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 51-57 of 57 results.

A328500 Number of inversion sequences of length n avoiding the consecutive pattern 102.

Original entry on oeis.org

1, 1, 2, 6, 22, 96, 492, 2902, 19350, 143918, 1181540, 10614698, 103589738, 1091367634, 12346368424, 149276823258, 1921099070062, 26220186000950, 378308908684300, 5753387612678314, 91988260677198002, 1542570178562361018, 27072325866355742048
Offset: 0

Views

Author

Vaclav Kotesovec and Juan S. Auli, Oct 17 2019

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, j, t) option remember; `if`(n=0, 1, add(
          `if`(i<=j or i>=t, b(n-1, i, j), 0), i=1..n))
        end:
    a:= n-> b(n, 0$2):
    seq(a(n), n=0..25);  # Alois P. Heinz, Oct 18 2019
  • Mathematica
    b[n_, j_, t_] := b[n, j, t] = If[n == 0, 1, Sum[If[i <= j || i >= t, b[n - 1, i, j], 0], {i, 1, n}]];
    a[n_] := b[n, 0, 0];
    a /@ Range[0, 25] (* Jean-François Alcover, Mar 01 2020, after Alois P. Heinz *)

Formula

a(n) ~ n! * c * d^n * n^alfa, where d = 1/A240885 = 1/(sqrt(2) * InverseErf(sqrt(2/Pi))), alfa = 0.294868853646259565..., c = 2.22826071050847602... - Vaclav Kotesovec, Oct 19 2019

A328501 Number of inversion sequences of length n avoiding the consecutive pattern 201.

Original entry on oeis.org

1, 1, 2, 6, 24, 118, 684, 4548, 34036, 282696, 2577936, 25589100, 274539856, 3164909164, 39006958856, 511759353776, 7120140764224, 104703385864788, 1622530610142744, 26425922582118000, 451264786489454168, 8062192403534869432, 150395837509736576208
Offset: 0

Views

Author

Vaclav Kotesovec and Juan S. Auli, Oct 17 2019

Keywords

Crossrefs

Formula

a(n) ~ n! * c * d^n * n^alfa, where d = 1/A240885 = 1/(sqrt(2) * InverseErf(sqrt(2/Pi))) = 0.783976931203547499124248654869812535747328200022..., alfa = 1.9218908815253415257398764962146978742409244378248756048362586275529..., c = 0.05831456121798260255226478044037424484656774525125436523149657... - Vaclav Kotesovec, Oct 18 2019

A328507 Number of inversion sequences of length n avoiding the consecutive pattern 101.

Original entry on oeis.org

1, 1, 2, 6, 23, 109, 619, 4113, 31352, 269841, 2589026, 27404677, 317265161, 3988181568, 54099618419, 787705115000, 12253696410675, 202831037178017, 3559585021719875, 66018657264425355, 1290284788431977106, 26505045303122642171, 570918508059059670322
Offset: 0

Views

Author

Vaclav Kotesovec and Juan S. Auli, Oct 17 2019

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n, j, t) option remember; `if`(n=0, 1, add(
          `if`(i<=j or i<>t, b(n-1, i, j), 0), i=1..n))
        end:
    a:= n-> b(n, 0$2):
    seq(a(n), n=0..25);  # Alois P. Heinz, Oct 18 2019
  • Mathematica
    b[n_, j_, t_] := b[n, j, t] = If[n == 0, 1, Sum[If[i <= j || i != t, b[n-1, i, j], 0], {i, 1, n}]];
    a[n_] := b[n, 0, 0];
    a /@ Range[0, 25] (* Jean-François Alcover, Mar 19 2020, after Alois P. Heinz *)

Formula

a(n) ~ n! * c / sqrt(n), where c = 2.48988835987151440021135203237... - Vaclav Kotesovec, Oct 19 2019

A324131 Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 1 and t = 123.

Original entry on oeis.org

1, 1, 2, 6, 24, 116, 657, 4260, 31144, 253400, 2271250, 22234380, 236042879, 2700973070, 33139335352, 433996381926, 6042468288640, 89124117755852, 1388234052651161, 22771513253008320, 392354340340237176, 7084700602143004688, 133785708212530414358, 2636998678988431607188
Offset: 0

Views

Author

N. J. A. Sloane, Feb 16 2019

Keywords

Examples

			From _Petros Hadjicostas_, Nov 01 2019: (Start)
In a permutation of [n] that contains the shuffle pattern s-k-t, where s = 1 and t = 123, k should be greater than the numbers in pattern s and the numbers in pattern t. (The numbers in each of the patterns s and t should be contiguous.) Clearly, for n = 0..4, all permutations of [n] avoid this shuffle pattern (since we need at least five numbers to get this pattern). Hence, a(n) = n! for n = 0..4.
For n = 5, the permutations of [n] that contain this shuffle pattern should have k = 5 and the last three numbers in these permutations (with pattern t) should be one of the choices 123, 124, 134, and 234. The corresponding permutations that contain this shuffle pattern are 45123, 35124, 25134, and 15234. Hence a(5) = 5! - 4 = 116. (End)
		

Crossrefs

Cf. A049774.

Formula

From Petros Hadjicostas, Oct 30 2019: (Start)
Let b(n) = A049774(n) = number of permutations avoiding a consecutive 123 pattern. Then a(n) = 2*a(n-1) - b(n-1) + Sum_{i = 1..n-1} binomial(n-1,i) * b(i) * a(n-1-i) for n >= 1 with a(0) = b(0) = 1. [See the recurrence for C_n on p. 220 of Kitaev (2005).]
E.g.f.: If A(x) is the e.g.f. of (a(n): n >= 0) and B(x) is the e.g.f. of (b(n): n >= 0), then A'(x) = (1 + B(x)) * A(x) - B(x) with A(0) = B(0) = 1. [Theorem 16, p. 219, in Kitaev (2005)] (End)

Extensions

More terms from Petros Hadjicostas, Oct 30 2019 using Kitaev's (2005) recurrence

A324135 Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 12 and t = 123.

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 710, 4815, 36650, 308778, 2850294, 28602468, 310041806, 3610879857, 44975227466, 596677473990, 8401332033264, 125140942951896, 1966223504686334, 32501786913873447, 563877339150924866, 10245134152041643818, 194553155073687332550, 3854328529787275833204
Offset: 0

Views

Author

N. J. A. Sloane, Feb 16 2019

Keywords

Crossrefs

Formula

Let b(n) = A049774(n) = number of permutations avoiding a consecutive 123 pattern. Then a(n) = Sum_{i = 0..n-1} binomial(n-1,i) (a(n-1-i) + b(i) * a(n-1-i) - b(n-1-i)) for n >= 1 with a(0) = b(0) = 1. [See the recurrence for C_n on p. 220 of Kitaev (2005).] - Petros Hadjicostas, Oct 30 2019

Extensions

More terms from Petros Hadjicostas, Oct 30 2019

A324137 Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 123 and t = 123.

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 720, 5020, 39790, 352470, 3445032, 36775404, 425282892, 5292245764, 70471602994, 999394962306, 15032677450752, 238984379214960, 4002966472631160, 70448661940661068, 1299478739106621670, 25067231918730741438, 504674373639695198712, 10584965637367018566180
Offset: 0

Views

Author

N. J. A. Sloane, Feb 16 2019

Keywords

Crossrefs

Formula

Let b(n) = A049774(n) = number of permutations avoiding a consecutive 123 pattern. Then a(n) = Sum_{i = 0..n-1} binomial(n-1,i) (2*b(i)*a(n-1-i) - b(i)*b(n-1-i)) for n >= 1 with a(0) = b(0) = 1. [See the recurrence for C_n on p. 220 of Kitaev (2005).] - Petros Hadjicostas, Oct 30 2019

Extensions

More terms from Petros Hadjicostas, Oct 31 2019

A324138 Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 123 and t = 132.

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 720, 5020, 39755, 351518, 3425572, 36419844, 419026188, 5182797757, 68535001302, 964404124479, 14383519018582, 226579159065496, 3758349089828472, 65466833442028670, 1194655878120996337, 22788580047064423474, 453513206778006345040
Offset: 0

Views

Author

N. J. A. Sloane, Feb 16 2019

Keywords

Examples

			From _Petros Hadjicostas_, Nov 01 2019: (Start)
In a permutation of [n] that contains the shuffle pattern s-k-t, where s = 123 and t = 132, k should be greater than the numbers in pattern s and the numbers in pattern t. (The numbers in each of the patterns s and t should be contiguous.) Clearly, for n = 0..6, all permutations of [n] avoid this shuffle pattern (since we need at least seven numbers to get this pattern). Hence, a(n) = n! for n = 0..6.
For n = 7, k should be equal to 7, and for the pattern s = 123 we have binomial(6,3) = 20 choices: 123, 124, 125, ..., 456. The corresponding permutations of [7] that contain this shuffle pattern are 1237465, 1247365, 1257364, ..., 4567132. Thus, a(7) = 7! - 20 = 5020. (End)
		

Crossrefs

Formula

Let b(n) = A049774(n) = number of permutations of [n] that avoid consecutive pattern s = 123 and c(n) = A111004(n) = number of permutations of [n] that avoid consecutive pattern t = 132. Then a(n) = Sum_{i = 0..n-1} binomial(n-1,i) * (b(i)*a(n-1-i) + c(i)*a(n-1-i) - b(i)*c(n-1-i)) for n >= 1 with a(0) = b(0) = c(0) = 1. [This follows from the recurrence for C_n on p. 220 in Kitaev (2005).] - Petros Hadjicostas, Nov 01 2019

Extensions

More terms from Petros Hadjicostas, Nov 01 2019
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