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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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A269124 Number of sequences with n copies each of 1,2,...,6 avoiding the pattern 12...6.

Original entry on oeis.org

1, 719, 7235651, 117108036719, 2169397142273171, 41489887418194798439, 791857392420720220446647, 14919265746950486383601562197, 276745258345325814650980975177139, 5055320871467502855510583170028558565, 91050643996464879303067913358954122530151
Offset: 0

Views

Author

Alois P. Heinz, Feb 19 2016

Keywords

Links

Crossrefs

Row n=6 of A269129.
Cf. A008979, A268842.

Formula

a(n) = A008979(n) - A268842(n).

A273628 a(n) = (7*n)!/((5*n)!*n!^2).

Original entry on oeis.org

1, 42, 6006, 1085280, 217567350, 46262007792, 10217700004512, 2317454130543552, 536022010184210550, 125863265857621191900, 29909151834298018538256, 7176685161839833601969280, 1735941935586019529116213920, 422752608090008019258722317800
Offset: 0

Views

Author

Peter Bala, Jul 15 2016

Keywords

Comments

This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n + k,n)*binomial(4*n - k,n) = (-1)^m*a(m) for n = 2*m. For similar results see A001451, A006480 and A273629. Note the related sums:
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n + k,n)*binomial(4*n + k,n) = (-1)^n*(2*n)!*(4*n)!/(n!^3*(3*n)!) = (-1)^n*binomial(2*n,n)*binomial(4*n,n) = (-1)^n*A000984(n)*A005810(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = (3*n)!/n!^3 = A006480(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n + k,n)*binomial(4*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n + k,n)*binomial(4*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)*binomial(4*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n).

Crossrefs

Cf. A000984, A001451, A005810, A006480, A008979, A245086, A273629.

Programs

  • Magma
    [Factorial(7*n) div (Factorial(5*n)*Factorial(n)^2): n in [0..15]]; // Vincenzo Librandi, Jul 16 2016
  • Maple
    seq((7*n)!/((5*n)!*n!^2), n = 0..20);
  • Mathematica
    Table[(7 n)!/((5 n)! n!^2), {n, 0, 13}] (* or *)
Table[Binomial[7 n, n] Binomial[6 n, n], {n, 0, 13}] (* Michael De Vlieger, Jul 15 2016 *)

Formula

a(n) = (7*n)!/((5*n)!*n!^2) = binomial(7*n,2*n)*binomial(2*n,n).
a(n) = binomial(7*n,n)*binomial(6*n,n) = [x^n](1 + x)^(7*n) * [x^n](1 + x)^(6*n).
It appears that a(n) = [x^n] F(x)^(42*n), where F(x) = 1 + x + 30*x^2 + 2280*x^3 + 232715*x^4 + 27800465*x^5 + 3661895341*x^6 + ... has all integer coefficients. Cf. A273629 and A008979.
Recurrence: 5*n^2*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)*a(n) = 7*(7*n - 1)*(7*n - 2)*(7*n - 3)*(7*n - 4)*(7*n - 5)*(7*n - 6)*a(n-1).
a(n) ~ 5^(-5*n-1/2)*7^(7*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 15 2016
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(6*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A273629 a(n) = (9*n)!/((7*n)!*n!^2).

Original entry on oeis.org

1, 72, 18360, 5920200, 2118223800, 803927196072, 316938365223480, 128313095514575400, 52976845635264939960, 22204947580777261872000, 9418997650746914743158360, 4034374193416822645489549632, 1741969558937890710303111545400
Offset: 0

Views

Author

Peter Bala, Jul 15 2016

Keywords

Comments

This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(4*n + k,n)*binomial(5*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. For similar results see A001451, A006480 and A273628.
Note the related sums:
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(4*n - k,n)*binomial(5*n - k,n) = binomial(2*n,n)*binomial(4*n,n) = A000984(n)*A005810(n);
Sum_{k = 0..2*n} (-1)^k*binomial(n,k)*binomial(4*n + k,n)*binomial(5*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(n,k)*binomial(4*n - k,n)*binomial(5*n - k,n) = binomial(2*n,n) = A000984(n).
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(4*n + k,n)*binomial(5*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(4*n - k,n)*binomial(5*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n).

Crossrefs

Cf. A000984, A001451, A004381, A005810, A006480, A008979, A169958, A245086, A273628.

Programs

  • Magma
    [Factorial(9*n)/(Factorial(7*n)*Factorial(n)^2): n in [0..40]]; // Vincenzo Librandi, Jul 17 2016
  • Maple
    seq((9*n)!/((7*n)!*n!^2), n = 0..20);
  • Mathematica
    Table[Factorial[9 n] / (Factorial[7 n] Factorial[n]^2), {n, 0, 20}] (* Vincenzo Librandi, Jul 17 2016 *)

Formula

a(n) = (9*n)!/((7*n)!*n!^2) = binomial(9*n,2*n)* binomial(2*n,n).
a(n) = binomial(8*n,n)*binomial(9*n,n) = A004381(n)*A169958(n).
a(n) = [x^n](1 + x)^(8*n) * [x^n] (1 + x)^(9*n).
It appears that a(n) = [x^n] F(x)^(72*n), where F(x) = 1 + x + 56*x^2 + 7700*x^3 + 1422008*x^4 + 307144278*x^5 + 73118586828*x^6 + ... has all integer coefficients. Cf. A273628 and A008979.
Recurrence: 7*n^2*(7*n - 1)*(7*n - 2)*(7*n - 3)*(7*n - 4)*(7*n - 5)*(7*n - 6)*a(n) = 9*(9*n - 1)*(9*n - 2)*(9*n - 3)*(9*n - 4)*(9*n - 5)*(9*n - 6)*(9*n - 7)*(9*n - 8)*a(n-1).
a(n) ~ 3^(18*n+1)*7^(-7*n-1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 15 2016
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(8*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024
Previous Showing 11-13 of 13 results.

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