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 1-5 of 5 results.

A001756 a(n) = A059366(n,n-3) = A059366(n,3) for n >= 3, where the triangle A059366 arises from the expansion of a trigonometric integral.

Original entry on oeis.org

15, 60, 450, 4500, 55125, 793800, 13097700, 243243000, 5016886875, 113716102500, 2808787731750, 75071235739500, 2158298027510625, 66409170077250000, 2177272076104125000, 75769068248423550000, 2789248824895091934375, 108288483790044745687500
Offset: 3

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Author

Keywords

Comments

Previous name was: Expansion of an integral.

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 166-167.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[3]==15,a[n]==a[n-1]n (2n-7)/(n-3)},a,{n,20}] (* Harvey P. Dale, Nov 08 2011 *)
    Join[{c = 15}, Table[c = c*n*(2*n - 7)/(n - 3), {n, 4, 20}]] (* T. D. Noe, Aug 10 2012 *)

Formula

a(n) = 5*A007531(n)*A001147(n-2)/(2*(2*n-5)). - Philippe Deléham, Jun 26 2006
a(3) = 15, a(n) = a(n-1)*n*(2*n-7)/(n-3). - Philippe Deléham, Sep 19 2006
From Petros Hadjicostas, May 12 2020: (Start)
a(n) = n! * Sum_{k=0..n-3} (-1)^k * 2^(2*k-n) * binomial(n-3, k) * binomial(2*n-2*k, n-3) * binomial(n-2*k+3, n-k) for n >= 3. [Special case of a formula by Comtet, but corrected]
a(n) = 20 * binomial(2*n-6, n-3) * n!/2^n for n >= 3. [Special case of a formula due to Reinhard Zumkeller]
a(n) = binomial(-1/2, 3) * binomial(-1/2, n-3) * (-1)^n * n! * 2^n for n >= 3. (End)
a(n) ~ sqrt(2)*(5/16)*(2*n/e)^n. - Peter Luschny, May 13 2020

Extensions

More terms from Philippe Deléham, Sep 19 2006
Corrected and extended by Harvey P. Dale, Nov 08 2011
New name by Petros Hadjicostas, May 12 2020

A001757 Expansion of an integral: central elements of rows of triangle in A059366.

Original entry on oeis.org

1, 1, 2, 9, 54, 450, 4500, 55125, 771750, 12502350, 225042300, 4538353050, 99843767100, 2410513805700, 62673358948200, 1762688220418125, 52880646612543750, 1698056319002793750, 57733914846094987500, 2084194325944029048750
Offset: 0

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Author

Keywords

Comments

For information about the trigonometric integral whose expansion involves the triangle A059366, see my comments and examples there. - Petros Hadjicostas, May 13 2020

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 166-167.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Formula

a(n) = A000984(floor(n/2))*A000984(ceiling(n/2))*A000142(n)/A000079(n). [Corrected by Petros Hadjicostas, May 13 2020]
From Petros Hadjicostas, May 13 2020: (Start)
a(n) = A059366(n, floor(n/2)) = A059366(n, ceiling(n/2)).
a(2*n) = A283678(n). (End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Feb 08 2001
Corrected and extended by Sean A. Irvine, Nov 19 2012
a(0) = 1 prepended by Petros Hadjicostas, May 13 2020

A001194 a(n) = A059366(n,n-2) = A059366(n,2) for n >= 2, where the triangle A059366 arises in the expansion of a trigonometric integral.

Original entry on oeis.org

3, 9, 54, 450, 4725, 59535, 873180, 14594580, 273648375, 5685805125, 129636356850, 3217338674550, 86331921100425, 2490343877896875, 76844896803675000, 2525635608280785000, 88081541838792376875, 3248654513701342370625
Offset: 2

Views

Author

Keywords

Comments

Old name was: Expansion of an integral.

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 166-167.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Programs

  • Mathematica
    Table[3*n*(n-1)*(2*n-4)!/(2^(n-1)*(n-2)!),{n,2,20}] (* Vaclav Kotesovec, Jan 05 2014 *)

Formula

a(n) = (2*n - 1)*a(n-1) - 3*(n - 1)*(2*n - 7)!! for n > 3. - Sean A. Irvine, Mar 23 2012
a(n) = 3*n*(n-1)*(2*n-4)!/(2^(n-1)*(n-2)!) for n >= 2. - Vaclav Kotesovec, Jan 05 2014
a(n) = binomial(-1/2, 2) * binomial(-1/2, n-2) * (-1)^n * n! * 2^n for n >= 2. - Petros Hadjicostas, May 13 2020
a(n) ~ sqrt(2)*(3/8)*(2*n/e)^n. - Peter Luschny, May 14 2020

Extensions

More terms from Sean A. Irvine, Mar 22 2012
New name by Petros Hadjicostas, May 13 2020

A001193 a(n) = (n+1)*(2*n)!/(2^n*n!) = (n+1)*(2n-1)!!.

Original entry on oeis.org

1, 2, 9, 60, 525, 5670, 72765, 1081080, 18243225, 344594250, 7202019825, 164991726900, 4111043861925, 110681950128750, 3201870700153125, 99044533658070000, 3262279327362680625, 113987877673731311250, 4211218814057295665625, 164015890652757831187500
Offset: 0

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Author

Keywords

Comments

Solution to y' = A(x), y(0) = 0 satisfies 0 = x^2 + 2*y^2*x - y^2, where A(x) = e.g.f. - Michael Somos, Mar 11 2004

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 166-167.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

From Johannes W. Meijer, Nov 12 2009: (Start)
Equals the first right hand column of A167591.
Equals the first left hand column of A167594. (End)
Cf. A059366.

Programs

  • Maple
    f:= x-> x/sqrt(1-2*x): a:= n-> subs(x=0, (D@@(n+1))(f)(x)):
    seq(a(n), n=0..20); # Zerinvary Lajos, Apr 04 2009
    # second Maple program:
    a:= n-> (n+1)*doublefactorial(2*n-1):
    seq(a(n), n=0..23);  # Alois P. Heinz, May 13 2020
  • Mathematica
    Table[(n+1) (2*n-1)!!, {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Apr 14 2011 *)
  • PARI
    a(n)=if(n<0, 0, (n+1)*(2*n)!/(2^n*n!))

Formula

E.g.f.: (1-x)/(1-2*x)^(3/2) = d/dx (x/(1-2*x)^(1/2)).
a(n) = uppermost term in the vector (M(T))^n * [1,0,0,0,...], where T = Transpose and M = the production matrix:
1, 2;
1, 2, 3;
1, 2, 3, 4;
1, 2, 3, 4, 5;
...
- Gary W. Adamson, Jul 08 2011
G.f.: A(x) = 1 + 2*x/(G(0) - 2*x) ; G(k) = 1 + k + x*(k+2)*(2*k+1) - x*(k+1)*(k+3)*(2*k+3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 06 2011
G.f.: U(0)/2 where U(k) = 1 + (2*k+1)/(1 - x/(x + 1/U(k+1))) (continued fraction). - Sergei N. Gladkovskii, Sep 25 2012
From Peter Bala, Nov 07 2016 and May 14 2020: (Start)
a(n) = (n + 1)*(2*n - 1)/n * a(n-1) with a(0) = 1.
a(n) = 2*a(n-1) + (2*n - 3)*(2*n + 1)*a(n-2) with a(0) = 1, a(1) = 2.
(End)
a(n) = A059366(n+1, n) = A059366(n+1, 1). - Petros Hadjicostas, May 13 2020

Extensions

Better description from Wouter Meeussen, Mar 08 2001
More terms from James Sellers, May 01 2000

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

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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).

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 1-5 of 5 results.