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 1-4 of 4 results.

A007788 Number of augmented Andre 3-signed permutations: E.g.f. (1-sin(3*x))^(-1/3).

Original entry on oeis.org

1, 1, 4, 19, 136, 1201, 13024, 165619, 2425216, 40132801, 740882944, 15091932019, 336257744896, 8134269015601, 212309523595264, 5946914908771219, 177934946000306176, 5663754614516217601, 191097349696090537984, 6812679868133940475219, 255885704427935576621056
Offset: 0

Views

Author

R. Ehrenborg (ehrenbor(AT)lacim.uqam.ca) and M. A. Readdy (readdy(AT)lacim.uqam.ca)

Keywords

Comments

It appears that all members are of the form 3k+1. - Ralf Stephan, Nov 12 2007

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( (1-Sin(3*x))^(-1/3) ))); // G. C. Greubel, Mar 05 2020
    
  • Maple
    m:=20; S:=series( (1-sin(3*x))^(-1/3), x, m+1): seq(j!*coeff(S, x, j), j=0..m); # G. C. Greubel, Mar 05 2020
  • Mathematica
    With[{nn=20},CoefficientList[Series[(1-Sin[3x])^(-1/3),{x,0,nn}], x] Range[0,nn]!] (* Harvey P. Dale, Nov 23 2011 *)
  • PARI
    Vec(serlaplace( (1-sin(3*x))^(-1/3) +O('x^20) )) \\ G. C. Greubel, Mar 05 2020
    
  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a007559(n) = prod(k=0, n-1, 3*k+1);
    a(n) = sum(k=0, n, a007559(k)*(3*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025
    
  • Sage
    m=20;
    def A007788_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( (1-sin(3*x))^(-1/3) ).list()
    a=A007788_list(m+1); [factorial(n)*a[n] for n in (0..m)] # G. C. Greubel, Mar 05 2020

Formula

E.g.f.: (1-sin(3*x))^(-1/3).
a(n) ~ n! * 2*6^n/(Pi^(n+2/3)*n^(1/3)*Gamma(2/3)). - Vaclav Kotesovec, Jun 25 2013
a(n) = Sum_{k=0..n} A007559(k) * (3*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A227544 Expansion of e.g.f. 1/(1 - sin(6*x))^(1/6).

Original entry on oeis.org

1, 1, 7, 55, 721, 11761, 240247, 5801095, 162512161, 5171130721, 184337942887, 7275081518935, 314918762166001, 14834964193292881, 755507853144691927, 41362173671901329575, 2422478811455080626241, 151132171549872325122241, 10006051653759338150151367, 700695219796759105368529015
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2013

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)). - Vaclav Kotesovec, Jan 03 2014

Examples

			E.g.f.: A(x) = 1 + x + 7*x^2/2! + 55*x^3/3! + 721*x^4/4! + 11761*x^5/5! + ...
where A(x)^3 = 1 + 3*x + 27*x^2/2! + 297*x^3/3! + 4617*x^4/4! + 87723*x^5/5! + ...
and 1/A(x)^3 = 1 - 3*x - 9*x^2/2! + 27*x^3/3! + 81*x^4/4! - 243*x^5/5! + ...
which illustrates 1/A(x)^3 = cos(3*x) - sin(3*x).
O.g.f.: 1/(1-x - 6*1*1*x^2/(1-7*x - 6*2*4*x^2/(1-13*x - 6*3*7*x^2/(1-19*x - 6*4*10*x^2/(1-25*x - 6*5*13*x^2/(1-...)))))), a continued fraction.
		

Crossrefs

Cf. A001586 (p=2), A007788 (p=3), A144015 (p=4), A230134 (p=5), A235128 (p=7), A230114 (p=8).

Programs

  • Mathematica
    CoefficientList[Series[1/(1-Sin[6*x])^(1/6), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jan 03 2014 *)
  • PARI
    {a(n)=local(X=x+x*O(x^n)); n!*polcoeff((cos(3*X)-sin(3*X))^(-1/3), n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n)); for(i=0, n, A=exp(intformal(A^3/subst(A^3, x, -x)))); n!*polcoeff(A, n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a008542(n) = prod(k=0, n-1, 6*k+1);
    a(n) = sum(k=0, n, a008542(k)*(6*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

Formula

E.g.f. A(x) satisfies:
(1) A(x) = (cos(3*x) - sin(3*x))^(-1/3).
(2) A(x)^3/A(-x)^3 = 1/cos(6*x) + tan(6*x).
(3) A(x) = exp( Integral A(x)^3/A(-x)^3 dx ).
O.g.f.: 1/G(0) where G(k) = 1 - (6*k+1)*x - 6*(k+1)*(3*k+1)*x^2/G(k+1) [continued fraction formula from A144015 due to Sergei N. Gladkovskii].
a(n) ~ n! * 2^(2*n+1/2) * 3^n / (Gamma(1/3) * n^(2/3) * Pi^(n+1/3)). - Vaclav Kotesovec, Jan 03 2014
a(n) = Sum_{k=0..n} A008542(k) * (6*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A230114 Expansion of e.g.f. 1/(1 - sin(8*x))^(1/8).

Original entry on oeis.org

1, 1, 9, 89, 1521, 32401, 869049, 27608489, 1019581281, 42824944801, 2017329504489, 105299243488889, 6032850630082641, 376363074361201201, 25396689469918450329, 1843101478742259481289, 143145930384321475601601, 11846611289341729822881601, 1040750126963789832859930569
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2013

Keywords

Comments

Generally, for e.g.f. 1/(1-sin(p*x))^(1/p) is a(n) ~ n! * 2^(n+3/p) * p^n / (Gamma(2/p) * n^(1-2/p) * Pi^(n+2/p)). - Vaclav Kotesovec, Jan 03 2014

Examples

			E.g.f.: A(x) = 1 + x + 9*x^2/2! + 89*x^3/3! + 1521*x^4/4! + 32401*x^5/5! + ...
where A(x)^4 = 1 + 4*x + 48*x^2/2! + 704*x^3/3! + 14592*x^4/4! + 369664*x^5/5! + ...
and 1/A(x)^4 = 1 - 4*x - 16*x^2/2! + 64*x^3/3! + 256*x^4/4! - 1024*x^5/5! + ...
which illustrates 1/A(x)^4 = cos(4*x) - sin(4*x).
O.g.f.: 1/(1-x - 8*1*1*x^2/(1-9*x - 8*2*5*x^2/(1-17*x - 8*3*9*x^2/(1-25*x - 8*4*13*x^2/(1-33*x - 8*5*17*x^2/(1-...)))))), a continued fraction.
		

Crossrefs

Cf. A001586 (p=2), A007788 (p=3), A144015 (p=4), A230134 (p=5), A227544 (p=6), A235128 (p=7).

Programs

  • Mathematica
    CoefficientList[Series[1/(1-Sin[8*x])^(1/8), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jan 03 2014 *)
  • PARI
    {a(n)=local(X=x+x*O(x^n)); n!*polcoeff((cos(4*X)-sin(4*X))^(-1/4), n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n)); for(i=0, n, A=exp(intformal(A^4/subst(A^4, x, -x)))); n!*polcoeff(A, n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a045755(n) = prod(k=0, n-1, 8*k+1);
    a(n) = sum(k=0, n, a045755(k)*(8*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

Formula

E.g.f. A(x) satisfies:
(1) A(x) = (cos(4*x) - sin(4*x))^(-1/4).
(2) A(x)^4/A(-x)^4 = 1/cos(8*x) + tan(8*x).
(3) A(x) = exp( Integral A(x)^4/A(-x)^4 dx ).
O.g.f.: 1/G(0) where G(k) = 1 - (8*k+1)*x - 8*(k+1)*(4*k+1)*x^2/G(k+1) [continued fraction formula from A144015 due to Sergei N. Gladkovskii].
a(n) ~ n! * 2^(4*n+3/8) / (Gamma(1/4) * n^(3/4) * Pi^(n+1/4)). - Vaclav Kotesovec, Jan 03 2014
a(n) = Sum_{k=0..n} A045755(k) * (8*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A385896 Array read by ascending antidiagonals: A(n, k) = k! * [x^k] (1 - sin(n*x))^(-1/n) for n > 0, A(0, k) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 5, 1, 1, 1, 4, 11, 16, 1, 1, 1, 5, 19, 57, 61, 1, 1, 1, 6, 29, 136, 361, 272, 1, 1, 1, 7, 41, 265, 1201, 2763, 1385, 1, 1, 1, 8, 55, 456, 3001, 13024, 24611, 7936, 1, 1, 1, 9, 71, 721, 6301, 42125, 165619, 250737, 50521, 1
Offset: 0

Views

Author

Peter Luschny, Jul 20 2025

Keywords

Examples

			Table starts:
  [0] 1, 1, 1,  1,    1,     1,      1, ... [A000012]
  [1] 1, 1, 2,  5,   16,    61,    272, ... [A000111]
  [2] 1, 1, 3, 11,   57,   361,   2763, ... [A001586]
  [3] 1, 1, 4, 19,  136,  1201,  13024, ... [A007788]
  [4] 1, 1, 5, 29,  265,  3001,  42125, ... [A144015]
  [5] 1, 1, 6, 41,  456,  6301, 108576, ... [A230134]
  [6] 1, 1, 7, 55,  721, 11761, 240247, ... [A227544]
  [7] 1, 1, 8, 71, 1072, 20161, 476288, ... [A235128]
  [8] 1, 1, 9, 89, 1521, 32401, 869049, ... [A230114]
     [A000027]  | [A187277] | [A385898].
            [A028387]   [A385897]
.
Seen as a triangle:
  [0] 1;
  [1] 1, 1;
  [2] 1, 1, 1;
  [3] 1, 1, 2,  1;
  [4] 1, 1, 3,  5,   1;
  [5] 1, 1, 4, 11,  16,    1;
  [6] 1, 1, 5, 19,  57,   61,    1;
  [7] 1, 1, 6, 29, 136,  361,  272,    1;
  [8] 1, 1, 7, 41, 265, 1201, 2763, 1385, 1;
		

Crossrefs

Programs

  • Maple
    MAX := 16: ser := n -> series((1 - sin(n*x))^(-1/n), x, MAX):
    A := (n, k) -> if n = 0 then 1 else k!*coeff(ser(n), x, k) fi:
    seq(lprint(seq(A(n, k), k = 0..8)), n = 0..8);
  • Mathematica
    T[n_, k_, m_] := T[n, k, m] =
      Which[
        n <  0 || k <  0, 0,
        n == 0 && k == 0, 1,
        k == 0, T[n - 1, n - 1, m],
        True, T[n, k - 1, m] + m*T[n - 1, n - k - 1, m]
    ];
    A[n_, k_] := T[k, k, n - k];
    Table[A[n, k], {n, 0, 10}, {k, 0, n}] // Flatten

Formula

A(n, k) = T(k, k, n - k) where T(n, k, m) = T(n, k-1, m) + m * T(n-1, n-k-1, m) for k > 0, T(n, 0, m) = T(n-1, n-1, m), and T(0, 0, m) = 1.
Column n is a linear recurrence with kernel [(-1)^k*A135278(n, k), k = 0..n].
Showing 1-4 of 4 results.