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 1-2 of 2 results.

A094796 Triangle read by rows giving coefficients of polynomials arising in successive differences of central binomial numbers.

Original entry on oeis.org

1, 3, 1, 9, 15, 6, 27, 108, 135, 42, 81, 594, 1539, 1530, 456, 243, 2835, 12555, 25245, 22122, 6120, 729, 12393, 83835, 281475, 482436, 383292, 101520, 2187, 51030, 489888, 2466450, 6916833, 10546200, 7786692, 1980720
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Let D_0(n)=binomial(2*n,n) and D_{k+1}(n)=D_{k}(n+1)-D_{k}(n); then D_{k}(n)*(n+1)*(n+2)*...*(n+k) = binomial(2*n,n)*P_{k}(n) where P_{k} is a polynomial with integer coefficients of degree k.

Examples

			The third differences of the central binomial numbers are given by D_3(n) = binomial(2*n,n)*(n+1)*(n+2)*(n+3)*(27*n^3 + 108*n^2 + 135*n + 42) and the fourth row of the triangle is 27, 108, 135, 42.
From _M. F. Hasler_, Nov 15 2019: (Start)
The table reads:
  n  |  row(n)
  0  |    1
  1  |    3      1
  2  |    9     15       6
  3  |   27    108     135       42
  4  |   81    594    1539     1530      456
  5  |  243   2835   12555    25245    22122      6120
  6  |  729  12393   83835   281475   482436    383292    101520
  7  | 2187  51030  489888  2466450  6916833  10546200   7786692   1980720
  8  | 6561 201204 2602530 18329976 75981969 186899076 260520300 181218384 44634240
(End)

Crossrefs

Cf. A000984 (central binomial coefficients), A163771 (square array of central binomial coefficients and higher differences), A000244 (column k=0).
Main diagonal gives A098461.

Programs

  • Maple
    Dnk := proc(n,k)
    option remember;
    if k < 0 then
        0 ;
    elif k = 0 then
        binomial(2*n,n) ;
    else
        procname(n+1,k-1)-procname(n,k-1) ;
    end if;
end proc:
A094796 := proc(n,k)
    local xyvec,i,x ;
    xyvec := [] ;
    for i from 0 to n do
        xyvec := [op(xyvec),[i,Dnk(i,n)*mul(i+j,j=1..n)/Dnk(i,0)]] ;
    end do:
    CurveFitting[PolynomialInterpolation](xyvec,x) ;
    coeff(%,x,n-k) ;
end proc: # R. J. Mathar, Nov 19 2019
  • Mathematica
    Dnk[n_, k_] := Dnk[n, k] = Which[k < 0, 0, k == 0, Binomial[2*n, n], True, Dnk[n + 1, k - 1] - Dnk[n, k - 1]];
T[n_, k_] := Module[{xyvec, i, x , ip}, xyvec = {}; For[i = 0, i <= n, i++, AppendTo[xyvec, {i, Dnk[i, n]*Product[i + j, {j, 1, n}]/Dnk[i, 0]}]]; ip = InterpolatingPolynomial[xyvec, x]; Coefficient[ip, x, n - k]];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 01 2024, after R. J. Mathar *)
  • PARI
    apply( {A094796_row(n,D(n,k)=if(k,D(n+1,k-1)-D(n,k-1),binomial(2*n,n)))=Vec(polinterpolate([0..n],vector(n+1,k,D(k--,n)*(n+k)!/k!/binomial(2*k,k))))}, [0..8]) \\ M. F. Hasler, Nov 15 2019

Formula

T(n,0) = 3^n. T(n,1) = A027472(n+2) + 6*A027472(n+1). T(n,2) = 3*(2*A036217(n-2) + 15*A036217(n-3) + 18*A036217(n-4)). - R. J. Mathar, Nov 19 2019

Extensions

Corrected and edited by M. F. Hasler, following observations by R. J. Mathar and Don Reble, Nov 15 2019
More terms from Don Reble, Nov 15 2019

A368235 Triangle read by rows: n-th row polynomial equals the numerator of the rational function (-1)^n*f(x) * (d/dx)^n (1/f(x)), where f(x) = sqrt(x + x^2).

Original entry on oeis.org

1, 1, 2, 3, 8, 8, 15, 54, 72, 48, 105, 480, 864, 768, 384, 945, 5250, 12000, 14400, 9600, 3840, 10395, 68040, 189000, 288000, 259200, 138240, 46080, 135135, 1018710, 3333960, 6174000, 7056000, 5080320, 2257920, 645120, 2027025, 17297280, 65197440, 142248960, 197568000, 180633600, 108380160, 41287680, 10321920
Offset: 0

Views

Author

Peter Bala, Dec 18 2023

Keywords

Comments

Unsigned row reverse of A123516.
The row polynomials also occur on repeated integration of 1/sqrt(x + x^2). See the example section.

Examples

			Triangle begins
 n\k |     0       1        2        3        4        5       6
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0  |     1
  1  |     1       2
  2  |     3       8        8
  3  |    15      54       72       48
  4  |   105     480      864      768      384
  5  |   945    5250    12000    14400     9600     3840
  6  | 10395   68040   189000   288000   259200   138240   46080
 ...
Repeated integration of 1/f(x), where f(x) = sqrt(x + x^2):
Let I denote the integral operator h(x) -> Integral_{t = 0..x} h(t) dt.
Let g(x) = I(1/f(x)) = log(2*x + 1 + 2*f(x)). Then
(2^1 * 1!^2) * I^(2)(1/f(x)) = (2*x + 1)*g(x) - 2*f(x).
(2^2 * 2!^2) * I^(3)(1/f(x)) = (8*x^2 + 8*x + 3)*g(x) - 6*(2*x + 1)*f(x).
(2^3 * 3!^2) * I^(4)(1/f(x)) = (48*x^3 + 72*x^2 + 54*x + 15)*g(x) - 2*(44*x^2 + 44*x + 15)*f(x).
(2^4 * 4!^2) * I^(5)(1/f(x)) = (384*x^4 + 768*x^3 + 864*x^2 + 480*x + 105)*g(x) - 10*(2*x + 1)*(40*x^2 + 40*x + 21)*f(x).

Crossrefs

Cf. A001147 (column 1), 2*A161120 (column 2), A000165 (main diagonal) A014479 (first subdiagonal), 3*A286725 (second subdiagonal).
Cf. A059366, A098461, A123516, A156919, A177145, A286724, A331817.

Programs

  • Maple
    # sequence in triangular form
T := (n, k) -> n! * 2^(2*k-n) * binomial(n, k)*binomial(2*n-2*k, n-k):
for n from 0 to 8 do seq(T(n, k), k = 0..n) od;

Formula

T(n, k) = n! * 2^(2*k-n) * binomial(n, k) * binomial(2*n-2*k, n-k).
k*T(n, k) = (2*n^2)*T(n-1, k-1) for k >= 1 with T(n, 0) = (2*n - 1)!! = A001147(n).
T(n, 1) = 2*A161120(n).
T(n, n) = 2^n * n! = A000165(n); T(n+1, n) = 2^n * n! * (n+1)^2 = A014479(n);
T(n+2, n) = 3 * 2^(n-1)*(n+2)!*binomial(n+2, 2) = 3 * A286725(n).
More generally, T(n+r, n) = (2*r - 1)!! * A286724(n+r, r).
E.g.f.: Sum_{k >= 0} (1/2^k)*binomial(2*k, k)*t^k/(1 - 2*t*x)^(k+1) = 1 + (1 + 2*x)*t + (3 + 8*x + 8*x^2)*t^2/2! + (15 + 54*x + 72*x^2 + 48*x^3)*t^3/3! + ....
n-th row polynomial R(n, x) = (-2)^n*(x + x^2)^(n+1/2)*(d/dx)^n (1/sqrt(x + x^2)).
Recurrence for row polynomials:
R(n+1, x) = (2*x + 1)*(2*n + 1)*R(n, x) - 4*x*(x + 1)*n^2*R(n-1, x), with R(0, x) = 1.
R'(n, x) = 2*n^2 * R(n-1, x) for n >= 1.
Functional equation: R(n, -1 - x) = (-1)^n * R(n, x).
Conjecture: the zeros of the polynomial R(n, -x) lie on the vertical line Re(x) = 1/2 in the complex plane.
(-1)^n * x^n * R(n, (- 1 - x)/x) equals the n-th row polynomial of A123516.
(1 - x)^n * R(n, x/(1 - x)) equals the n-th row polynomial of A059366.
Let D denote the operator (1/x)*d/dx. Then D^(n+1)( arcsinh(x) ) = (-1)^n*R(n, x^2)/(x*sqrt(1 + x^2))^(2*n+1).
R(n, 1/2) = A331817(n); R(n, -1/2) = A177145(n+1);
(2^n) * R(n, 1/4) = A098461(n).
Alternating row sums R(n, -1) = (-1)^n * A001147(n).
Showing 1-2 of 2 results.

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