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.

Showing 1-5 of 5 results.

A318614 Scaled g.f. S(u) = Sum_{n>0} a(n)*16*(u/16)^n satisfies T(u) = d/du S(u), with T(u) as defined by A318417; sequence gives a(n).

Original entry on oeis.org

1, 6, 76, 1260, 24276, 515592, 11721072, 280020312, 6945369860, 177358000248, 4635276570288, 123449340098448, 3339525750984528, 91535631253610400, 2537277723600799680, 71015600640006437040, 2004523477053308685540, 57003431104378084982040
Offset: 1

Views

Author

Bradley Klee, Aug 30 2018

Keywords

Comments

Area interior to the central loop of u = 2*H = x^2 + y^2 - (1/2)*(x^4 + y^4) equals to Pi*S(u), when u in [0,1/2].

Examples

			Singular Value: S(1/2) = 1/sqrt(2).
N=4, h=1/sqrt(2) Quantization: S(u) = (n+1/2)*h/N.
  n  |                  u
==================================================
  0  |  0.08544689553344134756293807606337...
  1  |  0.23840989875904155311088418238272...
  2  |  0.36638282702449450473835851051425...
  3  |  0.46595506694324457665483887176081...
		

References

  • E. Heller, The Semiclassical Way to Dynamics and Spectroscopy, Princeton University Press, 2018, page 204.

Crossrefs

Programs

  • GAP
    a:=[1,6];; for n in [3..20] do a[n]:=(1/(n*(n-1)^2))*(12*(n-1)*(2*n-3)^2*a[n-1]-(128*(n-2)*(2*n-5)*(2*n-3)*a[n-2])); od; a; # Muniru A Asiru, Sep 24 2018
  • Mathematica
    RecurrenceTable[{(n-1)^2*n*a[n] - 12*(n-1)*(2*n-3)^2*a[n-1] + 128*(n-2)*(2*n-5)*(2*n-3)*a[n-2] == 0, a[1] == 1, a[2] == 6}, a, {n, 1, 1000}]

Formula

(n-1)^2*n*a(n) - 12*(n-1)*(2*n-3)^2*a(n-1) + 128*(n-2)*(2*n-5)*(2*n-3)*a(n-2) == 0.
a(n) = A000108(n-1)*A098410(n-1).

A186375 a(n) equals the sum of the squares of the expansion coefficients for (x + y + 2*z)^n.

Original entry on oeis.org

1, 6, 54, 588, 7110, 91476, 1224636, 16849944, 236523078, 3371140740, 48630906324, 708412918824, 10403176168476, 153813188724552, 2287366047735480, 34185974267420208, 513159651195396678, 7732530110414488932
Offset: 0

Views

Author

Paul D. Hanna, Feb 19 2011

Keywords

Comments

Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials: (1 + 2*x^k + x^p)^n, (2 + x^k + x^p)^n, or (1 + x^k + 2*x^p)^n, for all p > n*k and fixed k > 0.
Rescaling the g.f. G(x) to T(u)=G(3*u/16) moves the singular point x=1/16 to u=1/3. Period function T(u) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_3 symmetry. For precise definitions, pictures, a proof certificate, and more information, see A318245. - Bradley Klee, Aug 22 2018

Examples

			G.f.: A(x) = 1 + 6*x + 54*x^2/2!^2 + 588*x^3/3!^2 + 7110*x^4/4!^2 + ...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2]^2 *[Sum_{n>=0} (4x)^n/n!^2] where
[Sum_{n>=0} x^n/n!^2]^2 = 1 + 2*x + 6*x^2/2!^2 + 20*x^3/3!^2 + 70*x^4/4!^2 + ... + (2n)!/n!^2 *x^n/n!^2 + ...
a(4) =   256
       + 1024 + 1024
       + 576  + 2304 + 576
       + 64   + 576  + 576 + 64
       + 1    + 16   + 36  + 16 + 1  = 7110.
		

Crossrefs

Programs

  • Maple
    A186375 := n -> 4^n*hypergeom([1/2,-n,-n], [1,1], 1):
    seq(simplify(A186375(n)), n=0..17); # Peter Luschny, May 24 2017
  • Mathematica
    Table[Sum[Binomial[n,k]^2*Binomial[2k,k]*4^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
    (* From Bradley Klee, Aug 22 2018: Start *)PyramidLevel[n_]:=If[n==0, {{1}}, Table[Coefficient[(2*x+y+z)^n,x^j*y^k*z^(n-j-k)]^2, {j,0,n}, {k,0,n-j}]]; a1[n_]:= Total[Flatten[PyramidLevel[n]]];
    a1 /@ Range[0, 10]
    RecurrenceTable[{4*(4*n-5)*(4*n-3)*a[n-2]-2*(10*n^2-10*n+3)*a[n-1]+n^2*a[n]==0, a[0]==1, a[1]==6},a,{n,0,1000}] (*  End *)
    a[ n_] := If[ n < 0, 0, Block[ {x, y, z}, Expand[ (x + y + 2 z)^n] /. {t_Integer -> t^2, x -> 1, y -> 1, z -> 1}]]; (* Michael Somos, Aug 27 2018 *)
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*binomial(2*(n-k),n-k)*4^k)}
    
  • PARI
    {a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)^2*sum(m=0,n,(2^2*x)^m/m!^2),n)}
    
  • PARI
    {a(n)=local(V=Vec((1+2*x+x^(n+2))^n));V*V~}

Formula

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*4^(n-k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x)^2 * B(2^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: n^2*a(n) = 2*(10*n^2-10*n+3)*a(n-1) - 4*(4*n-5)*(4*n-3)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(4*n+1/2)/(Pi*n). - Vaclav Kotesovec, Oct 20 2012
a(n) = 4^n*hypergeom([1/2,-n,-n], [1,1], 1). - Peter Luschny, May 24 2017
G.f.: G(x)=Sum_{n>=0}a(n)x^n, 6*(10*x-1)*G + (192*x^2-40*x+1)*G' + x*(16*x-1)*(4*x-1)*G''=0. - Bradley Klee, Aug 22 2018

Extensions

Name edited by Bradley Klee, Aug 22 2018

A318245 Scaled g.f. T(v) = Sum_{n>=0} a(n)*(3*v/64)^n satisfies 9*(5*v-4)*T + d/dv(16*v*(v-1)*(3*v-4)*T') = 0, and a(0)=1; sequence gives a(n).

Original entry on oeis.org

1, 12, 180, 2928, 49860, 875952, 15754704, 288722880, 5373771876, 101334517680, 1932405892560, 37208369165760, 722497419680400, 14132680228175040, 278236490874120000, 5508974545258860288, 109624581377872629156, 2191185332414847848880, 43971545517545956240464
Offset: 0

Views

Author

Bradley Klee, Aug 22 2018

Keywords

Comments

The linked document "Proof Certificate" explains that period function T(v) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_4 symmetry (also cf. Examples and A186375).

Examples

			Period function T_{O}(w): Take T_{C3}(u) and T_{C4}(v) from A186375 and A318245 respectively. Set (u,v)=(w-2/3,2-w), with u in [0,1/3], v in [0,1], and w in [2/3,2]. Define piecewise function T_{O}(w) = T_{C3}(w-2/3) if w in [2/3,1] or T_{O}(w) = T_{C4}(2-w) if w in [1,2].
Geometric Singular Points: Construct a family of algebraic sphere curves by intersecting a sphere 1=X^2+Y^2+Z^2 with the octahedral surface w=2*(X^4+Y^4+Z^4). Four cube vertex axes--(x+y+z, -x+y+z, x-y+z, x+y-z)--intersect the sphere in eight circular points with w=2/3. Three octahedron vertex axes--(x, y, z)--intersect the sphere in six circular points with w=2. Six cuboctahedron vertex axes--(x+y, x-y, y+z, y-z, z+x, z-x)--intersect the sphere in twelve hyperbolic points with w=1.
		

Crossrefs

Periods: A186375, A318417.

Programs

  • Mathematica
    CoefficientList[Expand[Normal@Series[Divide[Sqrt[S],Sqrt[1-4*S*x]*Sqrt[S-8 + 8*Sqrt[1-4*S*x]]], {x, 0, 13}]/.{S->12+4*Q^2}]/.{Q^n_:>(1/2)^n*Binomial[n, n/2]} /.{x->1/3*x}, x]
    RecurrenceTable[{3*n^2*a[n] - 4*(28*n^2-28*n+9)*a[n-1] + 64*(4*n-5)*(4*n-3)*a[n-2] == 0, a[0]==1, a[1]==12}, a, {n,0,1000}]

Formula

3*n^2*a(n) - 4*(28*n^2-28*n+9)*a(n-1) + 64*(4*n-5)*(4*n-3)*a(n-2) = 0.
For n > 0, a(n) mod 3 = 0 (conjecture, tested up to n=3*10^6).
From Bradley Klee, May 30 2023: (Start)
The defining ODE can be derived from the following Weierstrass data:
g2 = (3/16)*(256 - 576*x + 405*x^2 - 81*x^3);
g3 = (1/64)*(4096 - 13824*x + 17496*x^2 - 9963*x^3 + 2187*x^4);
which determine an elliptic surface with four singular fibers. (End)

A318495 Scaled g.f. T(u) = Sum_{n>=0} a(n)*(u/16)^n satisfies 5*(21*u-16)*T + d/du( 4*u*(u-1)*(27*u-32)*T') = 0, and a(0)=1; sequence gives a(n).

Original entry on oeis.org

1, 10, 120, 1540, 20500, 279480, 3876600, 54496200, 774468900, 11107261000, 160553895040, 2336799457200, 34219387524400, 503846306168800, 7455357525594000, 110811908027490960, 1653792126235140900, 24774309852363829800, 372404448149589213600
Offset: 0

Views

Author

Bradley Klee, Aug 27 2018

Keywords

Comments

The linked document "Proof Certificate" gives data for verifying that period function T(u) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_5 symmetry (also cf. Examples and A318496).

Examples

			Period function T_{I}(w): Take T_{C5}(u) and T_{C3}(v) from A318495 and A318496 respectively. Set (u,v)=(1-w,w+5/27), with u in [0,1], v in [0,5/27], and w in [-5/27,1]. Define piecewise function T_{I}(w) = T_{C5}(1-w) if w in [0,1] or T_{I}(w) = T_{C3}(w+5/27) if w in [-5/27,0].
Geometric Singular Points: Construct a family of algebraic sphere curves by intersecting a sphere 1=X^2+Y^2+Z^2 with the icosahedral surface w=Z^6 - 5*(X^2+Y^2)*Z^4 + 5*(X^2+Y^2)^2*Z^2 - 2*(X^4-10*X^2*Y^2+5*Y^4)*X*Z. Six icosahedron vertex axes intersect the sphere in twelve circular points with w=1. Ten dodecahedron vertex axes intersect the sphere in twenty circular points with w=-5/27. Fifteen icosidodecahedron vertex axes intersect the sphere in thirty hyperbolic points with w=0.
		

Crossrefs

Cf. A318496. Periods: A186375, A318245, A318417.

Programs

  • GAP
    a:=[1,10];; for n in [3..20] do a[n]:=(1/(2*(n-1)^2))*(( (59*(n^2-3*n+2)+20)*a[n-1]-(12*(6*n-13)*(6*n-11))*a[n-2])); od; a; # Muniru A Asiru, Sep 24 2018
  • Mathematica
    RecurrenceTable[{2 n^2 a[n] - (59 n^2 - 59 n + 20) a[n - 1] + 12 (6 n - 7) (6 n - 5) a[n - 2] == 0, a[0] == 1, a[1] == 10}, a, {n, 0, 1000}]

Formula

2*n^2*a(n) - (59*n^2-59*n+20)*a(n-1) + 12*(6*n-7)*(6*n-5)*a(n-2) = 0.
For n > 0, a(n) mod 10 = 0 (conjecture, tested up to n=10^6).
From Bradley Klee, May 30 2023: (Start)
The defining ODE can be derived from the following Weierstrass data:
g2 = (243/256)*(256-640*x+520*x^2-135*x^3);
g3 = (729/8192)*(8192-30720*x+44160*x^2-29680*x^3+8775*x^4-729*x^5);
which determine an elliptic surface with four singular fibers. (End)
G.f.: hypergeom([1/12, 5/12],[1],1728*x^5*(27*x-2)^3*(16*x-1)^2/(2160*x^3-520*x^2+40*x-1)^3)/(1-40*x+520*x^2-2160*x^3)^(1/4). - Mark van Hoeij, Dec 13 2024

A318496 Scaled g.f. T(v) = Sum_{n>=0} a(n)*(v/16)^n satisfies 15*(189*v-80)*T + d/dv(4*v*(27*v-5)*(27*v-32)*T') = 0, and a(0)=1; sequence gives a(n).

Original entry on oeis.org

1, 30, 1440, 85260, 5606100, 391231080, 28360117800, 2110794125400, 160187289344100, 12339496371120600, 961855480344860640, 75700880007230883600, 6005580964527420946800, 479651805879329497831200, 38529018420812424368031600, 3110295017383730347887664560
Offset: 0

Views

Author

Bradley Klee, Aug 27 2018

Keywords

Comments

Period function T(v) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_3 symmetry. For precise definitions, pictures, a proof certificate, and more information, see A318495.

Crossrefs

Cf. A318495. Periods: A186375, A318245, A318417.

Programs

  • GAP
    a:=[1,30];; for n in [3..20] do a[n]:=(1/(10*(n-1)^2))*(3*(333*(n^2-3*n+2)+100)*a[n-1]-(324*(6*n-13)*(6*n-11)*a[n-2])); od; a; # Muniru A Asiru, Sep 24 2018
  • Mathematica
    RecurrenceTable[{10 n^2 a[n] - 3 (333 n^2 - 333 n + 100) a[n-1] + 324 (6*n - 7) (6 n - 5) a[n-2] == 0, a[0] == 1, a[1] == 30}, a, {n, 0, 15}]

Formula

10*n^2*a(n) - 3*(333*n^2-333*n+100)*a(n-1) + 324*(6*n-7)*(6*n-5)*a(n-2) = 0.
For n > 0, a(n) mod 30 = 0 (conjecture, tested up to n=10^6).
G.f.: hypergeom([1/12, 5/12],[1],-1728*(432*x-5)^2*x^3*(27*x-2)^5/(1-120*x+3240*x^2+174960*x^3)^3)/(1-120*x+3240*x^2+174960*x^3)^(1/4). - Mark van Hoeij, Dec 13 2024
Showing 1-5 of 5 results.