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 1-3 of 3 results.

A000828 E.g.f. cos(x)/(cos(x) - sin(x)).

Original entry on oeis.org

1, 1, 2, 8, 40, 256, 1952, 17408, 177280, 2031616, 25866752, 362283008, 5535262720, 91620376576, 1633165156352, 31191159799808, 635421069967360, 13753735117275136, 315212388819402752, 7625476699018231808
Offset: 0

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Comments

For a refinement of these numbers see A185896.
A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...|x_n|} = {1,2...,n}. Let x_1,...,x_n be a signed permutation. Then we say 0,x_1,...,x_n,0 is a snake of type S(n;0,0) when 0 < x_1 > x_2 < ... 0. For example, 0 4 -3 -1 -2 0 is a snake of type S(4;0,0). Then a(n) equals the cardinality of S(n;0,0) [Verges]. An example is given below. - Peter Bala, Sep 02 2011
Original name was: E.g.f. cos(x)*(cos(x)+sin(x)) /cos(2*x). - Arkadiusz Wesolowski, Jul 25 2012
Number of plane (that is, ordered) increasing 0-1-2 trees on n vertices where the vertices of outdegree 1 or 2 come in two colors. An example is given below. - Peter Bala, Oct 10 2012

Examples

			a(3) = 8: The eight snakes of type S(3;0,0) are
0 1 -2 3 0, 0 1 -3 2 0, 0 2 1 3 0, 0 2 -1 3 0, 0 2 -3 1 0,
0 3 1 2 0, 0 3 -1 2 0, 0 3 -2 1 0.
1 + x + 2*x^2 + 8*x^3 + 40*x^4 + 256*x^5 + 1952*x^6 + 17408*x^7 + ...
a(3) = 8: The eight increasing 0-1-2 trees on 3 vertices are
..1o (x2 colors)......1o (x2 colors)......1o (x2 colors).....
...|................./.\................./.\.................
..2o (x2 colors)...2o...o3.............3o...o2...............
...|
..3o
Totals.......................................................
...4......+...........2.........+.........2....=...8.........
		

Crossrefs

Programs

  • Maple
    A000828 := n -> (-1)^((n-1)*n/2)*4^n*(Euler(n,1/2)+Euler(n,1))/2: # Peter Luschny, Nov 25 2010
  • Mathematica
    a[n_] := (-1)^((n-1)*n/2)*4^n*(EulerE[n, 1/2] + EulerE[n, 1])/2; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Nov 22 2012, after Peter Luschny *)
  • Maxima
    a(n):=sum(if evenp(n+k) then (-1)^((n+k)/2)*sum(j!/n!*stirling2(n,j)*2^(n-j)*(-1)^(n+j-k)*binomial(j-1,k-1),j,k,n),k,1,n); /* Vladimir Kruchinin, Aug 18 2010 */
    
  • PARI
    my(x='x + O('x^30)); Vec(serlaplace(cos(x)/(cos(x)-sin(x)))) \\ Michel Marcus, Nov 21 2020

Formula

E.g.f.: 1/(1- tan(x)). - Emeric Deutsch, Sep 10 2001
a(n) = A000831(n)/2 for n>0. - Peter Luschny, Nov 25 2010
a(n) = Sum_{k=1..n, n+k is even} (-1)^((n+k)/2)*Sum_{j=k..n} j!/n!*Stirling2(n,j)*2^(n-j)*(-1)^(n+j-k)*binomial(j-1,k-1), n>0. - Vladimir Kruchinin, Aug 18 2010
a(n) = (-1)^((n^2-n)/2)*4^n*(E_{n}(1/2)+E_{n}(1))/2 for n >= 0, where E_{n}(x) is an Euler polynomial. - Peter Luschny, Nov 25 2010
From Peter Bala, Sep 02 2011: (Start)
a(n) = (2*i)^(n-1)*Sum_{k = 1..n} (-1)^(n-k)*k!* Stirling2(n,k) * ((1-i)/2)^(k-1), where i = sqrt(-1).
a(n) = 2^(n-1)*A000111(n) for n >= 1.
Let f(x) = 1+x^2 and define the effect of the operator D on a function g(x) by D(g(x)) = d/dx(f(x)*g(x)). Then for n >= 0, a(n+1) = D^n(1) evaluated at x = 1. (End)
From Sergei N. Gladkovskii, Dec 09 2011 - Dec 23 2013: (Start) Continued fractions:
E.g.f.: 1 + x/(G(0)-x); G(k) = 2*k + 1 - (x^2)/G(k+1).
E.g.f.: 1 + x/(U(0)-2*x) where U(k) = 4*k+1 + x/(1+x/(4*k+3 - x/(1- x/U(k+1)))).
E.g.f.: 1 + x/(U(0)-x) where U(k) = 2*k+1 - x^2/U(k+1).
G.f.: 1 + x/G(0) where G(k) = 1 - x*(2*k+2) - 2*x^2*(k+1)*(k+2)/G(k+1).
E.g.f.: 1 + x/T(0) where T(k) = 4*k+1 - x/(1 - x/(4*k+3 + x/(1 + x/T(k+1)))).
G.f.: 1 + x/Q(0) where Q(k) = 1 - 2*x*(2*k+1) - 2*x^2*(2*k+1)*(2*k+2)/(1 - 2*x*(2*k+2) - 2*x^2*(2*k+2)*(2*k+3)/Q(k+1)).
G.f.: 1 + x/(1-2*x)*T(0) where T(k) = 1 - 2*x^2*(k+1)*(k+2)/( 2*x^2*(k+1)*(k+2) - (1 - 2*x*(k+1))*(1 - 2*x*(k+2))/T(k+1)).
E.g.f.: T(0) where T(k) = 1 + x/(4*k+1 - x/(1 - x/( 4*k+3 + x/T(k+1)))). (End)
G.f.: 1 /(1 - 1*x /(1 - 1*x /(1 - 4*x /(1 - 2*6*x^2 /(1 - 6*x /(1 - 4*x /(1 - 4*x /(1 - 10*x /(1 - 5*12*x^2 /(1 - 12*x / ...)))))))))). - Michael Somos, May 12 2012
a(n) ~ n! * 2^(2*n+1)/Pi^(n+1). - Vaclav Kotesovec, Jun 21 2013
a(0) = a(1) = 1; a(n) = 2 * Sum_{k=1..n-1} binomial(n-1,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Nov 21 2020
From Peter Bala, Dec 04 2021: (Start)
F(x) = exp(2*x)*(exp(2*x) - 1)/(exp(4*x) + 1) = x + 2*x^2/2! - 8*x^3/3! - 40*x^4/4! + 256*x^5/5! + 1952*x^6/6! - - + + ... is the e.g.f. for the sequence [1, 2, -8, -40, 256, 1952, ...], a signed version of this sequence without the first term.
Let G(x) = x + 2*x^2 - 8*x^3 - 40*x^4 + 256*x^5 + 1952*x^6 - - + + ... be the corresponding o.g.f. We have the continued fraction representation G(x) = x/(1 - 2*x + 12*x^2/(1 + 20*x^2/(1 - 2*x + 56*x^2/(1 + 72*x^2/(1 - 2*x + ... + 4*n*(4*n-1)*x^2/(1 + 4*n*(4*n+1)*x^2/(1 - 2*x + ... ))))))).
The inverse binomial transform 1/(1 + x)*G(x/(1 + x)) = x - 11*x^3 + 361*x^5 - 24611*x^7 + - ... is a g.f. for a signed and aerated version of A000464. (End)

Extensions

Name changed by Arkadiusz Wesolowski, Jul 25 2012

A000281 Expansion of cos(x)/cos(2x).

Original entry on oeis.org

1, 3, 57, 2763, 250737, 36581523, 7828053417, 2309644635483, 898621108880097, 445777636063460643, 274613643571568682777, 205676334188681975553003, 184053312545818735778213457, 193944394596325636374396208563
Offset: 0

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Comments

a(n) is (2n)! times the coefficient of x^(2n) in the Taylor series for cos(x)/cos(2x).

Examples

			cos x / cos 2*x = 1 + 3*x^2/2 + 19*x^4/8 + 307*x^6/80 + ...
		

References

  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x / cos 2x and sin x / cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
  • I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
  • 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

  • Maple
    a := n -> (-1)^n*2^(6*n+1)*(Zeta(0,-2*n,1/8)-Zeta(0,-2*n,5/8)):
    seq(a(n), n=0..13); # Peter Luschny, Mar 11 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Cos[x]/Cos[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Oct 06 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n*=2; n! * polcoeff( cos(x + x * O(x^n)) / cos(2*x + x * O(x^n)), n))}; /* Michael Somos, Feb 09 2006 */

Formula

a(n) = Sum_{k=0..n} (-1)^k*binomial(2n, 2k)*A000364(n-k)*4^(n-k). - Philippe Deléham, Jan 26 2004
E.g.f.: Sum_{k>=0} a(k)x^(2k)/(2k)! = cos(x)/cos(2x).
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
G.f.: 1 / (1 - 1*3*x / (1 - 4*4*x / (1 - 5*7*x / (1 - 8*8*x / (1 - 9*11*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/E(0) where E(k) = 1 - 3*x - 16*x*k*(2*k+1) - 16*x^2*(k+1)^2*(4*k+1)*(4*k+3)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
G.f.: T(0)/(1-3*x), where T(k) = 1 - 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2/( 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2 - (32*x*k^2+16*x*k+3*x-1 )*(32*x*k^2+80*x*k+51*x -1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
From Peter Bala, Mar 09 2015: (Start)
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(4*k + 1)^2) = 1 + 3*x + 57*x^2 + 2763*x^3 + ....
We appear to have the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + .... See A093954.
Bisection of A001586. See also A188458 and A212435. Second row of A235605 (read as a square array).
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255883. (End)
From Peter Luschny, Mar 11 2015: (Start)
a(n) = ((-64)^n/((n+1/2)))*(B(2*n+1,7/8)-B(2*n+1,3/8)), B(n,x) Bernoulli polynomials.
a(n) = 2*(-16)^n*LerchPhi(-1, -2*n, 1/4).
a(n) = (-1)^n*Sum_{0..2*n} 2^k*C(2*n,k)*E(k), E(n) the Euler secant numbers A122045.
a(n) = (-4)^n*SKP(2*n,1/2) where SKP are the Swiss-Knife polynomials A153641.
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/8) - Zeta(-2*n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. (End)
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + x - 4*x/(1 - 12*x/(1 + x - 40*x/(1 - 56*x/(1 + x - ... - 4*n(4*n - 3)*x/(1 - 4*n(4*n - 1)*x/(1 + x - ...
G.f.: 1/(1 + 9*x - 12*x/(1 - 4*x/(1 + 9*x - 56*x/(1 - 40*x/(1 + 9*x - ... - 4*n(4*n - 1)*x/(1 - 4*n(4*n - 3)*x/(1 + 9*x - .... (End)
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L-series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)

A253165 a(n) = (-1)^n*2^(6*n+3)*(zeta(-2*n-1,1/2) - zeta(-2*n-1,1)), where zeta(a,z) is the generalized Riemann zeta function.

Original entry on oeis.org

1, 8, 256, 17408, 2031616, 362283008, 91620376576, 31191159799808, 13753735117275136, 7625476699018231808, 5192022022552652087296, 4258996468871236847403008, 4142655008190840426050093056, 4714505177821257067736657297408, 6206008749802659037752564348092416
Offset: 0

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Author

Peter Luschny, Mar 11 2015

Keywords

Crossrefs

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+3)*(Zeta(0,-2*n-1,1/2)-Zeta(0,-2*n-1, 1)):
    seq(a(n), n=0..14);
  • Mathematica
    f[n_] := (-1)^n*2^(6 n + 3) (Zeta[-2 n - 1, 1/2] - Zeta[-2 n - 1, 1]); Array[f, 15, 0] (* Robert G. Wilson v, Mar 11 2015 *)
    max = 20; Clear[g]; g[max + 2] = 1; g[k_] := g[k] = 1 - 4*x*(k+1)*(k+2)/(4*x*(k+1)*(k+2) - 1/g[k+1]); gf = g[0]; CoefficientList[Series[gf, {x, 0, max}], x] (* Vaclav Kotesovec, Jun 01 2015, after Sergei N. Gladkovskii *)

Formula

a(n) = (-1)^n*2^(4*n+1)*(E(2*n+1,1/2)-E(2*n+1,0)), where E(n,x) are the Euler polynomials.
a(n) = A000825(2*n+1).
a(n) = A000828(2*n+1).
a(n) = A000831(2*n+1)/2.
a(n) = A012393(n+1)/2.
G.f.: S(0), where S(k)= 1 - 4*x*(k+1)*(k+2)/(4*x*(k+1)*(k+2) - 1/S(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 28 2015
a(n) ~ (2*n+1)! * 2^(4*n+3) / Pi^(2*n+2). - Vaclav Kotesovec, Jun 01 2015
Showing 1-3 of 3 results.