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 1-10 of 45 results. Next

A007559 Triple factorial numbers (3*n-2)!!! with leading 1 added.

Original entry on oeis.org

1, 1, 4, 28, 280, 3640, 58240, 1106560, 24344320, 608608000, 17041024000, 528271744000, 17961239296000, 664565853952000, 26582634158080000, 1143053268797440000, 52580450364682240000, 2576442067869429760000, 133974987529210347520000, 7368624314106569113600000
Offset: 0

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Author

Keywords

Comments

a(n) is the number of increasing quaternary trees on n vertices. (See A001147 for ternary and A000142 for binary trees.) - David Callan, Mar 30 2007
a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 1. - Peter Luschny, Jun 23 2011
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
Partial products of A016777. - Reinhard Zumkeller, Sep 20 2013
For n > 2, a(n) is a Zumkeller number. - Ivan N. Ianakiev, Jan 28 2020
a(n) is the number of generalized permutations of length n related to the degenerate Eulerian numbers (see arXiv:2007.13205), cf. A336633. - Orli Herscovici, Jul 28 2020

Examples

			G.f. = 1 + x + 4*x^2 + 28*x^3 + 280*x^4 + 3640*x^5 + 58240*x^6 + ...
a(3) = 28 and a(4) = 280; with top row of M^3 = (28, 117, 108, 27), sum = 280.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n)= A035469(n, 1), n >= 1, (first column of triangle A035469(n, m)).
Cf. A107716. - Gary W. Adamson, Oct 22 2009
Cf. A095660. - Gary W. Adamson, Jul 19 2011
Subsequence of A007661. A007696, A008548.
a(n) = A286718(n,0), n >= 0.
Row sums of A336633.

Programs

  • GAP
    List([0..20], n-> Product([0..n-1], k-> 3*k+1 )); # G. C. Greubel, Aug 20 2019
  • Haskell
    a007559 n = a007559_list !! n
    a007559_list = scanl (*) 1 a016777_list
    -- Reinhard Zumkeller, Sep 20 2013
    
  • Magma
    b:= func< n | (n lt 2) select n else (3*n-2)*Self(n-1) >;
    [1] cat [b(n): n in [1..20]]; // G. C. Greubel, Aug 20 2019
    
  • Maple
    A007559 := n -> mul(k, k = select(k-> k mod 3 = 1, [$1 .. 3*n])): seq(A007559(n), n = 0 .. 17); # Peter Luschny, Jun 23 2011
    # second Maple program:
    b:= proc(n) option remember; `if`(n<1, 1, n*b(n-3)) end:
    a:= n-> b(3*n-2):
    seq(a(n), n=0..20);  # Alois P. Heinz, Dec 18 2024
  • Mathematica
    a[ n_] := If[ n < 0, 1 / Product[ k, {k, - 2, 3 n - 1, -3}],
      Product[ k, {k, 1, 3 n - 2, 3}]]; (* Michael Somos, Oct 14 2011 *)
    FoldList[Times,1,Range[1,100,3]] (* Harvey P. Dale, Jul 05 2013 *)
    Range[0, 19]! CoefficientList[Series[((1 - 3 x)^(-1/3)), {x, 0, 19}], x] (* Vincenzo Librandi, Oct 08 2015 *)
  • Maxima
    a(n):=if n=1 then 1 else (n)!*(sum(m/n*sum(binomial(k,n-m-k)*(-1/3)^(n-m-k)* binomial (k+n-1,n-1),k,1,n-m),m,1,n)+1); /* Vladimir Kruchinin, Aug 09 2010 */
    
  • PARI
    {a(n) = if( n<0, (-1)^n / prod(k=0,-1-n, 3*k + 2), prod(k=0, n-1, 3*k + 1))}; /* Michael Somos, Oct 14 2011 */
    
  • PARI
    my(x='x+O('x^33)); Vec(serlaplace((1-3*x)^(-1/3))) /* Joerg Arndt, Apr 24 2011 */
    
  • Sage
    def A007559(n) : return mul(j for j in range(1,3*n,3))
    [A007559(n) for n in (0..17)]  # Peter Luschny, May 20 2013
    

Formula

a(n) = Product_{k=0..n-1} (3*k + 1).
a(n) = (3*n - 2)!!!.
a(n) = A007661(3*n-2).
E.g.f.: (1-3*x)^(-1/3).
a(n) ~ sqrt(2*Pi)/Gamma(1/3)*n^(-1/6)*(3*n/e)^n*(1 - (1/36)/n - ...). - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = 3^n*Pochhammer(1/3, n).
a(n) = Sum_{k=0..n} (-3)^(n-k)*A048994(n, k). - Philippe Deléham, Oct 29 2005
a(n) = n!*(1+Sum_{m=1..n} (m/n)*Sum_{k=1..n-m} binomial(k, n-m-k)*(-1/3)^(n-m-k)*binomial(k+n-1, n-1)), n>1. - Vladimir Kruchinin, Aug 09 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term in M^n, M = a variant of Pascal (1,3) triangle (Cf. A095660); as an infinite square production matrix:
1, 3, 0, 0, 0,...
1, 4, 3, 0, 0,...
1, 5, 7, 3, 0,...
...
a(n+1) = sum of top row terms of M^n. (End)
a(n) = (-2)^n*Sum_{k=0..n} (3/2)^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+1)/( 1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 21 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k+1)/(x*(3*k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
E.g.f.: E(0)/2, where E(k)= 1 + 1/(1 - x*(3*k+1)/(x*(3*k+1) + (k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
Let D(x) = 1/sqrt(1 - 2*x) be the e.g.f. for the sequence of double factorial numbers A001147. Then the e.g.f. A(x) for the triple factorial numbers satisfies D( Integral_{t=0..x} A(t) dt ) = A(x). Cf. A007696 and A008548. - Peter Bala, Jan 02 2015
O.g.f.: hypergeom([1, 1/3], [], 3*x). - Peter Luschny, Oct 08 2015
a(n) = 3^n * Gamma(n + 1/3)/Gamma(1/3). - Artur Jasinski, Aug 23 2016
a(n) = (-1)^n / A008544(n), 0 = a(n)*(+3*a(n+1) -a(n+2)) +a(n+1)*a(n+1) for all n in Z. - Michael Somos, Sep 30 2018
D-finite with recurrence: a(n) +(-3*n+2)*a(n-1)=0, n>=1. - R. J. Mathar, Feb 14 2020
Sum_{n>=1} 1/a(n) = (e/9)^(1/3) * (Gamma(1/3) - Gamma(1/3, 1/3)). - Amiram Eldar, Jun 29 2020

Extensions

Better description from Wolfdieter Lang

A004988 a(n) = (3^n/n!) * Product_{k=0..n-1} (3*k + 2).

Original entry on oeis.org

1, 6, 45, 360, 2970, 24948, 212058, 1817640, 15677145, 135868590, 1182056733, 10316131488, 90266150520, 791564704560, 6954461332920, 61199259729696, 539318476367946, 4758692438540700, 42035116540442850, 371678925199705200, 3289358488017391020
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Crossrefs

Programs

  • GAP
    List([0..20], n-> 3^n*Product([0..n-1], k-> 3*k+2)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
  • Magma
    [1] cat [3^n*&*[3*k+2: k in [0..n-1]]/Factorial(n): n in [1..20]]; // G. C. Greubel, Aug 22 2019
    
  • Maple
    A004988 := proc(n)
            binomial(-2/3,n)*(-9)^n ;
    end proc: # R. J. Mathar, Sep 16 2012
  • Mathematica
    Table[FullSimplify[9^n*Gamma[n+2/3]/(Gamma[2/3]*Gamma[n+1])],{n,0,20}] (* Vaclav Kotesovec, Feb 09 2014 *)
    CoefficientList[Series[(1-9x)^(-2/3), {x, 0, 20}], x] (* Vincenzo Librandi, Feb 10 2014 *)
    Table[9^n*Pochhammer[2/3, n]/n!, {n,0,20}] (* G. C. Greubel, Aug 22 2019 *)
  • PARI
    a(n)=if(n<0,0,prod(k=0,n-1,3*k+2)*3^n/n!)
    
  • Sage
    [9^n*rising_factorial(2/3, n)/factorial(n) for n in (0..20)] # G. C. Greubel, Aug 22 2019
    

Formula

G.f.: (1-9*x)^(-2/3).
a(n) = 9^n*Gegenbauer_C(n,1/3,1). - Paul Barry, Apr 21 2009
a(n) = Product_{k=1..n} (9 - 3/k). - Michel Lagneau, Sep 16 2012
a(n) = (-9)^n*binomial(-2/3, n). - R. J. Mathar, Sep 16 2012
D-finite with recurrence: n*a(n) +3*(-3*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 03 2012
a(n) = 9^n * Gamma(n+2/3) / (Gamma(2/3) * Gamma(n+1)). - Vaclav Kotesovec, Feb 09 2014
Sum_{n>=0} 1/a(n) = 9/8 + sqrt(3)*Pi/32 - 3*log(3)/32. - Amiram Eldar, Dec 02 2022
Representation as the n-th moment of a positive function on (0, 9): a(n) = Integral_{x = 0..9} x^n * w(x) dx, n >= 0, where w(x) = sqrt(3)/(2*Pi) * 1/(x*(9 - x)^2)^(1/3). The weight function w(x) is the solution of the Hausdorff moment problem on (0, 9) with moments equal to a(n). As a consequence this representation is unique. Cf. A004987. - Peter Bala, Oct 13 2024
a(n) ~ c * 9^n / n^(1/3), where c = 1/Gamma(2/3) = 1/A073006 = 0.738488... . - Amiram Eldar, Aug 17 2025

A078532 Coefficients of power series that satisfies A(x)^3 - 9*x*A(x)^4 = 1, A(0)=1.

Original entry on oeis.org

1, 3, 27, 315, 4158, 59049, 880308, 13586859, 215233605, 3479417370, 57168561996, 951892141473, 16026585711660, 272383068872700, 4666865660812044, 80521573261807755, 1397858693681272230, 24398716826612190447, 427921056863230599900, 7537621933880388620010
Offset: 0

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Author

Paul D. Hanna, Nov 28 2002

Keywords

Comments

If A(x) = Sum_{k>=1} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2) (conjecture).
If A(x) = Sum_{k>=1} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(k)=n^(2k)*binomial(k/n+1/n+k-1,k)/(k+1) and, consequently, a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2). - Emeric Deutsch, Dec 10 2002
A generalization of the Catalan sequence (A000108) since for n = 1 the equation A(x)^n -(n^2)*x*A(x)^(n+1) = 1 reduces to A(x)=1+xA(x)^2. - Emeric Deutsch, Dec 10 2002
Radius of convergence of g.f. A(x) is r = 1/(3*4^(4/3)) where A(r) = 4^(1/3). - Paul D. Hanna, Jul 24 2012
Self-convolution cube yields A214668.

Examples

			A(x)^3 - 9x*A(x)^4 = 1 since A(x)^3 = 1 +9x +108x^2 +1458x^3 +21060x^4 +... and A(x)^4 = 1 +12x +162x^2 +2340x^3 +... also a(2)=3^3, a(5)=3^10.
		

Crossrefs

Programs

  • Mathematica
    Table[3^(2n) Binomial[(4n-2)/3,n]/(n+1),{n,0,20}] (* Harvey P. Dale, Nov 03 2011 *)
  • PARI
    for(n=0,25, print1(9^n * binomial((4*n-2)/3, n)/(n+1), ", ")) \\ G. C. Greubel, Jan 26 2017

Formula

a(n) = 3^(2n)*binomial(4n/3-2/3, n)/(n+1). - Emeric Deutsch, Dec 10 2002
Sequence with offset 1 is expansion of reversion of g.f. x*(1-9*x)^(1/3), which equals x times the g.f. of A004990.
a(n) ~ 2^(8*n/3-5/6) * 3^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
D-finite with recurrence n*(n-1)*(n+1)*a(n) -216*(4*n-5)*(2*n-1)*(4*n-11)*a(n-3)=0. - R. J. Mathar, Mar 24 2023
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^5). - Seiichi Manyama, Jun 20 2025

Extensions

More terms from Harvey P. Dale, Nov 03 2011

A376636 G.f. A(x) satisfies A(x) = (1 + 9*x*A(x)^2)^(1/3).

Original entry on oeis.org

1, 3, 9, 18, 0, -162, -567, 0, 8019, 31590, 0, -520506, -2160756, 0, 38480265, 164549880, 0, -3072083274, -13390246485, 0, 258054995016, 1139882486490, 0, -22474826957232, -100257845970825, 0, 2011064804461548, 9039247392729582, 0, -183769714890451800
Offset: 0

Views

Author

Seiichi Manyama, Oct 23 2024

Keywords

Crossrefs

Programs

  • Mathematica
    A376636[n_] := 9^n*Binomial[(2*n + 1)/3, n]/(2*n + 1);
    Array[A376636, 35, 0] (* Paolo Xausa, Aug 04 2025 *)
  • PARI
    a(n) = 9^n*binomial(2*n/3+1/3, n)/(2*n+1);

Formula

a(n) = 9^n * binomial(2*n/3 + 1/3,n)/(2*n+1).
From Seiichi Manyama, Jun 20 2025: (Start)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)).
a(3*n+1) = 0 for n > 0. (End)
D-finite with recurrence n*(n-2)*a(n) +54*(2*n-5)*(n-4)*a(n-3)=0. - R. J. Mathar, Jul 30 2025

A034171 Related to triple factorial numbers A007559(n+1).

Original entry on oeis.org

1, 6, 42, 315, 2457, 19656, 160056, 1320462, 11003850, 92432340, 781473420, 6642524070, 56716936290, 486145168200, 4180848446520, 36059817851235, 311811366125385, 2702365173086670, 23467908082068450, 204170800313995515, 1779202688450532345, 15527587099204645920
Offset: 0

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Author

Keywords

Comments

Working with an offset of 1, we conjecture a(p*n) = a(n) (mod p^2) for prime p = 1 (mod 3) and all positive integers n except those n of the form n = m*p + k for 0 <= m <= (p-1)/3 and 1 <= k <= (p-1)/3. Cf. A298799, A004981 and A004982. - Peter Bala, Dec 23 2019

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(-1 + (1 - 9 x)^(-1/3))/(3 x), {x, 0, 19}], x] (* Michael De Vlieger, Oct 13 2019 *)

Formula

a(n) = 3^n*A007559(n+1)/(n+1)! where A007559(n+1)=(3*n+1)!!!.
G.f.: (-1+(1-9*x)^(-1/3))/(3*x).
a(n) = A035529(n+1, 1) (first column of triangle).
Convolution of A004987(n) with A025748(n+1), n >= 0.
From R. J. Mathar, Jan 28 2020: (Start)
D-finite with recurrence: (n+1)*a(n) + 3*(-3*n-1)*a(n-1) = 0.
G.f.: (1F0(1/3;;9*x)-1)/(3*x). (End)
Sum_{n>=0} 1/a(n) = 3/8 + 3*sqrt(3)*Pi/32 + 9*log(3)/32. - Amiram Eldar, Dec 22 2022
a(n) ~ 3^(2*n+1) * n^(-2/3) / Gamma(1/3). - Amiram Eldar, Aug 19 2025

A245114 G.f. A(x) satisfies A(x)^3 = 1 + 9*x*A(x)^5.

Original entry on oeis.org

1, 3, 36, 585, 10935, 221697, 4740120, 105225318, 2402040420, 56029889025, 1329627118248, 31998624800220, 779102941714461, 19157195459506230, 475034438632316400, 11865382635213387504, 298265217964573747095, 7539795161286074350785, 191548870595159091038640, 4888023169106780049244275
Offset: 0

Views

Author

Paul D. Hanna, Jul 31 2014

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 36*x^2 + 585*x^3 + 10935*x^4 + 221697*x^5 +...
where A(x)^3 = 1 + 9*x*A(x)^5:
A(x)^3 = 1 + 9*x + 135*x^2 + 2430*x^3 + 48195*x^4 + 1015740*x^5 +...
A(x)^5 = 1 + 15*x + 270*x^2 + 5355*x^3 + 112860*x^4 + 2480058*x^5 +...
		

Crossrefs

Programs

  • Maple
    rec:= 2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13):
    f:= gfun:-rectoproc({rec,a(0)=1,a(1)=3,a(2)=36},a(n),remember):
    map(f, [$0..30]); # Robert Israel, Jan 30 2018
  • Mathematica
    nmax = 19; sol = {a[0] -> 1};
    Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - (1 + 9 x A[x]^5) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
    sol /. HoldPattern[a[n_] -> k_] :> Set[a[n], k];
    a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)
  • PARI
    /* From A(x)^3 = 1 + 9*x*A(x)^5 : */
    {a(n) = local(A=1+x);for(i=1,n,A=(1 + 9*x*A^5 +x*O(x^n))^(1/3));polcoeff(A,n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n+1)}
    for(n=0,20,print1(a(n),", "))

Formula

a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n + 1).
2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13). - Robert Israel, Jan 30 2018
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7). - Seiichi Manyama, Jun 20 2025
a(n) ~ 3^n * 5^(5*n/3-1/6) / (sqrt(Pi) * 2^(2*(n+2)/3) * n^(3/2)). - Amiram Eldar, Sep 02 2025

A361375 Expansion of 1/(1 - 9*x/(1 - x))^(1/3).

Original entry on oeis.org

1, 3, 21, 165, 1380, 11982, 106626, 965442, 8854725, 82022115, 765787773, 7195638909, 67973370618, 644991134880, 6143707229880, 58714212503784, 562741793028282, 5407273475087934, 52074626299010130, 502513862912425650, 4857975310180620720
Offset: 0

Views

Author

Seiichi Manyama, Mar 28 2023

Keywords

Crossrefs

Programs

  • Maple
    a := n -> if n = 0 then 1 else 3*hypergeom([1 - n, 4/3], [2], -9) fi:
    seq(simplify(a(n)), n = 0..20); # Peter Luschny, Mar 30 2023
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/(1-9*x/(1-x))^(1/3))

Formula

a(n) = Sum_{k=0..n} (-9)^k * binomial(-1/3,k) * binomial(n-1,n-k).
a(0) = 1; a(n) = (3/n) * Sum_{k=0..n-1} (n+2*k) * a(k).
n*a(n) = (11*n-8)*a(n-1) - 10*(n-2)*a(n-2) for n > 1.
a(n) ~ 3^(2/3) * 10^(n - 1/3) / (Gamma(1/3) * n^(2/3)). - Vaclav Kotesovec, Mar 28 2023
a(n) = 3*hypergeom([1 - n, 4/3], [2], -9) for n >= 1. - Peter Luschny, Mar 30 2023

A008931 Expansion of (2/(1+sqrt(1-36*x)))^(1/3).

Original entry on oeis.org

1, 3, 45, 936, 22572, 592515, 16434495, 473825700, 14058408519, 426438391743, 13164565835421, 412255067017248, 13064028812911440, 418149414542496168, 13498863325944967656, 439006511643775469856, 14369623854340007790108, 473027210589699351461700
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • GAP
    a:=[1];; for n in [2..20] do a[n]:=6*(5-21*(n-1)+18*(n-1)^2)*a[n-1]/((n-1)*(3*n-2)); od; a; # G. C. Greubel, Sep 13 2019
  • Magma
    I:=[1]; [n le 1 select I[n] else 6*(5-21*(n-1)+18*(n-1)^2)*Self(n-1)/((n-1)*(3*n-2)): n in [1..20]]; // G. C. Greubel, Sep 13 2019
    
  • Maple
    seq(9^n*binomial(2*n +1/3, n)/(6*n+1), n=0..20); # G. C. Greubel, Sep 13 2019
  • Mathematica
    CoefficientList[Series[Surd[2/(1+Sqrt[1-36x]),3],{x,0,20}],x] (* Harvey P. Dale, Aug 12 2016 *)
    Table[9^n Binomial[2 n + 1/3, n]/(6 n + 1), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 12 2016 *)
  • PARI
    my(x='x+O('x^20)); Vec((2/(1+sqrt(1-36*x)))^(1/3)) \\ G. C. Greubel, Apr 11 2017
    
  • Sage
    [9^n*binomial(2*n +1/3, n)/(6*n+1) for n in (0..20)] # G. C. Greubel, Sep 13 2019
    

Formula

From Vladimir Reshetnikov, Oct 12 2016: (Start)
a(n) = 9^n*binomial(2*n + 1/3, n)/(6*n + 1).
D-finite with recurrence: n*(3*n+1)*a(n) = 6*(18*n^2-21*n+5)*a(n-1). (End)
a(n) ~ 2^(2*n-2/3)*3^(2*n-1)/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Oct 13 2016
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^9). - Seiichi Manyama, Jun 20 2025
G.f.: 2F1(1/6, 2/3 ; 4/3 ; 36*x). - R. J. Mathar, Jul 30 2025

A214763 G.f. satisfies: A(x) = 1/A(-x*A(x)^3).

Original entry on oeis.org

1, 2, 8, 40, 224, 1280, 7168, 40000, 231296, 1436928, 9773056, 72242176, 563679232, 4491707904, 35735001088, 280941652992, 2178641254400, 16710771339264, 127402021142528, 970887186407424, 7436390169329664, 57531833133899776, 451525691751628800, 3608174274928951296
Offset: 0

Views

Author

Paul D. Hanna, Jul 27 2012

Keywords

Comments

Compare to: G(x) = 1/G(-x*G(x)^3) when G(x) = 1 + x*G(x)^2 (A000108).
Compare to: B(x) = 1/B(-x*B(x)^3) when B(x) = 1/(1-9*x)^(1/3) = g.f. of A004987.
An infinite number of functions G(x) satisfy (*) G(x) = 1/G(-x*G(x)^3); for example, (*) is satisfied by G(x) = C(m*x) = (1-sqrt(1-4*m*x))/(2*m*x) for all m, where C(x) is the Catalan function.

Examples

			G.f.: A(x) = 1 + 2*x + 8*x^2 + 40*x^3 + 224*x^4 + 1280*x^5 + 7168*x^6 +...
A(x)^2 = 1 + 4*x + 20*x^2 + 112*x^3 + 672*x^4 + 4096*x^5 + 24640*x^6 +...
A(x)^3 = 1 + 6*x + 36*x^2 + 224*x^3 + 1440*x^4 + 9312*x^5 + 59456*x^6 +...
1/A(x) = A(-x*A(x)^3) = 1 - 2*x - 4*x^2 - 16*x^3 - 80*x^4 - 384*x^5 - 1664*x^6 - 7360*x^7 - 40832*x^8 - 304128*x^9 - 2667008*x^10 -...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+2*x);for(i=0,n,A=(A+1/subst(A,x,-x*A^3+x*O(x^n)))/2);polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

The g.f. of this sequence is the limit of the recurrence:
(*) G_{n+1}(x) = (G_n(x) + 1/G_n(-x*G_n(x)^3))/2 starting at G_0(x) = 1+2*x.

A004117 Numerators of expansion of (1-x)^(-1/3).

Original entry on oeis.org

1, 1, 2, 14, 35, 91, 728, 1976, 5434, 135850, 380380, 1071980, 9111830, 25933670, 74096200, 637227320, 1832028545, 5280552865, 137294374490, 397431084050, 1152550143745, 10043651252635, 29217894553120, 85112997176480
Offset: 0

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Keywords

Comments

For n >= 1, a(n) is also the numerator of beta(n+1/3,2/3)*sqrt(27)/(2*Pi). - Groux Roland, May 17 2011

Crossrefs

Programs

  • Mathematica
    Table[Numerator[Binomial[-1/3,n] (-1)^n],{n,0,40}] (* Vincenzo Librandi, Jun 13 2012 *)
  • PARI
    a(n)=prod(k=1,n,3*k-2)/n!*3^sum(k=1,n,valuation(k,3))

Formula

(1/n!) * 3^A054861(n) * Product_{k=0..n-1} (3k+1). - Ralf Stephan, Mar 13 2004
Numerators in (1-3t)^(-1/3) = 1 + t + 2*t^2 + (14/3)*t^3 + (35/3)*t^4 + (91/3)*t^5 + (728/9)*t^6 + (1976/9)*t^7 + (5434/9)*t^8 + ... = 1 + t + 4*t^2/2! + 28*t^3/3! + 280*t^4/4! + 3640*t^5/5! + 58240*t^6/6! + ... = e.g.f. for triple factorials A007559 (cf. A094638). - Tom Copeland, Dec 04 2013

Extensions

Typo in formula fixed by Pontus von Brömssen, Nov 25 2008
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