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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.

Showing 1-5 of 5 results.

A361030 a(n) = 20160*(3*n)!/(n!*(n+3)!^2).

Original entry on oeis.org

560, 210, 504, 2352, 15840, 135135, 1361360, 15519504, 194699232, 2636552100, 38003792400, 577037174400, 9155656500480, 150853746558690, 2568167588473200, 44990491457326800, 808333317429976800, 14853124707775823700, 278470827854627007600, 5316261259042879236000
Offset: 0

Views

Author

Peter Bala, Mar 01 2023

Keywords

Comments

Row 2 of square array A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 60*A000984(n) is divisible by (n + 1)*(n + 2)*(n + 3) and the result 60*(2*n)!/(n!*(n+3)!) is the super ballot number A007272(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 20160*A006480(n) is divisible by ((n + 1)*(n + 2)*(n + 3))^2.
Equivalently, the central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that (1*3*5)*A000984(n+3) is divisible by (2*n + 1)*(2*n + 3)*(2*n + 5). The result is always an even integer. In fact, (1/2)*(1*3*5)/((2*n + 1)*(2n + 3)*(2n + 5))*A000984(n+3) = A007272(n).
Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that (1*2*4*5*7*8)*A006480(n+3) is divisible by (3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n + 5)*(3*n + 7)*(3*n + 8). The result is always an integer divisible by 3.

Links

Crossrefs

Row 2 of A361027. Cf. A006480, A007272, A361028, A361029, A361031, A361038.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 560 else 3*(3*n-1)*(3*n-2)/(n+3)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Formula

a(n) = 20160/((n+1)*(n+2)*(n+3))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5*7*8) * A006480(n+3)/((3*n + 1)*(3*n + 2)*(3*n + 4)*
(3*n + 5)*(3*n + 7)*(3*n + 8)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/5)*A007272(n)*A361038(n). Using this it can be shown that a(n) is always an integer.
a(n) = (1/3)*27^(n+3)*binomial(7/3, n+3)*binomial(8/3, n+3).
a(n) ~ sqrt(3)*10080*(27^n)/(Pi*n^7).
P-recursive: (n + 3)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 560.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(7 - 54*x)*A'(x) + (9 - 6*x)*A(x) - 5040 = 0, with A(0) = 560 and A'(0) = 210.

A361027 Table of generalized de Bruijn's numbers (A006480) read by ascending antidiagonals.

Original entry on oeis.org

2, 30, 3, 560, 20, 20, 11550, 210, 75, 210, 252252, 2772, 504, 504, 2772, 5717712, 42042, 4620, 2352, 4620, 42042, 133024320, 700128, 51480, 15840, 15840, 51480, 700128, 3155170590, 12471030, 656370, 135135, 81675, 135135, 656370, 12471030, 75957810500, 233716340, 9237800
Offset: 0

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Author

Peter Bala, Feb 28 2023

Keywords

Comments

The Catalan numbers A000108 are given by the formula Catalan(n) = (2*n)!/(n!*(n + 1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(r,n) = J(r) * (2*n)!/(n!*(n + r + 1)!), where J(r) = (2*r + 2)!/(2*(r + 1)!) = (2^r)*Product_{j = 0..r} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.
For this table we carry out an analogous construction using the de Bruijn numbers B(n) = (3*n)!/n!^3 = A006480(n) in place of the central binomial numbers. We define the generalized de Bruijn number B(r,n), r = 0, 1, 2, ..., by B(r,n) = F(r) * (3*n)!/(n!*(n + r + 1)!^2), where choosing F(r) = (3*r + 3)!/(3*(r + 1)!) = (3^r)*Product_{j = 0..r} (3*j + 1)*(3*j + 2) appears to produce integer values for these quantities. We have verified this for rows 0, 1, 2 and 3 of the table.
An alternative expression for the generalized de Bruijn numbers is B(r,n) = G(r,n) * B(n+r+1), where G(r) = (1/3)*Product_{j = 0..r} ( (3*j + 1)*(3*j + 2)/((3*n + 3*j + 1)*(3*n + 3*j + 2)) ).
The rows of the square array below are the sequences of generalized de Bruijn numbers {B(0,k)}, {B(1,k)}, {B(2,k)}, ....

Examples

			The square array with rows n >= 0 and columns k >= 0 begins:
  n\k|       0       1       2        3        4         5         6 ...
  ----------------------------------------------------------------------
   0 |       2       3      20      210     2772     42042    700128 ...
   1 |      30      20      75      504     4620     51480    656370 ...
   2 |     560     210     504     2352    15840    135135   1361360 ...
   3 |   11550    2772    4620    15840    81675    550550   4492488 ...
   4 |  252252   42042   51480   135135   550550   3006003  20271888 ...
   5 | 5717712  700128  656370  1361360  4492488  20271888  ...
  ...
As a triangle:
 Row
  0 |        2
  1 |       30       3
  2 |      560      20     20
  3 |    11550     210     75    210
  4 |   252252    2772    504    504   2772
  5 |  5717712   42042   4620   2352   4620   42042
  ...

References

  • N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.

Links

Crossrefs

A208881 (column 1), A361028(row 0), A361029(row 1), A361030(row 2), A361031(row 3).
Cf. A006480, A024488, A135573.

Programs

  • Maple
    # as a square array
T := proc (n,k) (1/3)*27^(n+k+1)*binomial(n+1/3, n+k+1)*binomial(n+2/3,
n+k+1); end proc:
for n from 0 to 10 do seq(T(n,k), k = 0..10); end do;
# as a triangle
T := proc (n,k) (1/3)*27^(n+k+1)*binomial(n+1/3, n+k+1)*binomial(n+2/3,
n+k+1); end proc:
for n from 0 to 10 do seq(T(n-k,k), k = 0..n); end do;

Formula

T(n,k) = (3*n + 3)!/(3*(n + 1)!) * (3*k)!/(k!*(k + n + 1)!^2), n, k >= 0.
T(n,k) = (1/3)*27^(n+1+k)*binomial(n+1/3, n+1+k)*binomial(n+2/3, n+1+k).
T(n,k) = (1/(2*Pi))^2 * 1/27^(n+k+1) * Integral_{x = 0..27} (27 - x)^(n+2/3)*x^(k-2/3) dx * Integral_{x = 0..27} (27 - x)^(n+1/3)*x^(k-1/3) dx.
P-recursive: (n + k + 1)^2*T(n,k) = 3*(3*k - 1)*(3*k - 2)*T(n,k-1) with T(n,0) = 1/(n+1)!^2 * (3*n + 3)!/(3*(n + 1)!).
(n + k + 1)^2*T(n,k) = 3*(3*n + 1)*(3*n + 2)*T(n-1,k) with T(0,k) = 2*(k + 1)*(3*k)!/(k + 1)!^3.
T(n,0) = A208881(n+1).

A361029 a(n) = 120*(3*n)!/(n!*(n+2)!^2).

Original entry on oeis.org

30, 20, 75, 504, 4620, 51480, 656370, 9237800, 140229804, 2259901800, 38230005450, 673210036800, 12262039956000, 229872375708480, 4417859720647350, 86767376381987400, 1736954330906081100, 35364582637561485000, 730985923118395894950, 15315895532004485418000
Offset: 0

Views

Author

Peter Bala, Mar 01 2023

Keywords

Comments

Row 1 of the square array A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result 6*(2n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 120*A006480(n) is divisible by ((n + 1)*(n + 2))^2, leading to the present sequence. Do these numbers have a combinatorial interpretation?

Crossrefs

Row 1 of A361027. Cf. A006480, A007054, A361028, A361030, A361031, A361037.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 30 else 3*(3*n-1)*(3*n-2)/(n+2)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);
  • Mathematica
    Table[120 (3n)!/(n!(n+2)!^2),{n,0,20}] (* Harvey P. Dale, Jul 02 2023 *)

Formula

a(n) = 120/((n+1)*(n+2))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5) * A006480(n+2)/((3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n +
5)), where A006480(n) = (3*n)!/n!^3.
a(n) = (10*binomial(3*n,n) - 7*binomial(3*n,n+1) + binomial(3*n,n+2)) * (3*binomial(2*n,n) - 4*binomial(2*n,n+1) + binomial(2*n,n+2)), shows that a(n) is an integer for all n.
a(n) = (1/3)*27^(n+2)*binomial(4/3, n+2)*binomial(5/3, n+2).
a(n) ~ sqrt(3)*60*27^n/(Pi*n^5).
P-recursive: (n + 2)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 30.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(5 - 54*x)*A'(x) + (4 - 6*x)*A(x) - 120 = 0, with A(0) = 30 and A'(0) = 20.

A361031 a(n) = (3^3)*(1*2*4*5*7*8*10*11)*(3*n)!/(n!*(n+4)!^2).

Original entry on oeis.org

11550, 2772, 4620, 15840, 81675, 550550, 4492488, 42325920, 446185740, 5148297000, 63985977000, 846321189120, 11802213457650, 172255143129300, 2615726247519000, 41127042052404000, 666874986879730860, 11114583114662181000, 189866473537245687000, 3316382259894423720000
Offset: 0

Views

Author

Peter Bala, Mar 01 2023

Keywords

Comments

Row 3 of A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 840*A000984(n) is divisible by (n + 1)*(n + 2)*(n + 3)*(n + 4) and the result 840*(2*n)!/(n!*(n+4)!) is the super ballot number A348893(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 6652800 * A006480(n) is divisible by ((n + 1)*(n + 2)*(n + 3)*(n + 4))^2.

Crossrefs

Row 3 of A361027. Cf. A006480, A348893, A361028, A361029, A361030.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 11550 else
3*(3*n-1)*(3*n-2)/(n+4)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Formula

a(n) = (3^3)*(1*2*4*5*7*8*10*11)/((n+1)*(n+2)*(n+3)*(n+4))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5*7*8*10*11) * A006480(n+4)/((3*n + 1)*(3*n + 2)*(3*n + 4)* (3*n + 5)*(3*n + 7)*(3*n + 8)*(3*n + 10)*(3*n + 11)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/3)*27^(n+4)*binomial(10/3, n+4)*binomial(11/3, n+4).
a(n) = (1/7)*A348893(n)*A361039(n). It can be shown from this that a(n) is always an integer.
a(n) ~ sqrt(3)*3326400*(27^n)/(Pi*n^9).
P-recursive: (n + 4)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 11550.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(9 - 54*x)*A'(x) + (16 - 6*x)*A(x) - 184800 = 0, with A(0) = 11550 and A'(0) = 2772.

A361033 a(n) = 3*(4*n)!/(n!*(n+1)!^3).

Original entry on oeis.org

3, 9, 280, 17325, 1513512, 162954792, 20193091776, 2768662192725, 409716429837000, 64358256798795960, 10605621798062141760, 1817833036248401270280, 321997225483126007438400, 58649494641569379926280000, 10941649720331183519046796800, 2084191938036600263793119045925
Offset: 0

Views

Author

Peter Bala, Mar 01 2023

Keywords

Comments

Row 0 of A361032.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that A000984(n) is divisible by n + 1 and the result (2*n)!/(n!*(n+1)!) is the n-th Catalan number A000108(n). Similarly, the numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 3*A008977(n) is divisible by (n + 1)^3, leading to the present sequence. Cf. A361028. Do these numbers have a combinatorial interpretation?
Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.

Crossrefs

Cf. A000108, A007226, A007228, A008977, A361028, A361032, A361034, A361035.

Programs

  • Maple
    seq(3*(4*n)!/(n!*(n+1)!^3), n = 0..20);
  • Mathematica
    Table[3 (4n)!/(n! ((n+1)!)^3),{n,0,15}] (* Harvey P. Dale, Jul 30 2024 *)

Formula

a(n) = 3*A008977(n)/(n+1)^3.
a(n) = (3/4)*A008977(n+1)/((4*n+1)*(4*n+2)*(4*n+3)).
a(n) = (1/2)*A007228(n)*A007226(n)*A000108(n).
P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+1)^3 * a(n-1) with a(0) = 3.
The o.g.f. A(x) satisfies the differential equation
x^3*(1 - 256*x)*A(x)''' + x^2*(6 - 1152*x)*A(x)'' + x*(7 - 816*x)*A(x)' + (1 - 24*x)*A(x) - 3 = 0 with A(0) = 3, A'(0) = 9 and A''(0) = 560.
a(n) ~ 3*sqrt(1/(2*Pi^3)) * 2^(8*n)/n^(9/2).
Showing 1-5 of 5 results.

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