cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A295870 a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.

Original entry on oeis.org

1, 12, 660, 48720, 4005540, 349260912, 31626298704, 2940502593600, 278788387440420, 26831860080682800, 2613367831568654160, 257012469788428710720, 25479526081439438845200, 2543092744417831625342400, 255292245777771431285140800, 25755871314484468746363582720
Offset: 0

Views

Author

Bradley Klee, Feb 23 2018

Keywords

Comments

Compare with EllipticK A002894 and the Ramanujan period-energy functions A113424, A006480, A000897. The series expansion "T(x) = 2*Pi*Sum_{n>=0} a_n*x^n" determines the real period T of elliptic curves in the family "x=p^2+q^2-4*(q^2-p^2)*q, 0 < x < 1/108". This sequence serves as a counterexample to the naive idea that elliptic integrals will always evaluate to a hypergeometric function such as 2F1(a,b;c;x).
A300058 is the complex period-energy function, after scaling energy and time dimensions such that all a(n) are integers and a(0)=1. The Picard-Fuchs equation is "(12-288*x+9216*x^2)*T(x) + (-1+232*x-8160*x^2+82944*x^3)*T'(x) + (-x+164*x^2-6432*x^3+41472*x^4)*T''(x)".
Although the sequence is not generated by a hypergeometric function, it can be formulated in terms of Hypergeometric numbers, specifically the binomial coefficients. Then Zeilberger's algorithm outputs a second order recurrence with polynomial coefficients.
The contour plot is nice to look at, with reflection symmetry, three critical points, and two separatrices dividing the phase plane into eight distinct regions.
Hyperbolic Critical points are located at (q,p) locations (1/6,0) and (-1/4,sqrt(5)/4) and (-1/4,-sqrt(5)/4). Is it possible to use chord-and-tangent addition rules to produce an exponentially-convergent Diophantine approximation to sqrt(5) that moves along the upper separatrix x=1/8?
Does there exist a period-preserving transformation that takes any one of the curves with 0 < x < 1/108 into a particular Weierstrass curve from the L-function and Modular Forms Database?

References

  • D. Husemöller, Elliptic Curves, 2nd ed., New York: Springer, 2004.
  • J. H. Silverman, The Arithmetic of Elliptic Curves, 2nd ed., New York: Springer, 2009.

Crossrefs

Factors: A005190, A005809, A005721.
Complex Period: A300058.

Programs

  • Mathematica
    b[NN_]:=Total/@Table[((-1)^k)*Binomial[3*n,n]*Binomial[2*n,k]*Binomial[5*n-4*k-1,3*n-4*k],{n,0,NN},{k,0,Floor[3*n/4]}];
    c1=8*(-30+201*n-319*n^2+145*n^3);c2=-8640*(n-5/3)*(n-4/3)*(n-1/5);c3=10*(n-6/5)*n^2;a[0]=1;a[1]=12;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],{n->n0}];
    ({#,SameQ[a/@Range[0, 15],#]}&@b[15])[[1]]

Formula

a(n) = A005809(n)*A005721(n).
a(n) = Sum_{k=0..floor(3n/4)} ((-1)^k)*binomial(3*n,n)*binomial(2 *n, k)*binomial(5*n - 4*k - 1, 3*n - 4*k).
c1 = 8 *(-30 + 201*n - 319*n^2 + 145*n^3); c2 = -8640*(n - 5/3)*(n - 4/3)*(n - 1/5); c3 = 10*(n - 6/5)*n^2; a(0)=1; a(1)=12; a(n) = (c1/c3)*a(n-1) + (c2/c3)*a(n-2).

A300057 Coefficient of z^(3*n) in the expansion of (1 + 9*z + 9*z^2 + z^3)^(2*n).

Original entry on oeis.org

1, 164, 47148, 15454820, 5361965980, 1919987703504, 701459496193236, 259867456921970040, 97260263038893462300, 36686877800581349096240, 13924013746979490475444528, 5311128944356277793155688612, 2034235241375650519750351973188
Offset: 0

Views

Author

Bradley Klee, Feb 23 2018

Keywords

Crossrefs

Programs

  • GAP
    List([0..15],n->6^(2*n)Sum([0..2*n],k->Binomial(2*n,k)*Binomial(2*(n+k),n+k)*(1/6)^k)); # Muniru A Asiru, Apr 07 2018
  • Mathematica
    c1=16*(n-1/2)*(-230+2259*n-3933*n^2+1863*n^3);c2=1036800*(n-1)*(n-3/2)*(n-1/2)*(n-1/9);c3=81*n*(n-2/3)*(n-1/3)*(n-10/9);a[0]=1;a[1]=164;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0]
    b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[2*n,k2]*Binomial[2*n,k1]*Binomial[2*n, 3*n-k1-k2]*(4 + Sqrt[15])^(2*n-k1)*(4-Sqrt[15])^(2*n-k2),{n,0,NN},{k1,0,2*n},{k2,0,2*n}]]
    ({#,SameQ[Coefficient[(1+9*z+9*z^2+z^3)^(2*#),z,3*#]&/@Range[0,10],#],SameQ[a/@Range[0,10],#]}&@b[10])[[1]]
    Table[SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* Vaclav Kotesovec, Apr 18 2018 *)
  • PARI
    a(n) = polcoeff((1 + 9*z + 9*z^2 + z^3)^(2*n), 3*n); \\ Michel Marcus, Mar 06 2018
    

Formula

a(n) = 1/(2*Pi)*Integral_{0..2*Pi}(12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.
a(n) = Sum_{k1=0..2*n} Sum_{k2=0..2*n} binomial(2*n,k1)*binomial(2*n,k2)*binomial(2*n,3*n-k1-k2)*((4-sqrt(15))^(2*n-k1))*((4+sqrt(15))^(2*n-k2)).
a(n) = (c1/c3)*a(n-1)+(c2/c3)*a(n-2); with a(0)=1; a(1)=164; and
c1=16*(n-1/2)*(-230+2259*n-3933*n^2+1863*n^3);
c2=1036800*(n-1)*(n-3/2)*(n-1/2)*(n-1/9);
c3=81*n*(n-2/3)*(n-1/3)*(n-10/9).
From Wolfdieter Lang, Apr 06 2018: (Start)
a(n) = 4^(2*n)*(2/Pi)*Integral_{0..Pi/2} sin(x)^(2*n)*(3 + 2*sin(x)^2)^(2*n) dx. With the binomial formula and integrals over even powers of sin(x) this becomes
a(n) = 6^(2*n)*Sum_{k=0..2*n} binomial(2*n, k)*binomial(2*(n+k), n+k)*(1/6)^k = 6^(2*n)*binomial(2*n, n)*hypergeometric([-2*n, n+1/2], [n+1], -2/3). (End)
a(n) ~ 2^(4*n) * 5^(2*n + 1/2) / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 18 2018
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