2

From (13)

b(n)=(n+1)!*sum(k=0,n,sin((k+1)*Pi/4)/(k+1)/2^((k-1)/2))

1, 3, 10, 40, 194, 1134, 7848, 62784, 567576, 5698440, 62796240, 753554880,

with b(n)/(n+1)! -> Pi/2

Combining both,

a(n)/b(n) -> sqrt(2)

[ b(k+1)-(k+2)*b(k) is 1, 1, 0, -6, -30, -90, 0, 2520, 22680, 113400, 0, -7484400, -97297200, a version of A009775. This is also related to A090932. ]

From (16)

a(n)=2^n*(n+1)!*sum(k=0,n,[1,1,0,-1,-1,0][k%6+1]/(k+1)/2^k)

1, 5, 30, 234, 2316, 27792, 389808, 6241968, 112355424, 2246745600, 49424774400, 1186194585600,

with a(n)/(2^n*(n+1)!) -> 2*Pi*sqrt(3)/9

From (13)

b(n)=(n+1)!*sum(k=0,n,sin((k+1)*Pi/4)/(k+1)/2^((k-1)/2))

1, 3, 10, 40, 194, 1134, 7848, 62784, 567576, 5698440, 62796240, 753554880,

with b(n)/(n+1)! -> Pi/2

Combining both,

9*a(n)/(4*b(n)*2^n) -> sqrt(3)

Changing sin by cos in b(n),

b(n)=(n+1)!*sum(k=0,n,cos((k+1)*Pi/4)/(k+1)/2^((k-1)/2))

1, 2, 5, 17, 79, 474, 3408, 27894, 253566, 2535660, 27778860, 332098920,

Two more variants:

b(n)=(n+1)!*sum(k=0,n,sin(k*Pi/4)/(k+1)/2^((k-2)/2))

0, 1, 5, 23, 115, 660, 4440, 34890, 314010, 3162780, 35017380, 421455960,

b(n)=(n+1)!*sum(k=0,n,cos(k*Pi/4)/(k+1)/2^((k-2)/2))

2, 5, 15, 57, 273, 1608, 11256, 90678, 821142, 8234100, 90575100, 1085653800,