A305998 -id:A305998 - cp's OEIS frontend

A123178 Define K(n) = Integral_{t=-1..1} (t^(2n)(1-t^2)^(2n)/(1+it)^(3n+1))dt, and write K(n) = a(n)Pi - b(n)/c(n) where a(n), b(n), c(n) are positive integers; the sequence gives a(n).

14, 968, 75920, 6288296, 537005664, 46764723632, 4128230266160, 368090979124960, 33073373083339904, 2989771785328137728, 271603565356722214784, 24774311300942501337728, 2267541753957311770329600

Offset: 1

Benoit Cloitre, Oct 03 2006

nonn

The integrals K(n) give us a sequence of approximation to Pi whose qualities exceed 1.0449 in the long run. a(n) is divisible by 2^floor(n/2).
The integral K(n) can be evaluated for large n using Hermite reduction. In the range n = 1...200, a quality less than 1.0449 occurs irregularly, for n = 11, 16, 19, 23, 24, 32, 38, 42, 46, 50, 51, 55, 63, 85, 91, 94, 95, 100, 101, 103. - Bradley Klee, Jun 16 2018
Comment from Michael Somos, Jun 23 2018: The "quality" of approximations is as given by Frits Beukers "A rational approach to pi" where he writes "... |22/7 - pi| = 1/7^3.439, |355/113-pi| = 1/113^3.201. The exponents 3.429 and 3.201 will be called the quality of the respective approximations."
(Pseudocode, from Bradley Klee, Jun 18 2018)
Function HermiteReduce(f(t),g(t),m)
1: If m>1:
2: {u(t),v(t)} <- Solve f(t)=u(t)*g(t)+v(t)*g'(t);
3: Return HermiteReduce(u(t)+1/(m-1)*v'(t),g(t),m-1)
4: ElseIf m=1:
5: Return f(t)/g(t)
Function a(n)
1: f(t) <- ((1-i*t)^(3*n+1)+(1+i*t)^(3*n+1))*t^(2*n)*(1-t^2)^(2*n);
2: g(t) <- (1+t^2);
3: [dx] <- HermiteReduce(f(t),g(t),3*n+1)*dt;  ( drop exact differentials )
4: Return (1/Pi)*Integral_{t=0..1} [dx]

			K(5) = -3618728790016/2145 + 537005664*Pi so a(5) = 537005664.

Manuel Bronstein, Symbolic Integration I: Transcendental Functions, Springer, 2000, pages 39-46.

Bradley Klee, Table of n, a(n) for n = 1..1000
Frits Beukers, A rational approach to Pi, Nieuw archief voor wiskunde 5/1 No. 4, December 2000, p. 378.
Sam Blake, Integration Using Hermite Reduction, Wolfram Demonstrations Project.

Cf. A305997, A305998.

Kn := proc(n) local a,l ; a := 0 : for l from 0 to (3*n+1)/2 do a := a+2*binomial(3*n+1,2*l)*(-1)^l* int(t^(2*n+2*l)*(1-t^2)^(2*n)/(1+t^2)^(3*n+1),t=0..1) ; od ; a := subs(Pi=x,a) ; RETURN(a) ; end: A123178 := proc(n) RETURN( coeftayl(Kn(n),x=0,1)) ; end: for n from 1 to 20 do printf("%d,",A123178(n)) ; od ; # R. J. Mathar, Oct 07 2006

f[n_] := CoefficientList[ Integrate[t^(2n)*(1 - t^2)^(2n)/(1 + I*t)^(3n + 1), {t, -1, 1}], Pi][[ -1]]; Array[f, 13] (* Robert G. Wilson v *)
HermiteReduce[num_, den_, m_] := If[m > 1, Module[{cl = CoefficientList[num, t], deg, u, v, sol}, If[Length[cl] == 1, cl = PadRight[cl, 3]];deg=Length[cl]-1; u = Total[c[#] t^(2*#) & /@ Range[0, deg/2 - 1]]; v = Plus[ Total[-c[#] (m - 1)/(2 # + 1) t^(2*# + 1) & /@ Range[0, deg/2 - 1]], c[-1] t]; sol = Solve@MapThread[Equal, {cl,CoefficientList[Expand[Dot[{1 + t^2, 2 t}, {u, v}]], t]}]; HermiteReduce[Expand@ReplaceAll[u + 1/(m - 1) D[v, t], sol[[1]]],den,m-1]],num/4]
HermiteReduce[t^(2*#)*(1-t^2)^(2*#)*((1 + I*t)^(3*#+1)+(1-I*t)^(3*#+1)),(1+t^2),3*#+1]&/@Range[20](* Bradley Klee, Jun 18 2018 *)
RecurrenceTable[{64*(1+n)*(2+n)*(1+2*n)*(3+2*n)*(5+2*n)*(816+755*n+165*n^2)*a[n]-48*(2+n)*(3+2*n)*(5+2*n)*(4+3*n)*(2039+4103*n+2595*n^2+495*n^3)*a[n+1]+6*(5+2*n)*(4+3*n)*(5+3*n)*(893628+2406908*n+2163923*n^2+803750*n^3+106095*n^4)*a[n+2]-9*(3+n)*(4+3*n)*(5+3*n)*(7+3*n)*(8+3*n)*(226+425*n+165*n^2)*a[n+3]==0,
a[0]==1/2,a[1]==14,a[2]==968},a,{n,1,5000}] (* Bradley Klee, Jun 24 2018 *)

64*(1+n)*(2+n)*(1+2*n)*(3+2*n)*(5+2*n)*(816 + 755*n + 165*n^2)*a(n) - 48*(2+n)*(3+2*n)*(5+2*n)*(4+3*n)*(2039 + 4103*n + 2595*n^2 + 495*n^3)*a(n+1) + 6*(5+2*n)*(4+3*n)*(5+3*n)*(893628 + 2406908*n + 2163923*n^2 + 803750*n^3 + 106095*n^4)*a(n+2) - 9*(3+n)*(4+3*n)*(5+3*n)*(7+3*n)*(8+3*n)*(226 + 425*n + 165*n^2)*a(n+3) = 0. - Bradley Klee, Jun 20 2018
Define F(x) the g.f. of these a(n), G(x)=1/2+F(x), and G^(n)(x)=d^n/dx^n G(x). Period G(x) satisfies a Picard-Fuchs type differential equation, 0=Sum_{m=0..9,n=0..5}M_{m,n} x^m G^(n)(x), with integer matrix:
M={{698544,-24948,0,0,0,0},
{-2344608,33884712,-224532,0,0,0},
{2305584,-34982100,787834836,-3255714,0,0},
{-3490848,65404872,-690185556,1319686128,-3031182,0},
{4487040,-85092672,973263876,-1454575542,508724631,-505197},
{0,89740800,-280713984,2717626800,-642933018,48807765},
{0,0,190699200,-235103952,1409057154,-67970205},
{0,0,0,109184640,-61373632,164264580},
{0,0,0,0,20939520,-4518080},
{0,0,0,0,0,1196544}} - Bradley Klee, Jun 24 2018

More terms from R. J. Mathar, Oct 07 2006

A305997 Define K(n) = Integral_{t=-1..1} t^(2n)(1-t^2)^(2n)/(1+it)^(3n+1)dt and write K(n) = d(n)Pi - a(n)/c(n) where a(n), d(n), c(n) are positive integers; sequence gives a(n).

Original entry on oeis.org

44, 45616, 1669568, 9778855936, 3618728790016, 10227537305460736, 439851024281337856, 283497572919345676288, 262217569855510830645248, 1411010811095175238386712576, 51605826449550157277271425024, 14612860454957563743068313616384

Offset: 1

Bradley Klee, Jun 16 2018

nonn

Bradley Klee, Table of n, a(n) for n = 1..1000
Frits Beukers, A rational approach to Pi, Nieuw archief voor wiskunde 5/1 No. 4, December 2000, p. 378.

Cf. A123178, A305998.

HermiteReduceRational[num_, den_, m_] := If[m > 1, Module[{cl = CoefficientList[num, t], deg, u, v, sol, c},If[Length[cl] == 1, cl = PadRight[cl, 3]]; deg = Length[cl] - 1; u = Total[c[#]*t^(2 #) & /@ Range[0, deg/2 - 1]]; v = Plus[Total[-c[#]*(m - 1)/(2*# + 1) t^(2*# + 1) & /@ Range[0, deg/2 - 1]], c[-1] t]; sol = Solve@ MapThread[Equal, {cl,CoefficientList[Expand[Dot[{1 + t^2, 2 t}, {u, v}]], t]}]; Plus[ ReplaceAll[v/(m - 1)/den^(m - 1), sol[[1]]] /. t -> 1, HermiteReduceRational[ Expand@ReplaceAll[u+1/(m-1)*D[v, t], sol[[1]]], den, m - 1]]],0]
Numerator[ HermiteReduceRational[ t^(2*#)*(1-t^2)^(2*#)*((1+I*t)^(3*#+1)+(1-I*t)^(3*#+1)), (1+t^2), 3*#+1]]&/@Range[20] (* Bradley Klee, Jun 18 2018 *)
Numerator@RecurrenceTable[{64*(1+n)*(2+n)*(1+2*n)*(3+2*n)*(5+2*n)*(816+755*n+165*n^2)*a[n]-48*(2+n)*(3+2*n)*(5+2*n)*(4+3*n)*(2039+4103*n+2595*n^2+495*n^3)*a[n+1]+6*(5+2*n)*(4+3*n)*(5+3*n)*(893628+2406908*n+2163923*n^2+803750*n^3+106095*n^4)*a[n+2]-9*(3+n)*(4+3*n)*(5+3*n)*(7+3*n)*(8+3*n)*(226+425*n+165*n^2)*a[n+3]==0, a[0]==0,a[1]==44,a[2]==45616/15},a,{n,1,5000}] (* Bradley Klee, Jun 25 2018 *)

Define G(x) = Sum_{n>0} A305997(n)/A305998(n)*x^n, and G^(n)(x) = d^n/dx^n G(x). Period G(x) satisfies a nonhomogeneous differential equation: -1097712 + 4292640*x + 3901584*x^2 - 224352*x^3 = Sum_{m=0..9, n=0..5} M_{m,n} x^m G^(n)(x), with integer matrix M as in A123178.

A316911 Define K(n) = Integral_{t=0..1} (-1/2)^n/(1+t)((1-t)^2t^2/(1+t))^ndt and write K(n) = d(n)log(2) - a(n)/c(n) where a(n), d(n), c(n) are positive integers; sequence gives a(n).

Original entry on oeis.org

0, 25, 1719, 143731, 64456699, 1846991851, 781688106621, 445837607665267, 611642484654021, 674842075634295726569, 9142845536119405749427, 38984536004906714808649, 80321414381403813427242343, 342487507476162248453574514441, 562411667990487545372378396727201

Offset: 0

Bradley Klee, Jul 16 2018

As n goes to infinity, integral value K(n) goes to zero. Given a rational approximant r(n)=a(n)/c(n)/d(n)=p(n)/q(n) to irrational number log(2), the quality M(n) is defined as, M(n)=-log(|r(n)-log(2)|)/log(q(n)) (Cf. Beukers Link). For this approximation, we can easily measure M(n) over n=5,000..20,000, and estimate that M(n)~1.14... to the 99% confidence level (Cf. Histogram Link).

			{a(10),c(10),d(10)}={9142845536119405749427,307660953600,42872967012}.
r(10)=a(10)/c(10)/d(10)=9142845536119405749427/13190337914573262643200.
r(10)=0.693147180559945309417232121402...
log(2)=0.693147180559945309417232121458...
M(10)=-log(|r(10)-log(2)|)/log(13190337914573262643200)=1.27...

F. Beukers, A rational approach to Pi, Nieuw archief voor wiskunde 5/1 No. 4, December 2000, p. 378.
Bradley Klee, Quality Histogram.

Integer Part: A190726. Denominators: A316912. Similar Pi approximation: A123178, A305997, A305998.

FracData[n0_]:=RecurrenceTable[{2*(n-1)*(2*n-3)*(2*n-1)*(33*n-8)*a[n-2]+ 9*(2*n-1)*(693*n^3-1554*n^2+989*n-160)*a[n-1] -3*n*(3*n-2)*(3*n-1)*(33*n-41)*a[n] == 0, a[0]==0, a[1]==25/6}, a, {n, 0, n0}]
Numerator[FracData[5000]]

Define G(x) = Sum_{n>0} A316911(n)/A316912(n)*x^n, and G^(k)(x) = d^k/dx^k G(x). Period G(x) satisfies a nonhomogeneous differential equation: -225+112*x = Sum_{j=0..5,k=0..3} M_{j,k} x^j G^(k)(x), with integer matrix M as in A190726.

A316912 Define K(n) = Integral_{t=0..1} (-1/2)^n/(1+t)((1-t)^2t^2/(1+t))^ndt and write K(n) = d(n)log(2) - b(n)/a(n) where a(n), d(n), b(n) are positive integers; sequence gives a(n).

Original entry on oeis.org

1, 6, 40, 288, 10560, 24024, 792064, 34728960, 3627008, 302356454400, 307660953600, 98050867200, 15038824120320, 4757532010463232, 577952036826644480, 26189033224273920, 358597702262241361920, 244498433360619110400, 143982410756809031680

Offset: 0

Bradley Klee, Jul 16 2018

Integer Part: A190726. Numerators: A316911. Similar Pi Approximation: A123178, A305997, A305998.

FracData[n0_]:=RecurrenceTable[{2*(n-1)*(2*n-3)*(2*n-1)*(33*n-8)*a[n-2]+ 9*(2*n-1)*(693*n^3-1554*n^2+989*n-160)*a[n-1] -3*n*(3*n-2)*(3*n-1)*(33*n-41)*a[n] == 0, a[0]==0, a[1]==25/6}, a, {n, 0, n0}]
Denominator[FracData[5000]]

cp's OEIS Frontend

A123178 Define K(n) = Integral_{t=-1..1} (t^(2n)*(1-t^2)^(2n)/(1+it)^(3n+1))dt, and write K(n) = a(n)*Pi - b(n)/c(n) where a(n), b(n), c(n) are positive integers; the sequence gives a(n).

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A305997 Define K(n) = Integral_{t=-1..1} t^(2n)*(1-t^2)^(2n)/(1+it)^(3n+1)dt and write K(n) = d(n)*Pi - a(n)/c(n) where a(n), d(n), c(n) are positive integers; sequence gives a(n).

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A316911 Define K(n) = Integral_{t=0..1} (-1/2)^n/(1+t)*((1-t)^2*t^2/(1+t))^n*dt and write K(n) = d(n)*log(2) - a(n)/c(n) where a(n), d(n), c(n) are positive integers; sequence gives a(n).

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A316912 Define K(n) = Integral_{t=0..1} (-1/2)^n/(1+t)*((1-t)^2*t^2/(1+t))^n*dt and write K(n) = d(n)*log(2) - b(n)/a(n) where a(n), d(n), b(n) are positive integers; sequence gives a(n).

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A123178 Define K(n) = Integral_{t=-1..1} (t^(2n)(1-t^2)^(2n)/(1+it)^(3n+1))dt, and write K(n) = a(n)Pi - b(n)/c(n) where a(n), b(n), c(n) are positive integers; the sequence gives a(n).

A305997 Define K(n) = Integral_{t=-1..1} t^(2n)(1-t^2)^(2n)/(1+it)^(3n+1)dt and write K(n) = d(n)Pi - a(n)/c(n) where a(n), d(n), c(n) are positive integers; sequence gives a(n).

A316911 Define K(n) = Integral_{t=0..1} (-1/2)^n/(1+t)((1-t)^2t^2/(1+t))^ndt and write K(n) = d(n)log(2) - a(n)/c(n) where a(n), d(n), c(n) are positive integers; sequence gives a(n).

A316912 Define K(n) = Integral_{t=0..1} (-1/2)^n/(1+t)((1-t)^2t^2/(1+t))^ndt and write K(n) = d(n)log(2) - b(n)/a(n) where a(n), d(n), b(n) are positive integers; sequence gives a(n).