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.

Previous Showing 21-26 of 26 results.

A115864 Legendre_P(n,2)*4^n.

Original entry on oeis.org

1, 8, 88, 1088, 14176, 190208, 2600704, 36030464, 504047104, 7104278528, 100726755328, 1435037302784, 20526579564544, 294599134674944, 4240277467168768, 61183611081064448, 884741809748967424
Offset: 0

Views

Author

Paul Barry, Feb 01 2006

Keywords

Comments

Central coefficients of (1+8x+12x^2)^n. In general, Jacobi_P(n,0,0,sqrt(m))(k*sqrt(m))^n=Legendre_P(n,sqrt(m))(k*sqrt(m))^n has g.f. 1/sqrt(1-2*k*m*x+k^2*x^2), e.g.f. exp(k*m*x)Bessel_I(0,k*sqrt(m(m-1))*x) and gives the central coefficients of (1+k*m*x+k^2*(m(m-1)/4)*x^2)^n.

Programs

Formula

G.f.: 1/sqrt(1-16x+16x^2);
E.g.f.: exp(8x)Bessel_I(0,2*sqrt(12)x);
a(n)=Jacobi_P(n,0,0,sqrt(4))*(2*sqrt(4))^n; a(n)=2^n*A069835(n).
D-finite with recurrence: n*a(n) +8*(1-2*n)*a(n-1) +16*(n-1)*a(n-2) =0. - R. J. Mathar, Nov 16 2011
a(n) ~ sqrt(18+12*sqrt(3))*(8+4*sqrt(3))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 19 2012
a(n) = 2^n*A069835(n). - R. J. Mathar, Jan 20 2020

A115865 a(n) = Legendre_P(n,2)*6^n.

Original entry on oeis.org

1, 12, 198, 3672, 71766, 1444392, 29623644, 615614256, 12918175974, 273112332552, 5808412280628, 124127223181776, 2663248527920124, 57334738304731536, 1237861064261885688, 26791929483836768352
Offset: 0

Views

Author

Paul Barry, Feb 01 2006

Keywords

Comments

Central coefficients of (1+12*x+27*x^2)^n. In general, Jacobi_P(n,0,0,sqrt(m))(k*sqrt(m))^n = Legendre_P(n,sqrt(m))(k*sqrt(m))^n has g.f. 1/sqrt(1-2*k*m*x+k^2*x^2), e.g.f. exp(k*m*x)Bessel_I(0,k*sqrt(m(m-1))*x) and gives the central coefficients of (1+k*m*x+k^2*(m(m-1)/4)*x^2)^n.

Programs

Formula

G.f.: 1/sqrt(1-24*x+36*x^2).
E.g.f.: exp(12*x)*Bessel_I(0,3*sqrt(12)x).
a(n) = Jacobi_P(n,0,0,sqrt(4))*(3*sqrt(4))^n.
a(n) = 3^n*A069835(n).
D-finite with recurrence: n*a(n) +12*(1-2*n)*a(n-1) +36*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 14 2011
a(n) ~ sqrt(18+12*sqrt(3))*(12+6*sqrt(3))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 19 2012

A363570 Expansion of (1 / sqrt(1 - 8*x + 4*x^2) - 1 / (1 - x)) / 3.

Original entry on oeis.org

0, 1, 7, 45, 295, 1981, 13545, 93829, 656311, 4625181, 32788657, 233567269, 1670457321, 11987269477, 86268665917, 622391877045, 4500029549911, 32598283556317, 236542093805025, 1719008077215205, 12509403045819505, 91143878730342021, 664816240262272237
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 17 2023

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[(1/Sqrt[1 - 8 x + 4 x^2] - 1/(1 - x))/3, {x, 0, nmax}], x]
    nmax = 22; CoefficientList[Series[Exp[x] (Exp[3 x] BesselI[0, 2 Sqrt[3] x] - 1)/3, {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[Binomial[n, k]^2 3^(k - 1), {k, 1, n}], {n, 0, 22}]
    Table[(2^n LegendreP[n, 2] - 1)/3, {n, 0, 22}]

Formula

E.g.f.: exp(x) * (exp(3*x) * BesselI(0,2*sqrt(3)*x) - 1) / 3.
a(n) = Sum_{k=1..n} binomial(n,k)^2 * 3^(k-1).
a(n) = (2^n * LegendreP(n,2) - 1) / 3.
a(n) = (A069835(n) - 1) / 3.

A382642 a(n) = Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k))^2 * 2^(n-k).

Original entry on oeis.org

1, 6, 112, 2784, 79716, 2478936, 81369856, 2774798592, 97345792804, 3490750940376, 127377525333312, 4714499194430592, 176563416839871504, 6678628406445775968, 254781841509308692992, 9791397137378344986624, 378713818451270226094884, 14731112080159997036570328
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 08 2025

Keywords

Comments

Diagonal of the rational function 1 / (1 - y - z - x*y - z*w - 2*x*z - x*y*w - 2*x*y*z*w).

Crossrefs

Programs

  • Mathematica
    Table[Sum[(Binomial[n, k] Binomial[n + k, k])^2 2^(n - k), {k, 0, n}], {n, 0, 17}]
    Table[2^n HypergeometricPFQ[{-n, -n, n + 1, n + 1}, {1, 1, 1}, 1/2], {n, 0, 17}]
    Table[SeriesCoefficient[1/(1 - y - z - x y - z w - 2 x z - x y w - 2 x y z w), {x, 0, n}, {y, 0, n}, {z, 0, n}, {w, 0, n}], {n, 0, 17}]

Formula

a(n) ~ sqrt(6 + 4*sqrt(2) + sqrt(137/2 + 97/sqrt(2))) * (10 + 8*sqrt(2) + 4*sqrt(14 + 10*sqrt(2)))^n / (4 * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Apr 08 2025

A157077 Triangle read by rows, coefficients of the Legendre polynomials P(n, x) times 2^n: T(n, k) = 2^n * [x^k] P(n, x), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 2, -2, 0, 6, 0, -12, 0, 20, 6, 0, -60, 0, 70, 0, 60, 0, -280, 0, 252, -20, 0, 420, 0, -1260, 0, 924, 0, -280, 0, 2520, 0, -5544, 0, 3432, 70, 0, -2520, 0, 13860, 0, -24024, 0, 12870, 0, 1260, 0, -18480, 0, 72072, 0, -102960, 0, 48620, -252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756
Offset: 0

Views

Author

Roger L. Bagula, Feb 22 2009

Keywords

Examples

			The term order is Q(x) = a_0 + a_1*x + ... + a_n*x^n. The coefficients of the first few polynomials in this order are:
{1},
{0, 2},
{-2, 0, 6},
{0, -12, 0, 20},
{6, 0, -60, 0, 70},
{0, 60, 0, -280, 0, 252},
{-20, 0, 420, 0, -1260, 0, 924},
{0, -280, 0, 2520, 0, -5544, 0, 3432},
{70, 0, -2520, 0, 13860, 0, -24024, 0, 12870},
{0, 1260, 0, -18480, 0, 72072, 0, -102960, 0, 48620},
{-252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756}.
.
From _Jon E. Schoenfield_, Jul 04 2022: (Start)
As a right-aligned triangle:
                                                             1;
                                                     0,      2;
                                                 -2, 0,      6;
                                         0,     -12, 0,     20;
                                      6, 0,     -60, 0,     70;
                              0,     60, 0,    -280, 0,    252;
                         -20, 0,    420, 0,   -1260, 0,    924;
                  0,    -280, 0,   2520, 0,   -5544, 0,   3432;
              70, 0,   -2520, 0,  13860, 0,  -24024, 0,  12870;
        0,  1260, 0,  -18480, 0,  72072, 0, -102960, 0,  48620;
  -252, 0, 13860, 0, -120120, 0, 360360, 0, -437580, 0, 184756. (End)
		

Crossrefs

Programs

  • Maple
    with(orthopoly):with(PolynomialTools): seq(print(CoefficientList (2^n*P(n, x), x,termorder=forward)),n=0..10); # Peter Luschny, Dec 18 2014
  • Mathematica
    Table[CoefficientList[2^n*LegendreP[n, x], x], {n, 0, 10}]; Flatten[%]
  • PARI
    tabl(nn) = for (n=0, nn, print(Vecrev(2^n*pollegendre(n)))); \\ Michel Marcus, Dec 18 2014
    
  • Sage
    def A157077_row(n):
        if n==0: return [1]
        T = [c[0] for c in (2^n*gen_legendre_P(n, 0, x)).coefficients()]
        return [0 if is_odd(n+k) else T[k//2] for k in (0..n)]
    for n in range(9): print(A157077_row(n)) # Peter Luschny, Dec 19 2014

Formula

Row sums are 2^n.
From Peter Luschny, Dec 19 2014: (Start)
T(n,0) = A126869(n).
T(n,n) = A000984(n).
T(n,1) = (-1)^floor(n/2)*A005430(floor(n/2)+1) if n is odd else 0.
Let Q(n, x) = 2^n*P(n, x).
Q(n,0) = (-1)^floor(n/2)*A126869(floor(n/2)) if n is even else 0.
Q(n,1) = A000079(n).
Q(n,2) = A069835(n).
Q(n,3) = A084773(n).
Q(n,4) = A098269(n).
Q(n,5) = A098270(n). (End)
From Fabián Pereyra, Jun 30 2022: (Start)
n*T(n,k) = 2*(2*n-1)*T(n-1,k-1) - 4*(n-1)*T(n-2,k).
T(n,k) = (-1)^floor((n-k)/2)*binomial(n+k,k)*binomial(n,floor((n-k)/2))*(1+(-1)^(n-k))/2.
O.g.f.: A(x,t) = 1/sqrt(1-4*x*t+4*x^2) = 1 + (2*t)*x + (-2+6*t^2)*x^2 + (-12*t+20*t^3)*x^3 + (6-60*t^2+70*t^4)*x^4 + .... (End)

Extensions

Name clarified and edited by Peter Luschny, Dec 18 2014

A382934 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+k,k) * binomial(n+2*k,k) * 2^(n-k).

Original entry on oeis.org

1, 8, 142, 3188, 79306, 2091128, 57251944, 1609275536, 46123258714, 1341870616928, 39505611952852, 1174352843125976, 35189447673190864, 1061579548438995776, 32210037668484980992, 982173609216589910528, 30079350892561552670554, 924711257106480733093616, 28524228913983070512002044
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 09 2025

Keywords

Comments

Diagonal of the rational function 1 / (1 - x - y - z - 2*x*y*z).

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] Binomial[n + k, k] Binomial[n + 2 k, k] 2^(n - k), {k, 0, n}], {n, 0, 18}]
    Table[2^n HypergeometricPFQ[{n/2 + 1/2, n/2 + 1, -n}, {1, 1}, -2], {n, 0, 18}]
    Table[SeriesCoefficient[1/(1 - x - y - z - 2 x y z), {x, 0, n}, {y, 0, n}, {z, 0, n}], {n, 0, 18}]

Formula

a(n) ~ sqrt(1/4 + sqrt(13)*cosh(arccosh(47/13^(3/2))/3)/6) * (1 + 2*cosh(arccosh(2)/3))^(3*n) / (Pi*n). Equivalently, a(n) ~ (1 + (2 - sqrt(3))^(1/3) + (2 + sqrt(3))^(1/3))^(3*n) / (sqrt(6*((2 - sqrt(3))^(1/3) + (2 + sqrt(3))^(1/3) - 2))*Pi*n). - Vaclav Kotesovec, Apr 17 2025
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