A123178 -id:A123178 - cp's OEIS frontend

A006139 na(n) = 2(2n-1)a(n-1) + 4(n-1)a(n-2) with a(0) = 1.

1, 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152, 11258054144, 52767312896, 247736643584, 1164829376512, 5484233814016, 25852072517632, 121997903495168

Offset: 0

N. J. A. Sloane

a(n) = number of Delannoy paths (A001850) from (0,0) to (n,n) in which every Northeast step is immediately preceded by an East step. - David Callan, Mar 14 2004
The Hankel transform (see A001906 for definition) of this sequence is A036442 : 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
In general, 1/sqrt(1-4*r*x-4*r*x^2) has e.g.f. exp(2rx)BesselI(0,2r*sqrt((r+1)/r)x), a(n) = Sum_{k=0..n} C(2k,k)*C(k,n-k)*r^k, gives the central coefficient of (1+(2r)x+r(r+1)x^2) and is the (2r)-th binomial transform of 1/sqrt(1-8*C(n+1,2)x^2). - Paul Barry, Apr 28 2005
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the H and U steps can have two colors. - N-E. Fahssi, Feb 05 2008
Self-convolution of a(n)/2^n gives Pell numbers A000129(n+1). - Vladimir Reshetnikov, Oct 10 2016
This sequence gives the integer part of an integral approximation to Pi, and also appears in Frits Beukers's "A Rational Approach to Pi" (cf. Links, Example). Despite quality M ~ 0.9058... reported by Beukers, measurements between n = 10000 and 30000 lead to a contentious quality estimate, M ~ 0.79..., at the 99% confidence level. In "Searching for Apéry-Style Miracles" Doron Zeilberger Quotes that M = 0.79119792... and also gives a closed form. The same rational approximation to Pi also follows from time integration on a quartic Hamiltonian surface, 2*H=(q^2+p^2)*(1-4*q*(q-p)). - Bradley Klee, Jul 19 2018, updated Mar 17 2019
Diagonal of rational function 1/(1 - (x + y + x*y^2)). - Gheorghe Coserea, Aug 06 2018

			G.f. = 1 + 2*x + 8*x^2 + 32*x^3 + 136*x^4 + 592*x^5 + 2624*x^6 + 11776*x^7 + ...
J_3 = Integral_{y=0..Pi/4} 4*(4*(sin(y)-cos(y))*sin(y))^3*dy = 32*Pi - (304/3), |J_3| < 1. - _Bradley Klee_, Jul 19 2018

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Frits Beukers, A rational approach to Pi, Nieuw archief voor wiskunde 5/1 No. 4, December 2000, p. 377.
Dario Castellanos, A generalization of Binet's formula and some of its consequences, Fib. Quart., 27 (1989), 424-438.
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv:1410.5747 [math.CO], 2014.
Shalosh B. Ekhad and Doron Zeilberger, Searching for Apéry-Style Miracles [Using, Inter-Alia, the Amazing Almkvist-Zeilberger Algorithm], arXiv:1405.4445 [math.NT], 2014.
Bradley Klee, Approximating Pi with Trigonometric-Polynomial Integrals, Wolfram Demonstrations, July 27, 2018.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A001850, A002426, A036442, A084600-A084606, A084608-A084615.
Cf. A106258, A106259, A106260, A106261.
First column of A110446. A higher-quality Pi approximation: A123178.

a:=[1,2];; for n in [3..25] do a[n]:=1/(n-1)*(2*(2*n-3)*a[n-1]+4*(n-2)*a[n-2]); od; a; # Muniru A Asiru, Aug 06 2018

seq(add(binomial(2*k, k)*binomial(k, n-k), k=0..n), n=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 08 2001
A006139 := n -> 2^n*hypergeom([-n/2, 1/2-n/2], [1], 2):
seq(simplify(A006139(n)), n=0..29); # Peter Luschny, Sep 18 2014

Table[SeriesCoefficient[1/(1-4x-4x^2)^(1/2),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 05 2012 *)
Table[Abs[LegendreP[n, I]] 2^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 22 2015 *)
Table[Sum[Binomial[2*k, k]*Binomial[k, n - k], {k,0,n}], {n,0,50}] (* G. C. Greubel, Feb 28 2017 *)
a[n_] := If[n == 0, 1, Coefficient[(1 + 2 x + 2 x^2)^n, x^n]] (* Emanuele Munarini, Aug 04 2017 *)
CoefficientList[Series[1/Sqrt[(-4 x^2 - 4 x + 1)], {x, 0, 24}], x] (* Robert G. Wilson v, Jul 28 2018 *)

a(n) := coeff(expand((1+2*x+2*x^2)^n),x,n);
makelist(a(n),n,0,12); /* Emanuele Munarini, Aug 04 2017 */

for(n=0,30,t=polcoeff((1+2*x+2*x^2)^n,n,x); print1(t","))

for(n=0,25, print1(sum(k=0,n, binomial(2*k,k)*binomial(k,n-k)), ", ")) \\ G. C. Greubel, Feb 28 2017

{a(n) = (-2*I)^n * pollegendre(n, I)}; /* Michael Somos, Aug 04 2018 */

a(n) = Sum_{k=0..n} C(2*k, k)*C(k, n-k). - Detlef Pauly (dettodet(AT)yahoo.de), Nov 08 2001
G.f.: 1/(1-4x-4x^2)^(1/2); also, a(n) is the central coefficient of (1+2x+2x^2)^n. - Paul D. Hanna, Jun 01 2003
Inverse binomial transform of central Delannoy numbers A001850. - David Callan, Mar 14 2004
E.g.f.: exp(2*x) * BesselI(0, 2*sqrt(2)*x). - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..floor(n/2)} C(n,2k) * C(2k,k) * 2^(n-k). - Paul Barry, Sep 19 2006
a(n) ~ 2^(n - 3/4) * (1 + sqrt(2))^(n + 1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Oct 05 2012, simplified Jan 31 2023
G.f.: 1/(1 - 2*x*(1+x)*Q(0)), where Q(k)= 1 + (4*k+1)*x*(1+x)/(k+1 - x*(1+x)*(2*k+2)*(4*k+3)/(2*x*(1+x)*(4*k+3)+(2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
a(n) = 2^n*hypergeom([-n/2, 1/2-n/2], [1], 2). - Peter Luschny, Sep 18 2014
0 = a(n)*(+16*a(n+1) + 24*a(n+2) - 8*a(n+3)) + a(n+1)*(+8*a(n+1) + 16*a(n+2) - 6*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Oct 13 2016
It appears that Pi/2 = Sum_{n >= 1} (-1)^(n-1)*4^n/(n*a(n-1)*a(n)). - Peter Bala, Feb 20 2017
G.f.: G(x) = (1/(2*Pi))*Integral_{y=0..2*Pi} 1/(1-x*(4*(sin(y)-cos(y))*sin(y)))*dy, also satisfies: (2+4*x)*G(x)-(1-4*x-4*x^2)*G'(x)=0. - Bradley Klee, Jul 19 2018
a(n) = Sum_{k=0..n} (1-i)^k * (1+i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit. - Seiichi Manyama, Aug 29 2025

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.

A305998 Let K(n) = integral(t=-1,1, t^(2n)(1-t^2)^(2n)/(1+it)^(3n+1) dt) and write K(n) = d(n)Pi - b(n)/a(n) where a(n), b(n), d(n) are positive integers; sequence gives a(n).

Original entry on oeis.org

1, 15, 7, 495, 2145, 69615, 33915, 245157, 2523675, 150225075, 60480225, 187751655, 26397882693, 7073942421, 221214302595, 26306570659215, 362711807574025, 315526198564395, 154309366568181825, 495353332454332275, 13575552922770178725

Offset: 1

Bradley Klee, Jun 16 2018

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, A305997.

HermiteReduceRational[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]}]; 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]
Denominator[ 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 *)
Denominator@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 *)

A190726 Central coefficients of Riordan matrix A118384.

Original entry on oeis.org

1, 6, 62, 720, 8806, 110916, 1423796, 18520788, 243289670, 3220011684, 42872967012, 573608356272, 7705343534716, 103857425975400, 1403902871946000, 19024773303675420, 258372666772083270, 3515644245559211172, 47918193512409831380

Offset: 0

Emanuele Munarini, May 17 2011

nonn

This sequence gives the integer part of an integral approximation to log(2), thus bears strong similarity to A123178. Quality of rational approximants appears entirely sufficient to prove irrationality. - Bradley Klee, Jun 29 2018

			From _Bradley Klee_, Jul 16 2018: (Start)
I_2 = Integral_{t=0..1} ((1-t)^4*t^4)/(4*(1+t)^3)*dt = 62*log(2) - 1719/40 < 10^(-3).
I_3 = Integral_{t=0..1} - ((1-t)^6*t^6)/(8*(1+t)^4)*dt = 720*log(2) - 143731/288 < 10^(-5). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..88
Wadim Zudilin, An essay on irrationality measures of pi and other logarithms, arXiv:math/0404523 [math.NT], 2004.

Cf. A118384, A123178.
Log(2) approximation rationals: A316911, A316912.
Cf. A123178.

Table[Sum[Binomial[2n,k]Binomial[2n,n-k]2^k,{k,0,n}],{n,0,100}]
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]==1,a[1]==6},a,{n,0,10}] (* Bradley Klee, Jun 29 2018 *)

makelist(sum(binomial(2*n,k)*binomial(2*n,n-k)*2^k,k,0,n),n,0,12);

a(n)=sum(k=0,n,binomial(2*n,k)*binomial(2*n,n-k)<Charles R Greathouse IV, Jun 29 2011

a(n) = T(2*n,n), where T(n,k) = A118384(n,k).
a(n) = Sum_{k=0..n} binomial(2*n, k)*binomial(2*n, n-k)*2^k.
a(n) = Sum_{k=0..n} binomial(2*n, k)*binomial(k, n-k)*2^(n-k)*3^(2*k-n).
From Bradley Klee, Jun 29 2018: (Start)
a(n)*log(2) - A316911(n)/A316912(n) = I_n = Integral_{t=0..1}(-1/2)^n/(1+t)*((1-t)^2*t^2/(1+t))^n*dt.
Lim_{n->oo} I_n = 0, therefore:
Lim_{n->oo} A316911(n)/A316912(n)/a(n) = log(2).
G.f. G(x) and derivatives G^(n)(x) = d^n/dx^n G(x) satisfy a Picard-Fuchs type differential equation, 0 = Sum_{m=0..5,n=0..3} M_{m,n} x^m G^(n)(x), with integer matrix: M = {{324,-54,0,0}, {-36,10842,-486,0}, {84,8352,14931,-243}, {0,756,19026,3024}, {0,0,672,5364}, {0,0,0,112}}.
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.
(End)

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.

cp's OEIS Frontend

A006139 n*a(n) = 2*(2*n-1)*a(n-1) + 4*(n-1)*a(n-2) with a(0) = 1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Maple

Mathematica

Maxima

PARI

PARI

PARI

Formula

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).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A305998 Let K(n) = integral(t=-1,1, t^(2n)*(1-t^2)^(2n)/(1+it)^(3n+1) dt) and write K(n) = d(n)*Pi - b(n)/a(n) where a(n), b(n), d(n) are positive integers; sequence gives a(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A190726 Central coefficients of Riordan matrix A118384.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Maxima

PARI

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

A006139 na(n) = 2(2n-1)a(n-1) + 4(n-1)a(n-2) with a(0) = 1.

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).

A305998 Let K(n) = integral(t=-1,1, t^(2n)(1-t^2)^(2n)/(1+it)^(3n+1) dt) and write K(n) = d(n)Pi - b(n)/a(n) where a(n), b(n), d(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).