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 1-3 of 3 results.

A278301 Number of permutations of length n in the class of juxtapositions of 321-avoiders with 21-avoiders.

Original entry on oeis.org

1, 1, 2, 6, 23, 98, 434, 1949, 8803, 39888, 181201, 825201, 3767757, 17249560, 79191480, 364585230, 1683208515, 7792546040, 36174065285, 168367375735, 785637327745, 3674914227120, 17230132657815, 80965662243526, 381275131584373, 1799105397745998
Offset: 0

Views

Author

Jakub Sliacan, Nov 17 2016

Keywords

Comments

a(n) is also the number of permutations of length n in the class of juxtapositions of 231-avoiders with 21-avoiders.

Examples

			There are 23 permutations of length 4 which can be expressed as a juxtaposition of a 321-avoider and a 21-avoider. Only 4321 is not expressable this way.
		

Crossrefs

The other two juxtapositions of Catalan and monotone classes are enumerated by A033321, A165538.

Programs

  • Mathematica
    e = ee /. Solve[ee == 1 + x/(1 - x) ee, ee][[1]];
    c = cc /. Solve[cc == 1 + x cc^2, cc][[1]];
    cb = ccb /. Solve[ccb == 1 + x/(1 - x) ccb^2, ccb][[2]];
    b = bb /. Solve[bb == x^2/(1 - x) + x c bb e, bb][[1]];
    m = mm /.
       Solve[mm ==
          x c mm cb + b e x/(1 - x) (cb - 1) + x^2/(1 - x) (cb - 1),
         mm][[1]];
    f = c + c m cb/(1 - x);
    CoefficientList[Series[f, {x, 0, 25}], x]
    Rest[CoefficientList[Series[(1 - (1 - 4 x)^(1/2) + x (-4 + (1 - 4 x)^(1/2) + ((-1 + 5 x)^(1/2)) / ((-1 + x)^(1/2))))/ (-2 x^2), {x, 0, 33}], x]] (* Vincenzo Librandi, Nov 18 2016 *)

Formula

G.f.: (1 - (1-4*x)^(1/2) + x*(-4 + (1-4*x)^(1/2) + ((-1+5*x)^(1/2)) / ((-1+x)^(1/2)))) / (-2*x^2).
a(n) ~ 5^(n+3/2) / (8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Nov 17 2016

A026876 T(2n+1, n+1), T given by A026758.

Original entry on oeis.org

1, 3, 11, 43, 174, 721, 3042, 13022, 56420, 246977, 1090846, 4856162, 21770665, 98217318, 445634370, 2032434475, 9313305870, 42861416044, 198038717055, 918368326483, 4273103732070, 19944292913477, 93356009312511, 438150836819153
Offset: 0

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Author

Keywords

Comments

The invert transform of this sequences (shifted) appears to be A165538 (conjectured). - Michael D. Weiner, Aug 04 2018

Crossrefs

Formula

G.f. f satisfies: x^6*f^4 + x^3*(4*x-1)*(x-1)*f^3 + x*(x-1)*(6*x^2-6*x+1)*f^2 + (4*x-1)*(x-1)^2*f + (x-1)^2 = 0 (conjectured). - Michael D. Weiner, Aug 04 2018

A326348 Number of permutations of length n in the class of juxtapositions of separable permutations with 21-avoiders.

Original entry on oeis.org

1, 1, 2, 6, 24, 115, 609, 3409, 19728, 116692, 701062, 4261581, 26146111, 161631115, 1005522262, 6289410686, 39525228204, 249427451071, 1579885391573, 10040587733693, 64004713573508, 409139527503760, 2622049900367018, 16843666877986873, 108438876033442579
Offset: 0

Views

Author

Robert Brignall, Sep 11 2019

Keywords

Examples

			There are a(5) = 115 permutations of length 5 which can be expressed as a juxtaposition of a separable permutation (avoiding 2413 and 3142) with an increasing permutation. These 5 cannot be expressed: 25143, 35142, 35241, 41532 and 42531.
		

Crossrefs

Other juxtapositions of algebraic classes with monotone ones are enumerated by A033321, A165538, and A278301.

Programs

  • Mathematica
    CoefficientList[Series[(Sqrt[1 - 6*x + x^2]*(2 - 4*x + x^2)*Sqrt[1 - 8*x + 8*x^2]) / (4*(1 - x)*(-2 + 7*x - 7*x^2 + x^3)) + (-10 + 54*x - 99*x^2 + 66*x^3 - 9*x^4 + Sqrt[1 - 6*x + x^2]*(-2 + 10*x - 15*x^2 + 7*x^3) + Sqrt[1 - 8*x + 8*x^2]*(2 - 6*x + x^2 + 6*x^3 - x^4))/(4*(1 - x)^2*(-2 + 7*x - 7*x^2 + x^3)), {x, 0, 30}], x] (* Vaclav Kotesovec, Jul 07 2024 *)

Formula

G.f.: (2-4*z+z^2)*x*y/(4*(1-z)*(-2+7*z-7*z^2+z^3)) + ((-2+10*z-15*z^2+7*z^3)*x + (2-6*z+z^2+6*z^3-z^4)*y - 10+54*z-99*z^2+66*z^3-9z^4)/(4*(1-z)^2*(-2+7*z-7*z^2+z^3)) where x=sqrt(1-6*z+z^2) and y=sqrt(1-8*z+8z^2).
a(n) ~ (63 + 8*sqrt(2) + 3*sqrt(41 + 40*sqrt(2))) * 2^(3*n/2 - 1) * (1 + sqrt(2))^(n - 1/2) / (sqrt(Pi) * (73 + 53*sqrt(2)) * n^(3/2)). - Vaclav Kotesovec, Jul 07 2024
Showing 1-3 of 3 results.