A068553 -id:A068553 - cp's OEIS frontend

A120101 Triangle T(n,k) = lcm(1,...,2n+2)/((k+1)binomial(2*k+2,k+1)).

1, 6, 1, 30, 5, 1, 420, 70, 14, 3, 1260, 210, 42, 9, 2, 13860, 2310, 462, 99, 22, 5, 180180, 30030, 6006, 1287, 286, 65, 15, 360360, 60060, 12012, 2574, 572, 130, 30, 7, 6126120, 1021020, 204204, 43758, 9724, 2210, 510, 119, 28, 116396280, 19399380, 3879876, 831402, 184756, 41990, 9690, 2261, 532, 126

Offset: 0

Paul Barry, Jun 09 2006

The rows give the coefficients of polynomials arising in the integration of x^(2m)/sqrt(4-x^2), m >= 0.

			Triangle begins:
       1;
       6,     1;
      30,     5,     1;
     420,    70,    14,    3;
    1260,   210,    42,    9,   2;
   13860,  2310,   462,   99,  22,   5;
  180180, 30030,  6006, 1287, 286,  65, 15;
  360360, 60060, 12012, 2574, 572, 130, 30, 7;

Muniru A Asiru, Table of n, a(n) for n = 0..5050

First column is A119634. Second column is A051538. Main diagonal is A068553. Subdiagonal is A119636. Inverse is A120113. Row sums are A120106. Diagonal sums are A120107.

Cf. A000984, A003418.

Flat(List([0..9],n->List([0..n],k->Lcm(List([1..2*n+2],i->i))/((k+1)*Binomial(2*k+2,k+1))))); # Muniru A Asiru, Feb 26 2019

[Lcm([1..2*n+2])/((k+1)*(k+2)*Catalan(k+1)): k in [0..n], n in [0..12]]; // G. C. Greubel, May 03 2023

T:=(n,k)-> ilcm(seq(q,q=1..2*n+2))/((k+1)*binomial(2*k+2,k+1)): seq(seq(T(n,k),k=0..n),n=0..9); # Muniru A Asiru, Feb 26 2019

Table[LCM@@Range[2*n+2]/((k+1)*Binomial[2*k+2,k+1]), {n,0,12}, {k,0, n}]//Flatten (* G. C. Greubel, May 03 2023 *)

def A120101(n,k):
    return lcm(range(1,2*n+3))/((k+1)*(k+2)*catalan_number(k+1))
flatten([[A120101(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, May 03 2023

Number triangle T(n,k) = [k<=n] * lcm(1,...,2n+2)/((k+1)*binomial(2k+2, k+1)).

A048619 a(n) = LCM(binomial(n,0), ..., binomial(n,n)) / binomial(n,floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 3, 4, 2, 10, 5, 30, 15, 7, 7, 56, 28, 252, 126, 60, 30, 330, 165, 396, 198, 286, 143, 2002, 1001, 15015, 15015, 7280, 3640, 1768, 884, 15912, 7956, 3876, 1938, 38760, 19380, 406980, 203490, 99484, 49742, 1144066, 572033, 1961256, 980628

Offset: 0

Labos Elemer

			If n=10 then A002944(10)=2520, A001405(10)=252, the quotient a(10)=10.

Index entries for sequences related to lcm's

Cf. A001405, A002944.

[Lcm([1..n+1]) div (Floor((n+3)/2)*Binomial(n+1,Floor((n+3)/2))): n in [0..50]]; // Vincenzo Librandi, Jul 10 2019

Table[Apply[LCM, Binomial[n, Range[0, n]]]/Binomial[n, Floor[n/2]], {n, 0, 48}] (* Michael De Vlieger, Jun 29 2017 *)

{A048619(n) = lcm(vector(n+1, i, i)) / binomial(n+1, (n+1)\2) / ((n+2)\2);}

a(n) = A002944(n)/A001405(n).
a(n) = lcm(1..n+1)/(floor((n+3)/2)*binomial(n+1,floor((n+3)/2))). - Paul Barry, Jul 03 2006
a(n) = lcm(1,2,...,n+1) / (ceiling((n+1)/2)*binomial(n+1,floor((n+1)/2))) = A003418(n+1) / A100071(n+1). - Max Alekseyev, Oct 23 2015
a(n) = A263673(n+1) / A110654(n+1) = A180000(n+1) / A152271(n). - Max Alekseyev, Oct 23 2015
a(2*n-1) = A068553(n) = A068550(n)/n.

Definition corrected and a(0)=1 prepended by Max Alekseyev, Oct 23 2015

A068550 a(n) = lcm{1, ..., 2n} / binomial(2n, n).

Original entry on oeis.org

1, 1, 2, 3, 12, 10, 30, 105, 56, 252, 1260, 330, 1980, 2574, 2002, 15015, 240240, 61880, 15912, 151164, 38760, 406980, 4476780, 1144066, 13728792, 24515700, 6249100, 84362850, 21474180, 5462730, 81940950, 1270084725, 645122400

Offset: 0

N. J. A. Sloane, Mar 23 2002

Known to be always an integer.

Hojoo Lee, Re: LCM [1,2,..,N] > 2^{N-1}, NMBRTHRY Mailing List, Feb 18 2002.
Daniel Ropp, Solution to Problem 31, Crux Mathematicorum, Vol. 13, No. 10 (1987), p. 314.

Bisection of A180000 and A263673.

a[0] = 1; a[n_] := (LCM @@ Range[2*n])/Binomial[2*n, n]; Array[a, 33, 0] (* Amiram Eldar, Mar 06 2022 *)

a(n) = lcm([1..2*n])/binomial(2*n, n); \\ Michel Marcus, Mar 06 2022

a(n) = A099996(n) / A000984(n) = A003418(2*n) / A001405(2*n) = A180000(2*n) = A263673(2*n).

a(n) = n * A068553(n) = n * A048619(2*n-1).

a(0)=1 prepended by Max Alekseyev, Oct 23 2015

A119636 a(n) = lcm(1,...,2n+4)/((n+1)*binomial(2n+2, n+1)).

Original entry on oeis.org

6, 5, 14, 9, 22, 65, 30, 119, 532, 126, 690, 825, 594, 4147, 62062, 15015, 3640, 32708, 7956, 79458, 833340, 203490, 2337874, 4004231, 980628, 12738550, 3124550, 766935, 11113830, 166613265, 81940950

Offset: 0

Paul Barry, Jun 09 2006

Subdiagonal of A120101.

Muniru A Asiru, Table of n, a(n) for n = 0..3000

Cf. A068553.

List([0..40],n->Lcm(List([1..2*n+4]))/((n+1)*Binomial(2*n+2,n+1))); # Muniru A Asiru, Mar 04 2019

[Lcm([1..2*n+4])/((n+1)*Binomial(2*n+2, n+1)): n in [0..40]]; // G. C. Greubel, Mar 04 2019

Table[LCM@@Range[2n+4]/((n+1)Binomial[2n+2,n+1]),{n,0,30}] (* Harvey P. Dale, Jun 08 2018 *)

[lcm(range(1, 2*(n+2)+1))/((n+1)*binomial(2*n+2, n+1)) for n in (0..40)] # G. C. Greubel, Mar 04 2019

A156744 Coefficients for estimation of derivative from equally spaced numerical data using the Lagrange interpolating polynomial.

Original entry on oeis.org

-1, 0, 1, 1, -8, 0, 8, -1, -1, 9, -45, 0, 45, -9, 1, 3, -32, 168, -672, 0, 672, -168, 32, -3

Offset: 1

Bruce Boghosian, Feb 14 2009

An effective way to approximate the derivative of equally spaced numerical data is to differentiate its Lagrange interpolating polynomial. If y[x] is equally spaced data from x = -n to +n, its Lagrange interpolating polynomial P(x) has degree 2*n+1. Then P'(0) may be expressed as a weighted sum over the y[x]. This is the triangle of coefficients C[n,m] such that P'(0) = (1/d[n]) * Sum_{m=-n}^n C[n,m] y[m]. The denominator d[n] is given by sequence A099996. This is very useful in numerical analysis. For example, when n=1, this gives the centered difference approximation to the derivative.

			Irregular triangle begins:
  -1, 0, 1;
  1, -8, 0, 8, -1;
  -1, 9, -45, 0, 45, -9, 1;
  3, -32, 168, -672, 0, 672, -168, 32, -3;
  -2, 25, -150, 600, -2100, 0, 2100, -600, 150, -25, 2;  ...

When divided by sequence A099996, this triangle gives the coefficients needed to estimate derivatives from equally spaced numerical data using Lagrange interpolation. The first and last entry of each row of the triangle has absolute value lcm{1, 2, ..., 2*n}/n*binomial(2n, n), as seen in A068553.

facs[x_, j_, n_] := Product[If[k == j, 1, x - k], {k, -n, n}]
coefs[x_, n_] := Table[facs[x, j, n]/facs[j, j, n], {j, -n, n}]
d[n_] := Apply[LCM, Table[j, {j, 1, 2*n}]]
dCoefs[x_, n_] := d[n] D[coefs[y, n], y] /. {y -> x}
MatrixForm[Table[dCoefs[0, n], {n, 1, 5}]]

cp's OEIS Frontend

A120101 Triangle T(n,k) = lcm(1,...,2*n+2)/((k+1)*binomial(2*k+2,k+1)).

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A048619 a(n) = LCM(binomial(n,0), ..., binomial(n,n)) / binomial(n,floor(n/2)).

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A068550 a(n) = lcm{1, ..., 2n} / binomial(2n, n).

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A119636 a(n) = lcm(1,...,2n+4)/((n+1)*binomial(2n+2, n+1)).

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Programs

GAP

Magma

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Sage

A156744 Coefficients for estimation of derivative from equally spaced numerical data using the Lagrange interpolating polynomial.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A120101 Triangle T(n,k) = lcm(1,...,2n+2)/((k+1)binomial(2*k+2,k+1)).