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A006139
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n*a(n) = 2*(2*n-1)*a(n-1) + 4*(n-1)*a(n-2) with a(0) = 1.
(Formerly M1849)
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39
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1, 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152, 11258054144, 52767312896, 247736643584, 1164829376512, 5484233814016, 25852072517632, 121997903495168
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OFFSET
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0,2
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COMMENTS
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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
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
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
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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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
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
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EXAMPLE
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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
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MAPLE
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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):
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MATHEMATICA
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Table[SeriesCoefficient[1/(1-4x-4x^2)^(1/2), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 05 2012 *)
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 *)
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PROG
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(PARI) for(n=0, 30, t=polcoeff((1+2*x+2*x^2)^n, n, x); print1(t", "))
(PARI) for(n=0, 25, print1(sum(k=0, n, binomial(2*k, k)*binomial(k, n-k)), ", ")) \\ G. C. Greubel, Feb 28 2017
(PARI) {a(n) = (-2*I)^n * pollegendre(n, I)}; /* Michael Somos, Aug 04 2018 */
(Maxima) a(n) := coeff(expand((1+2*x+2*x^2)^n), x, n);
(GAP) 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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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