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.

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A000464 Expansion of e.g.f. sin(x)/cos(2*x).

Original entry on oeis.org

1, 11, 361, 24611, 2873041, 512343611, 129570724921, 44110959165011, 19450718635716001, 10784052561125704811, 7342627959965776406281, 6023130568334172003579011, 5858598896811701995459355761, 6667317162352419006959182803611, 8776621742176931117228228227924441
Offset: 0

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Comments

From Peter Bala, Dec 22 2021: (Start)
Conjectures:
1) Taking the sequence (a(n))n>=1 modulo an integer k gives a purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 21 begins [11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, ...] with an apparent period of length 6, which divides phi(21) = 12.
2) For i >= 0, define a_i(n) = a(n+i). Then for each i the Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients.
3) a(m*n) == a(m)^n (mod 2^k) for k = 2*v_2(m) + 4, where v_p(i) denotes the p-adic valuation of i.
4)(i) a(2*m*n) == a(n)^(2*m) (mod 2^k) for k = v_2(m) + 4
(ii) a((2*m+1)*n) == a(n)^(2*m+1) (mod 2^k) for k = v_2(m) + 4. (End)

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag.
  • 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

Row 2 of A235606.
Cf. A064073. Bisection of A000822, A001586.

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+4)*(Zeta(0, -2*n-1, 5/8)-Zeta(0, -2*n-1, 7/8)):
    seq(a(n), n=0..12); # Peter Luschny, Oct 15 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Sin[x]/Cos[2x],{x,0,nn}],x] Range[0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Mar 23 2012 *)
    nmax = 15; km0 = 10; d[n_, km_] := Round[(2^(4n-1/2) (2n-1)! Sum[ JacobiSymbol[2, 2k+1]/(2k+1)^(2n), {k, 0, km}])/Pi^(2n)]; dd[km_] := dd[km] = Table[d[n, km], {n, 1, nmax}]; dd[km0]; dd[km = 2*km0]; While[dd[km] != dd[km/2, km = 2*km]]; A000464 = dd[km] (* Jean-François Alcover, Feb 08 2016 *)
  • PARI
    a(n)=if(n<0, 0, n+=n+1; n!*polcoeff(sin(x+x*O(x^n))/cos(2*x+x*O(x^n)),n)) /* Michael Somos, Feb 09 2006 */

Formula

E.g.f.: Sum_{k>=0} a(k)x^(2k+1)/(2k+1)! = sin(x)/cos(2x).
a(n) = (-1)^n*L(X,-2n+1) where L(X,z) is the Dirichlet L-function L(X,z) = Sum_{k>=0} X(k)/k^z and where X(k) is the Dirichlet character Legendre(k,2) which begins 1,0,-1,0,-1,0,1,0,1,0,-1,0,-1,0,1,0,1,0,-1,0.... - Benoit Cloitre, Mar 22 2009 [This Dirichlet character is A091337. - Jianing Song, Oct 22 2023]
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function:
2*exp(-t)*Sum_{n = 0..inf} (Product_{k = 1..n} (1-exp(-16*k*t))/Product_{k = 1..n+1} (1+exp(-(16*k-8)*t))) = 1 + 11*t + 361*t^2/2! + 24611*t^3/3! + .... For other sequences with generating functions of a similar type see A000364, A002105, A002439, A079144 and A158690.
a(n) = (-1)^(n+1)*L(-2*n-1), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s - 1/5^s + 1/7^s + - - + ... [Andrews et al., Theorem 5]. (End)
From Peter Bala, Jun 18 2009: (Start)
a(n) = (-1)^n*B_(2*n+2)(X)/(2*n+2), where B_n(X) denotes the X-Bernoulli number with X a Dirichlet character modulus 8 given by X(8*n+1) = X(8*n+7) = 1, X(8*n+3) = X(8*n+5) = -1 and X(2*n) = 0. See A161722 for the values of B_n(X).
For the theory and properties of the generalized Bernoulli numbers B_n(X) and the associated generalized Bernoulli polynomials B_n(X,x) see [Cohen, Section 9.4].
The present sequence also occurs in the evaluation of the finite sum of powers Sum_{i = 0..m-1} {(8*i+1)^n - (8*i+3)^n - (8*i+5)^n + (8*i+7)^n}, n = 1,2,... - see A151751 for details. (End)
G.f. 1/G(0) where G(k) = 1 + x - x*(4*k+3)*(4*k+4)/(1 - (4*k+4)*(4*k+5)*x/G(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Aug 11 2012
G.f.: 1/E(0) where E(k) = 1 - 11*x - 32*x*k*(k+1) - 16*x^2*(k+1)^2*(4*k+3)*(4*k+5)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
a(n) ~ (2*n+1)! * 2^(4*n+7/2) / Pi^(2*n+2). - Vaclav Kotesovec, May 03 2014
a(n) = (-1)^n*2^(6*n+4)*(Zeta(-2*n-1,5/8)-Zeta(-2*n-1,7/8)). - Peter Luschny, Oct 15 2015
From Peter Bala, May 11 2017: (Start)
G.f. A(x) = 1 + 11*x + 361*x^2 + ... = 1/(1 + x - 12*x/(1 - 20*x/(1 + x - 56*x/(1 - 72*x/(1 + x - ... - 4*n*(4*n - 1)*x/(1 - 4*n*(4*n + 1)*x/((1 + x) - ...))))))).
A(x) = 1/(1 + 9*x - 20*x/(1 - 12*x/(1 + 9*x - 72*x/(1 - 56*x/(1 + 9*x - ... - 4*n*(4*n + 1)*x/(1 - 4*n*(4*n - 1)*x/(1 + 9*x - ...))))))).
It follows that the first binomial transform of A(x) and the ninth binomial transform of A(x) have continued fractions of Stieltjes-type (S-fractions). (End)
a(n) = (-1)^(n+1)*4^(2*n+1)*E(2*n+1,1/4), where E(n,x) is the n-th Euler polynomial. Cf. A002439. - Peter Bala, Aug 13 2017
From Peter Bala, Dec 04 2021: (Start)
F(x) = exp(x)*(exp(2*x) - 1)/(exp(4*x) + 1) = x - 11*x^3/3! + 361x^5/5! - 24611*x^7/7! + ... is the e.g.f. for the sequence [1, 0, -11, 0, 361, 0, -24611, 0, ...], a signed and aerated version of this sequence.
The binomial transform exp(x)*F(x) = x + 2*x^2/2! - 8*x^3/3! - 40*x^4/4! + + - - is an e.g.f. for a signed version of A000828 (omitting the initial term). (End)
From Peter Bala, Dec 22 2021: (Start)
a(1) = 1, a(n) = (-1)^(n-1) - Sum_{k = 1..n} (-4)^k*C(2*n-1,2*k)*a(n-k).
a(n) == 1 (mod 10); a(5*n+1) == 0 mod(11);
a(n) == - 23^(n+1) (mod 108); a(n) == (7^2)*59^n (mod 144);
a(n) == 11^n (mod 240); a(n) == (11^2)*131^n (mod 360). (End)

Extensions

Better description, new reference, Aug 15 1995

A050970 Numerator of S(n)/Pi^n, where S(n) = Sum_{k=-inf..+inf} (4k+1)^(-n).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 61, 17, 277, 31, 50521, 691, 540553, 5461, 199360981, 929569, 3878302429, 3202291, 2404879675441, 221930581, 14814847529501, 4722116521, 69348874393137901, 56963745931, 238685140977801337, 14717667114151
Offset: 1

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Keywords

Comments

Reduced numerators of Favard constants.

Examples

			The first few values of S(n)/Pi^n are 1/4, 1/8, 1/32, 1/96, 5/1536, 1/960, ...
		

Crossrefs

Denominators: A068205. See also A050971.

Programs

  • Maple
    S := proc(n, k) option remember; if k = 0 then `if`(n = 0, 1, 0) else
    S(n, k - 1) + S(n - 1, n - k) fi end: EZ := n -> S(n, n)/(2^n * n!):
    A050970 := n -> numer(EZ(n-1)): seq(A050970(n), n=1..26); # Peter Luschny, Aug 02 2017
    # alternative
    A050970 := proc(n)
        if type(n,'even') then
            (-1)^(n/2)*2^(n-2)/(n-1)!*euler(n-1,0) ;
        else
            (-1)^((n-1)/2)*2^(n-2)/(n-1)!*euler(n-1,1/2) ;
        end if;
        %/2^n ;
        numer(%) ;
    end proc:
    seq(A050970(n),n=1..20) ; # R. J. Mathar, Jun 26 2024
  • Mathematica
    s[n_] := Sum[(4*k + 1)^(-n), {k, -Infinity, Infinity}]; a[n_] := Numerator[FullSimplify[s[n]/Pi^n]]; a[1] = 1; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Oct 25 2012 *)
    s[n_?EvenQ] := (-1)^(n/2-1)*(2^n-1)*BernoulliB[n]/(2*n!); s[n_?OddQ] := (-1)^((n-1)/2)*2^(-n-1)*EulerE[n-1]/(n-1)!; Table[s[n] // Numerator, {n, 1, 26}] (* Jean-François Alcover, May 13 2013 *)
    a[n_] := 4*Sum[((-1)^k/(2*k+1))^n, {k, 0, Infinity}] /. Pi -> 1 // Numerator; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Jun 20 2014 *)
    Table[4/(2 Pi)^n LerchPhi[(-1)^n, n, 1/2], {n, 21}] // Numerator (* Eric W. Weisstein, Aug 02 2017 *)
    Table[4/Pi^n If[Mod[n, 2] == 0, DirichletLambda, DirichletBeta][n], {n, 21}] // Numerator (* Eric W. Weisstein, Aug 02 2017 *)
  • PARI
    {a(n) = if( n<0, 0, numerator( polcoeff( 1 / (1 - tan(x/4 + x * O(x^n))), n)))}; /* Michael Somos, Nov 11 2014 */

Formula

There is a simple formula in terms of Euler and Bernoulli numbers.
a(2n) = A046976(n), a(2n+1) = A089171(n+1) (conjectured).
Numerator of coefficients of expansion of (sec(x/2) + tan(x/2) + 1)/2 in powers of x. - Sergei N. Gladkovskii, Nov 11 2014

Extensions

Entry revised by N. J. A. Sloane, Mar 24 2002

A296676 Expansion of e.g.f. 1/(1 - arctanh(x)).

Original entry on oeis.org

1, 1, 2, 8, 40, 264, 2048, 18864, 196992, 2330112, 30519552, 440998656, 6940852224, 118501542912, 2177222879232, 42886017982464, 900748014944256, 20107190510714880, 475167358873239552, 11854636521914695680, 311291779253770911744, 8583598112533040332800, 247944624171011289907200
Offset: 0

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Author

Ilya Gutkovskiy, Dec 18 2017

Keywords

Examples

			1/(1 - arctanh(x)) = 1 + x/1! + 2*x^2/2! + 8*x^3/3! + 40*x^4/4! + 264*x^5/5! + ...
		

Crossrefs

Programs

  • Maple
    S:= series(1/(1-arctanh(x)),x,41):
    seq(coeff(S,x,j)*j!,j=0..40); # Robert Israel, Dec 18 2017
    # second Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, add(`if`(j::odd,
          a(n-j)*binomial(n, j)*(j-1)!, 0), j=1..n))
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, Jun 22 2021
  • Mathematica
    nmax = 22; CoefficientList[Series[1/(1 - ArcTanh[x]), {x, 0, nmax}], x] Range[0, nmax]!
    nmax = 22; CoefficientList[Series[1/(1 + (Log[1 - x] - Log[1 + x])/2), {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    x='x+O('x^99); Vec(serlaplace(1/(1+(log(1-x)-log(1+x))/2))) \\ Altug Alkan, Dec 18 2017

Formula

E.g.f.: 1/(1 + (log(1 - x) - log(1 + x))/2).
a(n) ~ n! * 4*exp(2) * (exp(2)+1)^(n-1) / (exp(2)-1)^(n+1). - Vaclav Kotesovec, Dec 18 2017
a(n) = Sum_{k=0..n} k! * A111594(n,k). - Seiichi Manyama, Jun 30 2025

A320956 a(n) = A000110(n) * A000111(n). The exponential limit of sec + tan. Row sums of A373428.

Original entry on oeis.org

1, 1, 2, 10, 75, 832, 12383, 238544, 5733900, 167822592, 5859172975, 240072637440, 11388362495705, 618357843791872, 38057876106154882, 2632817442236631040, 203225803724876875315, 17390464322078045896704, 1640312648221489789841119, 169667967895669459925991424
Offset: 0

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Author

Peter Luschny, Nov 07 2018

Keywords

Comments

We say that the sequence S is the exponential limit of the function f relative to the kernel K if and only if the exponential generating functions
egf(n) = Sum_{k=0..n} K(n, k)*f(x*(n-k)) generate a family of sequences
T(n) = k -> (k!/n!)*[x^k] egf(n) which converge to S. Convergence here means that for every fixed k the terms T(n)(k) differ from S(k) only for finitely many indices.
The paradigmatic example is to set f(x) = exp(x), K(n, k) = !k*binomial(n, k) (!n is the subfactorial of n) and obtain for S the Bell numbers. This example is set forth in A320955.
Let D(f)(x) represent the derivative of f(x) with respect to x and (D^(n))(f) the n-th derivative of f. Then the exponential limit of f is B(n)*((D^(n))(f))(0) where B(n) is the n-th Bell number: ExpLim(f) = f(0), (D(f))(0), 2*((D^(2))(f))(0), 5*((D^(3))(f))(0), 15*((D^(4))(f))(0), 52*((D^(5))(f))(0), ... Since exp is a fixed point of D and exp(0) = 1 we have the identity ExpLim(exp)[n] = B(n). Similarly ExpLim(sin)[n] = B(n)*mod(n,2)*(-1)^binomial(n,2).
If we set f = sec + tan and K(n, k) = !k*binomial(n, k) the exponential limit is this sequence, a(n).

Examples

			Illustration of the convergence:
  [0] 1, 0, 0,  0,  0,   0,     0,      0,       0, ... A000007
  [1] 1, 1, 1,  2,  5,  16,    61,    272,    1385, ... A000111
  [2] 1, 1, 2,  8, 40, 256,  1952,  17408,  177280, ... A000828
  [3] 1, 1, 2, 10, 70, 656,  7442,  99280, 1515190, ... A320957
  [4] 1, 1, 2, 10, 75, 816, 11407, 194480, 3871075, ... A321394
  [5] 1, 1, 2, 10, 75, 832, 12322, 232560, 5325325, ...
  [6] 1, 1, 2, 10, 75, 832, 12383, 238272, 5693735, ...
  [7] 1, 1, 2, 10, 75, 832, 12383, 238544, 5732515, ...
  [8] 1, 1, 2, 10, 75, 832, 12383, 238544, 5733900, ...
		

Crossrefs

Cf. A000111 (n=1), A000828 (n=2), A320957 (n=3), A321394 (n=4).
Cf. A320955 (exp), A320962 (log(x+1)), this sequence (sec+tan), A320958 (arcsin), A320959 (arctanh).
Cf. A373428.

Programs

  • Maple
    ExpLim := proc(len, f) local kernel, sf, egf:
    sf := proc(n) option remember; `if`(n <= 1, 1 - n, (n-1)*(sf(n-1) + sf(n-2))) end:
    kernel := proc(n, k) option remember; binomial(n, k)*sf(k) end:
    egf := n -> add(kernel(n, k)*f(x*(n-k)), k=0..n):
    series(egf(len), x, len+2): seq(coeff(%, x, k)*k!/len!, k=0..len) end:
    ExpLim(19, sec + tan);
    # Alternative:
    explim := (len, f) -> seq(combinat:-bell(n)*((D@@n)(f))(0), n=0..len):
    explim(19, sec + tan);
    # Or:
    a := n -> A000110(n)*A000111(n): seq(a(n), n = 0..19);  # Peter Luschny, Jun 07 2024
  • Mathematica
    m = 20; CoefficientList[Sec[x] + Tan[x] + O[x]^m, x] * Range[0, m-1]! *
    BellB[Range[0, m-1]] (* Jean-François Alcover, Jun 19 2019 *)

Extensions

Name extended by Peter Luschny, Jun 07 2024

A185896 Triangle of coefficients of (1/sec^2(x))*D^n(sec^2(x)) in powers of t = tan(x), where D = d/dx.

Original entry on oeis.org

1, 0, 2, 2, 0, 6, 0, 16, 0, 24, 16, 0, 120, 0, 120, 0, 272, 0, 960, 0, 720, 272, 0, 3696, 0, 8400, 0, 5040, 0, 7936, 0, 48384, 0, 80640, 0, 40320, 7936, 0, 168960, 0, 645120, 0, 846720, 0, 362880, 0, 353792, 0, 3256320, 0, 8951040, 0, 9676800, 0, 3628800
Offset: 0

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Author

Peter Bala, Feb 07 2011

Keywords

Comments

DEFINITION
Define polynomials R(n,t) with t = tan(x) by
... (d/dx)^n sec^2(x) = R(n,tan(x))*sec^2(x).
The first few are
... R(0,t) = 1
... R(1,t) = 2*t
... R(2,t) = 2 + 6*t^2
... R(3,t) = 16*t + 24*t^3.
This triangle shows the coefficients of R(n,t) in ascending powers of t called the tangent number triangle in [Hodges and Sukumar].
The polynomials R(n,t) form a companion polynomial sequence to Hoffman's two polynomial sequences - P(n,t) (A155100), the derivative polynomials of the tangent and Q(n,t) (A104035), the derivative polynomials of the secant. See also A008293 and A008294.
COMBINATORIAL INTERPRETATION
A combinatorial interpretation for the polynomial R(n,t) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges].
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}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.
Let x_1,...,x_n be a signed permutation.
Then 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).
Let sc be the number of sign changes through a snake
... sc = #{i, 1 <= i <= n-1, x_i*x_(i+1) < 0}.
For example, the snake 0 4 -3 -1 -2 0 has sc = 1. The polynomial R(n,t) is the generating function for the sign change statistic on snakes of type S(n+1;0,0):
... R(n,t) = sum {snakes in S(n+1;0,0)} t^sc.
See the example section below for the cases n=1 and n=2.
PRODUCTION MATRIX
Define three arrays R, L, and S as
... R = superdiag[2,3,4,...]
... L = subdiag[1,2,3,...]
... S = diag[2,4,6,...]
with the indicated sequences on the main superdiagonal, the main subdiagonal and main diagonal, respectively, and 0's elsewhere. The array R+L is the production array for this triangle: the first row of (R+L)^n produces the n-th row of the triangle.
On the vector space of complex polynomials the array R, the raising operator, represents the operator p(x) - > d/dx (x^2*p(x)), and the array L, the lowering operator, represents the differential operator d/dx - see Formula (4) below.
The three arrays satisfy the commutation relations
... [R,L] = S, [R,S] = 2*R, [L,S] = -2*L
and hence give a representation of the Lie algebra sl(2).

Examples

			Table begins
  n\k|.....0.....1.....2.....3.....4.....5.....6
  ==============================================
  0..|.....1
  1..|.....0.....2
  2..|.....2.....0.....6
  3..|.....0....16.....0....24
  4..|....16.....0...120.....0...120
  5..|.....0...272.....0...960.....0...720
  6..|...272.....0..3696.....0..8400.....0..5040
Examples of recurrence relation
  T(4,2) = 3*(T(3,1) + T(3,3)) = 3*(16 + 24) = 120;
  T(6,4) = 5*(T(5,3) + T(5,5)) = 5*(960 + 720) = 8400.
Example of integral formula (6)
... Integral_{t = -1..1} (1-t^2)*(16-120*t^2+120*t^4)*(272-3696*t^2+8400*t^4-5040*t^6) dt = 2830336/1365 = -2^13*Bernoulli(12).
Examples of sign change statistic sc on snakes of type (0,0)
= = = = = = = = = = = = = = = = = = = = = =
.....Snakes....# sign changes sc.......t^sc
= = = = = = = = = = = = = = = = = = = = = =
n=1
...0 1 -2 0...........1................t
...0 2 -1 0...........1................t
yields R(1,t) = 2*t;
n=2
...0 1 -2 3 0.........2................t^2
...0 1 -3 2 0.........2................t^2
...0 2 1 3 0..........0................1
...0 2 -1 3 0.........2................t^2
...0 2 -3 1 0.........2................t^2
...0 3 1 2 0..........0................1
...0 3 -1 2 0.........2................t^2
...0 3 -2 1 0.........2................t^2
yields
R(2,t) = 2 + 6*t^2.
		

Crossrefs

Programs

  • Maple
    R = proc(n) option remember;
    if n=0 then RETURN(1);
    else RETURN(expand(diff((u^2+1)*R(n-1), u))); fi;
    end proc;
    for n from 0 to 12 do
    t1 := series(R(n), u, 20);
    lprint(seriestolist(t1));
    od:
  • Mathematica
    Table[(-1)^(n + 1)*(-1)^((n - k)/2)*Sum[j!*StirlingS2[n + 1, j]*2^(n + 1 - j)*(-1)^(n + j - k)*Binomial[j - 1, k], {j, k + 1, n + 1}], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Jul 22 2017 *)
  • PARI
    {T(n, k) = if( n<0 || k<0 || k>n, 0, if(n==k, n!, (k+1)*(T(n-1, k-1) + T(n-1, k+1))))};
    
  • PARI
    {T(n, k) = my(A); if( n<0 || k>n, 0, A=1; for(i=1, n, A = ((1 + x^2) * A)'); polcoeff(A, k))}; /* Michael Somos, Jun 24 2017 */

Formula

GENERATING FUNCTION
E.g.f.:
(1)... F(t,z) = 1/(cos(z)-t*sin(z))^2 = Sum_{n>=0} R(n,t)*z^n/n! = 1 + (2*t)*z + (2+6*t^2)*z^2/2! + (16*t+24*t^3)*z^3/3! + ....
The e.g.f. equals the square of the e.g.f. of A104035.
Continued fraction representation for the o.g.f:
(2)... F(t,z) = 1/(1-2*t*z - 2*(1+t^2)*z^2/(1-4*t*z -...- n*(n+1)*(1+t^2)*z^2/(1-2*n*(n+1)*t*z -....
RECURRENCE RELATION
(3)... T(n,k) = (k+1)*(T(n-1,k-1) + T(n-1,k+1)).
ROW POLYNOMIALS
The polynomials R(n,t) satisfy the recurrence relation
(4)... R(n+1,t) = d/dt{(1+t^2)*R(n,t)} with R(0,t) = 1.
Let D be the derivative operator d/dt and U = t, the shift operator.
(5)... R(n,t) = (D + DUU)^n 1
RELATION WITH OTHER SEQUENCES
A) Derivative Polynomials A155100
The polynomials (1+t^2)*R(n,t) are the polynomials P_(n+2)(t) of A155100.
B) Bernoulli Numbers A000367 and A002445
Put S(n,t) = R(n,i*t), where i = sqrt(-1). We have the definite integral evaluation
(6)... Integral_{t = -1..1} (1-t^2)*S(m,t)*S(n,t) dt = (-1)^((m-n)/2)*2^(m+n+3)*Bernoulli(m+n+2).
The case m = n is equivalent to the result of [Grosset and Veselov]. The methods used there extend to the general case.
C) Zigzag Numbers A000111
(7)... R_n(1) = A000828(n+1) = 2^n*A000111(n+1).
D) Eulerian Numbers A008292
The polynomials R(n,t) are related to the Eulerian polynomials A(n,t) via
(8)... R(n,t) = (t+i)^n*A(n+1,(t-i)/(t+i))
with the inverse identity
(9)... A(n+1,t) = (-i/2)^n*(1-t)^n*R(n,i*(1+t)/(1-t)),
where {A(n,t)}n>=1 = [1,1+t,1+4*t+t^2,1+11*t+11*t^2+t^3,...] is the sequence of Eulerian polynomials and i = sqrt(-1).
E) Ordered set partitions A019538
(10)... R(n,t) = (-2*i)^n*T(n+1,x)/x,
where x = i/2*t - 1/2 and T(n,x) is the n-th row po1ynomial of A019538;
F) Miscellaneous
Column 1 is the sequence of tangent numbers - see A000182.
A000670(n+1) = (-i/2)^n*R(n,3*i).
A004123(n+2) = 2*(-i/2)^n*R(n,5*i).
A080795(n+1) =(-1)^n*(sqrt(-2))^n*R(n,sqrt(-2)). - Peter Bala, Aug 26 2011
From Leonid Bedratyuk, Aug 12 2012: (Start)
T(n,k) = (-1)^(n+1)*(-1)^((n-k)/2)*Sum_{j=k+1..n+1} j! *stirling2(n+1,j) *2^(n+1-j) *(-1)^(n+j-k) *binomial(j-1,k), see A059419.
Sum_{j=i+1..n+1}((1-(-1)^(j-i))/(2*(j-i))*(-1)^((n-j)/2)*T(n,j))=(n+1)*(-1)^((n-1-i)/2)*T(n-1,i), for n>1 and 0
G.f.: 1/G(0,t,x), where G(k,t,x) = 1 - 2*t*x - 2*k*t*x - (1+t^2)*(k+2)*(k+1)*x^2/G(k+1,t,x); (continued fraction due to T. J. Stieltjes). - Sergei N. Gladkovskii, Dec 27 2013

A000816 E.g.f.: Sum_{n >= 0} a(n) * x^(2*n) / (2*n)! = sin(x)^2 / cos(2*x).

Original entry on oeis.org

0, 2, 40, 1952, 177280, 25866752, 5535262720, 1633165156352, 635421069967360, 315212388819402752, 194181169538675507200, 145435130631317935357952, 130145345400688287667978240, 137139396592145493713802493952
Offset: 0

Keywords

Programs

  • Mathematica
    Union[ Range[0, 26]! CoefficientList[ Series[ Sin[x]^2/Cos[ 2x], {x, 0, 26}], x]] (* Robert G. Wilson v, Apr 16 2011 *)
    Table[(-1)^(n + 1) 2^(2 n) I PolyLog[-2 n, I], {n, 1, 13}] (* Artur Jasinski, Mar 21 2022 *)
    With[{nn=30},Take[CoefficientList[Series[Sin[x]^2/Cos[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Oct 18 2024 *)
  • PARI
    {a(n) = local(m); if( n<0, 0, m = 2*n; m! * polcoeff( 1 / (2 - 1 / cos(x + x * O(x^m))^2) - 1, m))} /* Michael Somos, Apr 16 2011 */
    
  • Sage
    @CachedFunction
    def sp(n,x) :
        if n == 0 : return 1
        return -add(2^(n-k)*sp(k,1/2)*binomial(n,k) for k in range(n)[::2])
    def A000816(n) : return 0 if n == 0 else abs(sp(2*n,x)/2)
    [A000816(n) for n in (0..13)]   # Peter Luschny, Jul 30 2012

Formula

(1/2) * A002436(n), n > 0. - Ralf Stephan, Mar 09 2004
a(n) = 2^(2*n - 1) * A000364(n) except at n=0.
E.g.f.: sin(x)^2/cos(2x) = 1/Q(0) - 1/2; Q(k) = 1 + 1/(1-2*(x^2)/(2*(x^2)-(k+1)*(2k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2011
a(n) = A000819(n) unless n=0.
G.f.: (1/(G(0))-1)/2 where G(k) = 1 - 4*x*(k+1)^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 12 2013
G.f.: T(0)/2 - 1/2, where T(k) = 1 - 4*x*(k+1)^2/( 4*x*(k+1)^2 - 1/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 25 2013
E.g.f.: sin(x)^2/cos(2*x) = x^2/(1-2*x^2)*T(0), where T(k) = 1 - x^2*(2*k+1)*(2*k+2)/( x^2*(2*k+1)*(2*k+2) + ((k+1)*(2*k+1) - 2*x^2)*((k+2)*(2*k+3) - 2*x^2)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 25 2013
From Artur Jasinski, Mar 21 2022: (Start)
For n > 0:
a(n) = Pi^(2*n-1)*(-Psi(2*n,1/4) - (4^n)*(2^(2*n+1)-1)*Gamma(2*n+1)*Zeta(2*n+1)).
a(n) = (-1)^(n+1)*2^(2*n)*i*Li_(2*n,i) where i=sqrt(-1) and Li is polylogarithm function.
a(n) = (-64)^n*(zeta(-2*n,1/4)-zeta(-2*n,3/4)) where zeta is Hurwitz zeta function.
a(n) = (-16)^n*lerchphi(-1,-2*n,1/2). (End)

A012393 E.g.f. arctanh(tan(x)*tan(x)) (even powers only).

Original entry on oeis.org

0, 2, 16, 512, 34816, 4063232, 724566016, 183240753152, 62382319599616, 27507470234550272, 15250953398036463616, 10384044045105304174592, 8517992937742473694806016, 8285310016381680852100186112
Offset: 0

Author

Patrick Demichel (patrick.demichel(AT)hp.com)

Keywords

Comments

a(n) = 2*A000828(2*n-1). - corrected by Vaclav Kotesovec, Feb 08 2015

Examples

			arctanh(tan(x)*tan(x)) = (2/2!)*x^2 + (16/4!)*x^4 + (512/6!)*x^6 + (34816/8!)*x^8 + ...
		

Programs

  • Maple
    A012393 := n -> (-1)^(n*(2*n+1))*4^(2*n+1)*(euler(2*n+1,1/2) + euler(2*n+1,1)) end; # Peter Luschny, Nov 25 2010
  • Mathematica
    nn = 20; Table[(CoefficientList[Series[ArcTanh[Tan[x]^2], {x, 0, 2*nn}], x] * Range[0, 2*nn]!)[[n]], {n, 1, 2*nn+1, 2}] (* Vaclav Kotesovec, Feb 08 2015 *)

Formula

E.g.f.: log(sec(2*x))/2, cf. A000182. - Vladeta Jovovic, Jun 04 2005
a(n) = (-1)^(n(2n+1))4^(2n+1)(E{2n+1}(1/2)+E{2n+1}(1)); E_{n}(x) Euler polynomial. - Peter Luschny, Nov 25 2010
G.f.: 2*x/G(0) where G(k) = 1 - (2*k+2)*(2*k+4)*x/G(k+1) (continued fraction). - Sergei N. Gladkovskii, Oct 25 2012
G.f.: (2/G(0) - 1)*sqrt(-x), where G(k) = 1 + 1/(1 - 1/(1 + 1/(4*sqrt(-x)*(k+1)) - 1/G(k+1))) (continued fraction). - Sergei N. Gladkovskii, May 29 2013
G.f.: 2*x*T(0), where T(k) = 1 - x*(2*k+2)*(2*k+4)/(x*(2*k+2)*(2*k+4) - 1/T(k+1) ) (continued fraction). - Sergei N. Gladkovskii, Oct 12 2013
a(n) ~ 2^(4*n) * (2*n-1)! / Pi^(2*n). - Vaclav Kotesovec, Feb 08 2015

Extensions

a(0)=0 prepended by Joerg Arndt, Oct 26 2012

A000825 Expansion of cos x (1 + sin x ) /cos 2x.

Original entry on oeis.org

1, 1, 3, 8, 57, 256, 2763, 17408, 250737, 2031616, 36581523, 362283008, 7828053417, 91620376576, 2309644635483, 31191159799808, 898621108880097, 13753735117275136, 445777636063460643, 7625476699018231808
Offset: 0

Keywords

Crossrefs

Bisections are A000281 and (1/2) * A012393.

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[Cos[x] (1+Sin[x])/Cos[2x],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Nov 08 2013 *)

Formula

a(n) ~ n! * (sqrt(2) + 1 + (sqrt(2)-1)*(-1)^n) * 4^n / Pi^(n+1). - Vaclav Kotesovec, Jun 01 2015

A235132 E.g.f. 1/(1 - tan(3*x))^(1/3).

Original entry on oeis.org

1, 1, 4, 46, 568, 9976, 203104, 4995136, 140343808, 4493656576, 160609429504, 6356981099776, 275688520680448, 13008983675954176, 663382602064482304, 36360098005522825216, 2131554196360938815488, 133093201551208236875776, 8818123347826691244949504
Offset: 0

Author

Vaclav Kotesovec, Jan 03 2014

Keywords

Comments

Generally, for e.g.f. 1/(1-tan(p*x))^(1/p) is a(n) ~ n! * 2^(2*n+1/p) * n^((1-p)/p) * p^n / (Pi^(n+1/p) * Gamma(1/p)).

Crossrefs

Cf. A000828 (p=1), A235131 (p=2).

Programs

  • Mathematica
    CoefficientList[Series[1/(1 - Tan[3*x])^(1/3), {x, 0, 20}], x] * Range[0, 20]!

Formula

a(n) ~ n! * 2^(2*n+1/3) * 3^n / (Gamma(1/3) * Pi^(n+1/3) * n^(2/3)).

A331610 Expansion of e.g.f.: exp(1 / (1 - tan(x)) - 1).

Original entry on oeis.org

1, 1, 3, 15, 97, 777, 7379, 80983, 1007137, 13986289, 214383171, 3593224767, 65347120705, 1281151315641, 26928292883795, 603928982033863, 14392387319349697, 363135896514611041, 9669298448057196291, 270932711729869233903, 7967970654277850949025
Offset: 0

Author

Ilya Gutkovskiy, Jan 22 2020

Keywords

Programs

  • Maple
    S:= series(exp(1/(1-tan(x))-1), x, 31):
    seq(coeff(S,x,i)*i!, i=0..30); # Robert Israel, Dec 10 2024
  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[1/(1 - Tan[x]) - 1], {x, 0, nmax}], x] Range[0, nmax]!
    A000111[n_] := If[EvenQ[n], Abs[EulerE[n]], Abs[(2^(n + 1) (2^(n + 1) - 1) BernoulliB[n + 1])/(n + 1)]]; a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] 2^(k - 1) A000111[k] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]

Formula

E.g.f.: exp(sin(x) / (cos(x) - sin(x))).
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * 2^(k-1) * A000111(k) * a(n-k).
a(n) ~ 2^(2*n - 1/4) * exp(1/Pi - 1/2 + 2^(3/2)*sqrt(n/Pi) - n) * n^(n - 1/4) / Pi^(n + 1/4). - Vaclav Kotesovec, Jan 27 2020
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