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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A235134 Expansion of e.g.f. 1/(1 - sinh(2*x))^(1/2).

Original entry on oeis.org

1, 1, 3, 19, 153, 1561, 19563, 289339, 4932273, 95258161, 2055639123, 49019157859, 1280056939593, 36329281202761, 1113449691889083, 36651273215389579, 1289577677407798113, 48299079453732363361, 1918528841276621473443, 80559757274836073592499
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 03 2014

Keywords

Comments

Generally, for e.g.f. 1/(1-sinh(p*x))^(1/p) we have a(n) ~ n! * p^n / (Gamma(1/p) * 2^(1/(2*p)) * n^(1-1/p) * (arcsinh(1))^(n+1/p)).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/(1-Sinh[2*x])^(1/2), {x, 0, 20}], x] * Range[0, 20]!
  • PARI
    x='x+O('x^50); Vec(serlaplace(1/(sqrt(1-sinh(2*x))))) \\ G. C. Greubel, Apr 05 2017
    
  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a001147(n) = prod(k=0, n-1, 2*k+1);
    a(n) = sum(k=0, n, a001147(k)*2^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

Formula

a(n) ~ n! * 2^(n-1/4) / (sqrt(Pi*n) * (log(1+sqrt(2)))^(n+1/2)).
a(n) = Sum_{k=0..n} A001147(k) * 2^(n-k) * A136630(n,k). - Seiichi Manyama, Jun 24 2025

A117442 Number triangle read by rows, related to exp(x)/(cos(x) + sin(x)).

Original entry on oeis.org

1, -1, 1, 3, -2, 1, -11, 9, -3, 1, 57, -44, 18, -4, 1, -361, 285, -110, 30, -5, 1, 2763, -2166, 855, -220, 45, -6, 1, -24611, 19341, -7581, 1995, -385, 63, -7, 1, 250737, -196888, 77364, -20216, 3990, -616, 84, -8, 1, -2873041, 2256633, -885996, 232092, -45486, 7182, -924, 108, -9, 1
Offset: 0

Views

Author

Paul Barry, Mar 16 2006

Keywords

Examples

			Triangle begins
       1;
      -1,     1;
       3,    -2,     1;
     -11,     9,    -3,    1;
      57,   -44,    18,   -4,    1;
    -361,   285,  -110,   30,   -5,  1;
    2763, -2166,   855, -220,   45, -6,  1;
  -24611, 19341, -7581, 1995, -385, 63, -7, 1;
		

Crossrefs

Inverse of A117440.
Second column contains A161722 as subsequence.

Programs

  • Maple
    A117442_row := proc(n) 2^n*add(binomial(n,k)*euler(k)*((x+1)/2)^(n-k), k=0..n);
    seq((-1)^(n-j)*abs(coeff(%,x,j)),j=0..n) end:
    seq(print(A117442_row(n)),n=0..5);  # Peter Luschny, Jun 08 2013
  • Mathematica
    row[n_] := row[n] = 2^n Sum[Binomial[n, k] EulerE[k] ((x+1)/2)^(n-k), {k, 0, n}];
    T[n_, k_] := (-1)^(n-k) Abs[Coefficient[row[n], x, k]];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019, from Maple *)
    Table[(-1)^(n-k)*Binomial[n, k]*Abs[Numerator[EulerE[n-k, 1/4]]], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 02 2021 *)
  • PARI
    E(n) = 2^n*2^(n+1)*(subst(bernpol(n+1, x), x, 3/4) - subst(bernpol(n+1, x), x, 1/4))/(n+1); \\ A122045
    p(n) = 2^n*sum(k=0, n, binomial(n,k)*E(k)*((x+1)/2)^(n-k));
    row(n) = my(rp=p(n)); vector(n+1, k, k--; (-1)^(n-k)*abs(polcoeff(rp, k))); \\ Michel Marcus, Nov 16 2020
    
  • Sage
    def f(n): return (1/4)^n*sum( binomial(n, j)*2^j*euler_number(j) for j in (0..n)) # f(n) = Euler(n, 1/4)
    def A117442(n,k): return (-1)^(n+k)*binomial(n,k)*abs(numerator(f(n-k)))
    flatten([[A117442(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 02 2021

Formula

T(n, 0) = (-1)^n*A001586(n).
Sum_{k=0..n} T(n, k) = A117443(n).
Column k has e.g.f. (x^k/k!)/(cos(x) + sin(x)).
Apart from signs the T(n,k) are the coefficients of the polynomials p(n, x) = 2^n*Sum_{k=0..n} binomial(n,k)*euler(k)*((x+1)/2)^(n-k). - Peter Luschny, Jun 08 2013
From G. C. Greubel, Jun 02 2021: (Start)
T(n, k) = (-1)^(n+k) * binomial(n, k) * abs(numerator( Euler(n-k, 1/4) )), where Euler(n, x) is the Euler number polynomial.
T(n, n) = 1.
T(n, n-1) = -A000027(n) = -binomial(n+1, 1).
T(n, n-2) = A045943(n+1) = 3*binomial(n+2, 2).
T(n, n-3) = -A111080(n) = -11*binomial(n+3, 3).
T(j, k) = (-1)^k * binomial(j+k, k) * abs(numerator( Euler(k, 1/4) )) (columns).
T(n, n-j) = (-1)^n * binomial(n+j, j) * abs(numerator( Euler(n, 1/4) )) (downward diagonals). (End)
The pair of triangles P*((I + P^4)/2)^(-1) and P^3*((I + P^4)/2)^(-1), where P denotes Pascal's triangle A007318, give the present triangle but with a different pattern of signs. - Peter Bala, Mar 07 2024

A235128 Expansion of e.g.f. 1/(1 - sin(7*x))^(1/7).

Original entry on oeis.org

1, 1, 8, 71, 1072, 20161, 476288, 13315751, 432387712, 15959926081, 660372282368, 30265936565831, 1522069164439552, 83327826089289601, 4933286107483701248, 314052936209639958311, 21392225375507849838592, 1552501782546292090638721, 119588747474281844162428928
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 03 2014

Keywords

Comments

Generally, for e.g.f. 1/(1-sin(p*x))^(1/p) we have a(n) ~ n! * 2^(n+3/p) * p^n / (Gamma(2/p) * n^(1-2/p) * Pi^(n+2/p)).

Crossrefs

Cf. A001586 (p=2), A007788 (p=3), A144015 (p=4), A230134 (p=5), A227544 (p=6), A230114 (p=8).

Programs

  • Mathematica
    CoefficientList[Series[1/(1-Sin[7*x])^(1/7), {x, 0, 20}], x] * Range[0, 20]!
  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a045754(n) = prod(k=0, n-1, 7*k+1);
    a(n) = sum(k=0, n, a045754(k)*(7*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

Formula

a(n) ~ n! * 2^(n+3/7) * 7^n / (Gamma(2/7) * n^(5/7) * Pi^(n+2/7)).
a(n) = Sum_{k=0..n} A045754(k) * (7*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A318005 E.g.f.: A(x) satisfies: cos(A(x)) + sin(A(x)) = 1/(cos(x) - sin(x)).

Original entry on oeis.org

1, 4, 24, 224, 2880, 48064, 989184, 24218624, 687083520, 22151148544, 799546834944, 31934834253824, 1398132497448960, 66573473015578624, 3425078687463112704, 189331392774496845824, 11190654534195295027200, 704262689221037166690304, 47015904809670036594622464, 3318579148264602406039322624
Offset: 1

Views

Author

Paul D. Hanna, Aug 27 2018

Keywords

Examples

			E.g.f.: A(x) = x + 4*x^2/2! + 24*x^3/3! + 224*x^4/4! + 2880*x^5/5! + 48064*x^6/6! + 989184*x^7/7! + 24218624*x^8/8! + 687083520*x^9/9! + 22151148544*x^10/10! + ...
such that:
cos(A(x)) + sin(A(x)) = 1/( cos(x) - sin(x) ).
RELATED SERIES.
(a) cos(A(x)) + sin(A(x)) = 1/(cos(x) - sin(x)) = 1 + x + 3*x^2/2! + 11*x^3/3! + 57*x^4/4! + 361*x^5/5! + 2763*x^6/6! + ... + A001586(n)*x^n/n! + ...
(b) If F(F(x)) = A(x), then
F(x) = x + 2*x^2/2! + 6*x^3/3! + 40*x^4/4! + 360*x^5/5! + 4592*x^6/6! + 70896*x^7/7! + 1279360*x^8/8! + ... + A318006(n)*x^n/n! + ...
where F(x) = arcsin( 2*sin(2*x)/(2 - sin(2*x)) ) /2.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A = asin( sin(2*x +x*O(x^n))/(1 - sin(2*x +x*O(x^n))) )/2 ); n!*polcoeff(A,n)}
    for(n=1,20, print1(a(n),", "))

Formula

E.g.f. A(x) satisfies:
(1) A(-A(-x)) = x.
(2) 1 = Sum_{n>=0} (-1)^floor(n/2) * ( A(x) + (-1)^n*x )^n/n!.
(3a) 1 = cos(A(x) + x) + sin(A(x) - x).
(3b) 1 = ( cos(A(x)) + sin(A(x)) ) * ( cos(x) - sin(x) ).
(4) A(x) = arcsin( sin(2*x)/(1 - sin(2*x)) )/2.
a(n) = 2^(n-1) * A200560(n).

A385281 Expansion of e.g.f. 1/(1 - 2 * x * cosh(2*x))^(1/2).

Original entry on oeis.org

1, 1, 3, 27, 249, 2825, 41355, 708883, 13888497, 309267729, 7698772755, 211585744139, 6367841422569, 208299923870233, 7357493992966299, 279095125351544835, 11316313498670411745, 488403056864943302177, 22355228989851909617187, 1081663315375339026249211
Offset: 0

Views

Author

Seiichi Manyama, Jun 24 2025

Keywords

Crossrefs

Programs

  • PARI
    a185951(n, k) = binomial(n, k)/2^k*sum(j=0, k, (2*j-k)^(n-k)*binomial(k, j));
    a001147(n) = prod(k=0, n-1, 2*k+1);
    a(n) = sum(k=0, n, a001147(k)*2^(n-k)*a185951(n, k));

Formula

a(n) = Sum_{k=0..n} A001147(k) * 2^(n-k) * A185951(n,k), where A185951(n,0) = 0^n.
a(n) ~ 2^(n + 1/2) * n^n / (sqrt(1 + r*sqrt(1 - r^2)) * exp(n) * r^n), where r = A069814. - Vaclav Kotesovec, Jun 24 2025

A385306 Expansion of e.g.f. 1/(1 - 2 * sin(x))^(1/2).

Original entry on oeis.org

1, 1, 3, 14, 93, 796, 8343, 103424, 1479993, 24008656, 435364683, 8726775584, 191601310293, 4572794295616, 117871476051423, 3263515787807744, 96591500816346993, 3043368045293138176, 101702692426476460563, 3592948632452749243904, 133794496537591022166093
Offset: 0

Views

Author

Seiichi Manyama, Jun 24 2025

Keywords

Crossrefs

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[1/Sqrt[1-2Sin[x]],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 09 2025 *)
  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a001147(n) = prod(k=0, n-1, 2*k+1);
    a(n) = sum(k=0, n, a001147(k)*I^(n-k)*a136630(n, k));

Formula

a(n) = Sum_{k=0..n} A001147(k) * i^(n-k) * A136630(n,k), where i is the imaginary unit.
a(n) ~ 2^(n+1) * 3^(n + 1/4) * n^n / (exp(n) * Pi^(n + 1/2)). - Vaclav Kotesovec, Jun 28 2025

A079858 E.g.f. 1/(cos(2*x) - sin(2*x)).

Original entry on oeis.org

1, 2, 12, 88, 912, 11552, 176832, 3150208, 64188672, 1470996992, 37459479552, 1049279715328, 32063706796032, 1061443378552832, 37841217707753472, 1445427909919080448, 58892032991566036992, 2549444593020567683072
Offset: 0

Views

Author

Michael Somos, Jan 20 2003

Keywords

Crossrefs

Cf. A001586.
(-1)^(n*(n-1)/2)*a(n) gives the alternating row sums of A225118. - Wolfdieter Lang, Jul 12 2017

Programs

  • Mathematica
    CoefficientList[Series[1/(Cos[2*x]-Sin[2*x]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 07 2013 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(1/(cos(2*x+x*O(x^n))-sin(2*x+x*O(x^n))),n))
    
  • PARI
    x='x+O('x^66); v=Vec(serlaplace( 1/(cos(2*x)-sin(2*x)) ) ) \\ Joerg Arndt, Apr 27 2013
  • Sage
    from mpmath import mp, lerchphi
    mp.dps = 32; mp.pretty = True
    def A079858(n): return abs(2*8^n*lerchphi(-1,-n,1/4))
    [int(A079858(n)) for n in (0..17)]  # Peter Luschny, Apr 27 2013
    

Formula

E.g.f.: 1/(cos(2*x) - sin(2*x)).
a(n) = 2^n * A001586(n).
a(n) = | 2*8^n*lerchphi(-1,-n,1/4) |. - Peter Luschny, Apr 27 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - 8*x^2*(k+1)^2/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Sep 27 2013
a(n) ~ 4 * n^(n+1/2) * (8/Pi)^n / (sqrt(Pi)*exp(n)). - Vaclav Kotesovec, Oct 07 2013
E.g.f.: 1/(1-2*x)*(1 + 2*x^2/((1-2*x)*W(0) - x )), where W(k) = x + (k+1)/( 1 - 2*x/( 2*k+3 - x*(2*k+3)/W(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Dec 27 2013

A098432 Coefficients of polynomials S(n,x) related to Springer numbers.

Original entry on oeis.org

1, 8, 7, 128, 304, 177, 3072, 13952, 21080, 10199, 98304, 724992, 2016000, 2441056, 1051745, 3932160, 42762240, 187643904, 407505664, 428605352, 169913511, 188743680, 2839019520, 17974591488, 60428242944, 111985428352
Offset: 0

Views

Author

Ralf Stephan, Sep 07 2004

Keywords

Examples

			S(0,x) = 1,
S(1,x) = 8*x + 7,
S(2,x) = 128*x^2 + 304*x + 177,
S(3,x) = 3072*x^3 + 13952*x^2 + 21080*x + 10199.
		

Crossrefs

Cf. A001586. S(n, 1/2) = A000464(n+1), S(n, -1/2) = A000281(n).
Leading coefficients are A051189. Constant terms are in A098433.
Cf. A001586. S(n, 1/2) = A000464(n), S(n, -1/2) = A000281(n).

Programs

  • PARI
    S(n,x)=if(n<1,1,(2*x+2)*(2*x+4)*S(n-1,x+2)-(2*x+1)^2*S(n-1,x))

Formula

Recurrence: S(0, x)=1, S(n, x)=(2x+2)(2x+4)S(n-1, x+2)-(2x+1)^2S(n-1, x).
G.f.: Sum[n>=0, S(n, x)t^n] = 1/(1+t-4*2(x+1)t/(1-4*2(x+2)t/(1+t-4*4(x+3)t/(1-4+4(x+4)t/...)))).

A151751 Triangle of coefficients of generalized Bernoulli polynomials associated with a Dirichlet character modulus 8.

Original entry on oeis.org

2, 0, 6, -44, 0, 12, 0, -220, 0, 20, 2166, 0, -660, 0, 30, 0, 15162, 0, -1540, 0, 42, -196888, 0, 60648, 0, -3080, 0, 56, 0, -1771992, 0, 181944, 0, -5544, 0, 72, 28730410, 0, -8859960, 0, 454860, 0, -9240, 0, 90
Offset: 2

Views

Author

Peter Bala, Jun 17 2009

Keywords

Comments

Let X be a periodic arithmetical function with period m. The generalized Bernoulli polynomials B_n(X,x) attached to X are defined by means of the generating function
(1)... t*exp(t*x)/(exp(m*t)-1) * sum {r = 0..m-1} X(r)*exp(r*t)
= sum {n = 0..inf} B_n(X,x)*t^n/n!.
For the theory and properties of these polynomials see [Cohen, Section 9.4]. In the present case, X is chosen to be the Dirichlet character modulus 8 given by
(2)... X(8*n+1) = X(8*n+7) = 1; X(8*n+3) = X(8*n+5) = -1; X(2*n) = 0.
Cf. A153641.

Examples

			The triangle begins
n\k|........0.......1........2.......3......4.......5.......6
=============================================================
.2.|........2
.3.|........0.......6
.4.|......-44.......0.......12
.5.|........0....-220........0......20
.6.|.....2166.......0.....-660.......0......30
.7.|........0...15162........0...-1540.......0.....42
.8.|..-196888.......0....60648.......0...-3080......0......56
...
		

References

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

Crossrefs

Programs

  • Maple
    with(gfun):
    for n from 2 to 10 do
    Genbernoulli(n,x) := 8^(n-1)*(bernoulli(n,(x+1)/8)-bernoulli(n,(x+3)/8)-bernoulli(n,(x+5)/8)+bernoulli(n,(x+7)/8));
    seriestolist(series(Genbernoulli(n,x),x,10))
    end do;

Formula

TABLE ENTRIES
(1)... T(2*n,2*k+1) = 0, T(2*n+1,2*k) = 0;
(2)... T(2*n,2*k) = (-1)^(n-k-1)*C(2*n,2*k)*2*(n-k)*A000464(n-k-1);
(3)... T(2*n+1,2*k+1) = (-1)^(n-k-1)*C(2*n+1,2*k+1)*2*(n-k)*A000464(n-k-1);
where C(n,k) = binomial(n,k).
GENERATING FUNCTION
The e.g.f. for these generalized Bernoulli polynomials is
(4)... t*exp(x*t)*(exp(t)-exp(3*t)-exp(5*t)+exp(7*t))/(exp(8*t)-1)
= sum {n = 2..inf} B_n(X,x)*t^n/n! = 2*t^2/2! + 6*x*t^3/3! + (12*x^2 - 44)*t^4/4! + ....
In terms of the ordinary Bernoulli polynomials B_n(x)
(5)... B_n(X,x) = 8^(n-1)*{B_n((x+1)/8) - B_n((x+3)/8) - B_n((x+5)/8) + B_n((x+7)/8)}.
The B_n(X,x) are Appell polynomials of the form
(6)... B_n(X,x) = sum {j = 0..n} binomial(n,j)*B_j(X,0)*x*(n-j).
The sequence of generalized Bernoulli numbers
(7)... [B_n(X,0)]n>=2 = [2,0,-44,0,2166,0,...]
has the e.g.f.
(8)... t*(exp(t)-exp(3*t)-exp(5*t)+exp(7*t))/(exp(8*t)-1),
which simplifies to
(9)... t*sinh(t)/cosh(2*t).
Hence
(10)... B_(2*n)(X,0) = (-1)^(n+1)*2*n*A000464(n-1); B_(2*n+1)(X,0) = 0.
The sequence {B_(2*n)(X,0)}n>=2 is A161722.
RELATION WITH TWISTED SUMS OF POWERS
The generalized Bernoulli polynomials may be used to evaluate sums of k-th powers twisted by the function X(n). For the present case the result is
(11)... sum{n = 0..8*N-1} X(n)*n^k = 1^k-3^k-5^k+7^k- ... +(8*N-1)^k
= [B_(k+1)(X,8*N) - B_(k+1)(X,0)]/(k+1)
For the proof, apply [Cohen, Corollary 9.4.17 with m = 8 and x = 0].
MISCELLANEOUS
(12)... Row sums [2, 6, -32, ...] = (-1)^(1+binomial(n,2))*A109572(n)
= (-1)^(1+binomial(n,2))*n*A000828(n-1) = (-1)^(1+binomial(n,2))*n* 2^(n-2)*A000111(n-1).

A190392 E.g.f. A(x) satisfies A'(x) = sin(A(x)) + cos(A(x)).

Original entry on oeis.org

1, 1, 0, -4, -12, 4, 240, 1184, -1008, -59504, -401280, 643136, 38584128, 323581504, -848090880, -51666451456, -509739310848, 2004840714496, 123888658698240, 1386061762251776, -7721141999864832, -483475390212586496, -5974101514137292800, 45231727252157947904
Offset: 1

Views

Author

Vladimir Kruchinin, May 09 2011

Keywords

Comments

Let f(x) be a smooth function. The autonomous differential equation A'(x) = f(A(x)), with initial condition A(0) = 0, is separable and the solution is given by A(x) = inverse function of Integral_{t = 0..x} 1/f(t) dt. The inversion of the integral Integral_{t = 0..x} 1/f(t) dt is most conveniently found by applying [Dominici, Theorem 4.1]. The result is A(x) = Sum_{n>=1} D^(n-1)[f](0)*x^n/n!, where the nested derivative D^n[f](x)is defined recursively as D^0[f](x) = 1 and D^(n+1)[f](x) = (d/dx)(f(x)*D^n[f](x)) for n >= 0. See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x). In the present case we take f(x) = sin(x)+cos(x). - Peter Bala, Aug 27 2011

Crossrefs

Programs

  • Maple
    A := x ; for i from 1 to 35 do sin(A)+cos(A) ; convert(taylor(%,x=0,25),polynom) ; A := int(%,x) ; print(A) ; end do:
    for i from 1 to 25 do printf("%d,", coeftayl(A,x=0,i)*i!) ; end do: # R. J. Mathar, Jun 03 2011
  • Mathematica
    terms = 25; A[] = 0; Do[A[x] = Integrate[Sin[A[x]] + Cos[A[x]], x] + O[x]^terms, terms]; CoefficientList[A[x], x]*Range[0, terms-1]! (* Jean-François Alcover, Feb 21 2013, updated Jan 15 2018 *)
  • Maxima
    g(n):=(-1)^floor(n/2)*1/n!;
    a(n):=T190015_Solve(n,g);

Formula

E.g.f.: A(x) = inverse of Integral_{t = 0..x} 1/(sin(t)+cos(t)) dt = series reversion (x - x^2/2! + 3*x^3/3! - 11*x^4/4! + 57*x^5/5! - ...) = x + x^2/2! - 4*x^4/4! - 12*x^5/5! + .... a(n) = D^(n-1)[sin(x)+cos(x)](0), where the nested derivative operator D^n is defined above. Compare with A012244. -Peter Bala, Aug 27 2011
E.g.f.: A(x) = 2*arctan((sqrt(2)-1)*exp(sqrt(2)*x))-Pi/4. Compare with A028296. - Peter Bala, Sep 02 2011
G.f.: 1/G(0) where G(k) = 1 - 2*x*(k+1)/(1 + 1/(1 - 2*x*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 10 2013.
G.f.: -(1/x)/Q(0), where Q(k)= 2*k+1 - 1/x + (k+1)*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Apr 15 2013
G.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 + (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013
E.g.f. if offset 0: 2^(1/2)/(2^(1/2)*cosh(x*2^(1/2))-sinh(x*2^(1/2))). - Sergei N. Gladkovskii, Nov 10 2013
a(n) ~ -(n-1)! * 2^(3*n/2+1) * sin(n*arctan(Pi/log(3 - 2*sqrt(2)))) / (Pi^2 + log(3 - 2*sqrt(2))^2)^(n/2). - Vaclav Kotesovec, Jan 07 2014
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