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 14 results. Next

A003168 Number of blobs with 2n+1 edges.

Original entry on oeis.org

1, 1, 4, 21, 126, 818, 5594, 39693, 289510, 2157150, 16348960, 125642146, 976789620, 7668465964, 60708178054, 484093913917, 3884724864390, 31348290348086, 254225828706248, 2070856216759478, 16936016649259364
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of ways to dissect a convex (2n+2)-gon with non-crossing diagonals so that no (2m+1)-gons (m>0) appear. - Len Smiley
a(n) is the number of plane trees with 2n+1 leaves and all non-leaves having an odd number > 1 of children. - Jordan Tirrell, Jun 09 2017
a(n) is the number of noncrossing cacti with n+1 nodes. See A361242. - Andrew Howroyd, Mar 07 2023

Examples

			a(2)=4 because we may place exactly one diagonal in 3 ways (forming 2 quadrilaterals), or not place any (leaving 1 hexagon).
		

References

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

Crossrefs

Cf. A049124 (no 2m-gons).
Row sums of A102537, A243662. Column 2 of A336573.

Programs

  • Haskell
    import Data.List (transpose)
    a003168 0 = 1
    a003168 n = sum (zipWith (*)
       (tail $ a007318_tabl !! n)
       ((transpose $ take (3*n+1) a007318_tabl) !! (2*n+1)))
       `div` fromIntegral n
    -- Reinhard Zumkeller, Oct 27 2013
  • Maple
    Order := 40; solve(series((A-2*A^3)/(1-A^2),A)=x,A);
    A003168 := n -> `if`(n=0,1,A100327(n)/2): seq(A003168(n),n=0..20); # Peter Luschny, Jun 10 2017
  • Mathematica
    a[0] = 1; a[n_] = (2^(-n-1)*(3n)!* Hypergeometric2F1[-1-2n, -2n, -3n, -1])/((2n+1)* n!*(2n)!); Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 25 2011, after Vladimir Kruchinin *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse((x-2*x^3)/(1-x^2)+O(x^(2*n+2))),2*n+1))
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=(1+x*A)/(1-x*A)^2); sum(k=0,n,polcoeff(A^(n-k),k))} \\ Paul D. Hanna, Nov 17 2004
    
  • PARI
    seq(n) = Vec( 1 + serreverse(x/((1+2*x)*(1+x)^2) + O(x*x^n)) ) \\ Andrew Howroyd, Mar 07 2023
    

Formula

a(n) = Sum_{k=1..n} binomial(n, k)*binomial(2*n+k, k-1)/n.
G.f.: A(x) = Sum_{n>=0} a(n)*x^(2*n+1) satisfies (A-2*A^3)/(1-A^2)=x. - Len Smiley.
D-finite with recurrence 4*n*(2*n + 1)*(17*n - 22)*a(n) = (1207*n^3 - 2769*n^2 + 1850*n - 360)*a(n - 1) - 2*(17*n - 5)*(n - 2)*(2*n - 3)*a(n - 2). - Vladeta Jovovic, Jul 16 2004
G.f.: A(x) = 1/(1-G003169(x)) where G003169(x) is the g.f. of A003169. - Paul D. Hanna, Nov 17 2004
a(n) = JacobiP(n-1,1,n+1,3)/n for n > 0. - Mark van Hoeij, Jun 02 2010
a(n) = (1/(2*n+1))*Sum_{j=0..n} (-1)^j*2^(n-j)*binomial(2*n+1,j)*binomial(3*n-j,2*n). - Vladimir Kruchinin, Dec 24 2010
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 1
5, 5, 3, 1
7, 7, 5, 3, 1
9, 9, 7, 5, 3, 1
... (End)
a(n) ~ sqrt(14+66/sqrt(17)) * (71+17*sqrt(17))^n / (sqrt(Pi) * n^(3/2) * 2^(4*n+4)). - Vaclav Kotesovec, Jul 01 2015
From Peter Bala, Oct 05 2015: (Start)
a(n) = (1/n) * Sum_{i = 0..n} 2^(n-i-1)*binomial(2*n,i)* binomial(n,i+1).
O.g.f. = 1 + series reversion( x/((1 + 2*x)*(1 + x)^2) ).
Logarithmically differentiating the modified g.f. 1 + 4*x + 21*x^2 + 126*x^3 + 818*x^4 + ... gives the o.g.f. for A114496, apart from the initial term. (End)
G.f.: A(x) satisfies A = 1 + x*A^3/(1-x*A^2). - Jordan Tirrell, Jun 09 2017
a(n) = A100327(n)/2 for n>=1. - Peter Luschny, Jun 10 2017

A054514 Number of ways to place non-crossing diagonals in convex (n+4)-gon so as to create no triangles or quadrilaterals.

Original entry on oeis.org

1, 1, 1, 5, 10, 16, 45, 109, 222, 540, 1341, 3065, 7328, 18112, 43530, 105390, 260254, 639244, 1570257, 3893805, 9669236, 24014264, 59903650, 149806494, 374982790, 940835404, 2365679689, 5955973237, 15018854005, 37935575685, 95942896837, 242954350457, 616034170069, 1563810857705, 3974000543475
Offset: 1

Views

Author

Len Smiley, Apr 08 2000

Keywords

Examples

			a(4)=5 because the octagon has the null placement and four ways to place a single diagonal.
		

Crossrefs

Programs

  • Mathematica
    f[x_] = InverseSeries[Series[(y - y^2 - y^4)/(1 - y), {y, 0, 38}], x];
    CoefficientList[(f[x] - x)/x^4, x]
    (* Second program: *)
    a[n_] := Sum[Binomial[n-2j-1, n-3j-1] Binomial[n+3+j, n+2]/(n+3), {j, 0, (n-1)/3}]; Array[a, 35] (* Jean-François Alcover, Dec 08 2018, after David Callan *)
    Table[HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, 1 - n/3, 4 + n}, {2, 1/2 - n/2, 1 - n/2}, -27/4], {n, 1, 40}] (* Vaclav Kotesovec, Sep 16 2023 *)

Formula

a(n) = Sum_{j=0..(n-1)/3} binomial[n-2j-1, n-3j-1] binomial[n+3+j, n+2]/(n+3). This counts the polygon dissections above by number j of diagonals. - David Callan, Jul 15 2004

Extensions

More terms from Joerg Arndt, Jan 28 2014

A101785 G.f. satisfies: A(x) = 1 + x*A(x)/(1 - x^2*A(x)^2).

Original entry on oeis.org

1, 1, 1, 2, 5, 12, 30, 79, 213, 584, 1628, 4600, 13138, 37871, 110043, 321978, 947813, 2805104, 8341608, 24912004, 74686460, 224694128, 678143656, 2052640752, 6229616730, 18952875247, 57792705415, 176596786934, 540679385663
Offset: 0

Views

Author

Paul D. Hanna, Dec 16 2004

Keywords

Comments

Formula may be derived using the Lagrange Inversion theorem (cf. A049124).
a(n) = number of Dyck n-paths (A000108) all of whose descents have odd length. For example, a(3) counts UUUDDD, UDUDUD. - David Callan, Jul 25 2005
The number of noncrossing partitions of [n] with all blocks of odd size. E.g.: a(4)=5 with the five partitions being 123/4, 124/3, 134/2,1/234 and 1/2/3/4. - Louis Shapiro, Jan 07 2006
Number of ordered trees with n edges in which every non-leaf vertex has an odd number of children. - David Callan, Mar 30 2007
Number of valid hook configurations of permutations of [n] that avoid the patterns 312 and 321. - Colin Defant, Apr 28 2019

Examples

			Generated from Fibonacci polynomials (A011973) and
coefficients of odd powers of 1/(1-x):
a(1) = 1*1/1
a(2) = 1*1/1 + 0*1/3
a(3) = 1*1/1 + 1*3/3
a(4) = 1*1/1 + 2*6/3 + 0*1/5
a(5) = 1*1/1 + 3*10/3 + 1*5/5
a(6) = 1*1/1 + 4*15/3 + 3*15/5 + 0*1/7
a(7) = 1*1/1 + 5*21/3 + 6*35/5 + 1*7/7
a(8) = 1*1/1 + 6*28/3 + 10*70/5 + 4*28/7 + 0*1/9
This process is equivalent to the formula:
a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1,k)*C(n,2*k)/(2*k+1).
		

Crossrefs

Column k=2 of A212382.

Programs

  • Magma
    [n eq 0 select 1 else (&+[Binomial(n-k-1,k)*Binomial(n, 2*k)/(2*k+1): k in [0..Floor((n-1)/2)]]): n in [0..30]]; // G. C. Greubel, May 03 2019
    
  • Mathematica
    Flatten[{1,Table[Sum[Binomial[n-k-1,k]*Binomial[n,2*k]/(2*k+1),{k,0,Floor[(n-1)/2]}],{n,1,20}]}] (* Vaclav Kotesovec, Sep 17 2013 *)
    CoefficientList[InverseSeries[Series[x*(1-x^2)/(1+x-x^2), {x, 0, 30}], x]/x, x] (* G. C. Greubel, May 03 2019 *)
  • PARI
    {a(n)=if(n==0,1,sum(k=0,(n-1)\2,binomial(n-k-1,k)*binomial(n,2*k)/(2*k+1)))}
    for(n=1, 40, print1(a(n), ", "))
    
  • PARI
    N=66; Vec(serreverse(x/(1+sum(k=1,N,x^(2*k-1)))+O(x^N))/x) /* Joerg Arndt, Aug 19 2012 */
    
  • Sage
    [1]+[sum(binomial(n-k-1, k)*binomial(n, 2*k)/(2*k+1) for k in (0..floor((n-1)/2))) for n in (1..30)] # G. C. Greubel, May 03 2019

Formula

a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1, k)*C(n, 2*k)/(2*k+1) for n>0, with a(0)=1.
G.f.: (1/x) * Series_Reversion( x*(1-x^2)/(1+x-x^2) ).
Recurrence: 4*n*(n+1)*(91*n^2 - 379*n + 360)*a(n) = 6*n*(182*n^3 - 849*n^2 + 1075*n - 264)*a(n-1) - 2*(182*n^4 - 1122*n^3 + 2011*n^2 - 603*n - 648)*a(n-2) + 6*(n-3)*(364*n^3 - 1698*n^2 + 2267*n - 696)*a(n-3) - 5*(n-4)*(n-3)*(91*n^2 - 197*n + 72)*a(n-4). - Vaclav Kotesovec, Sep 17 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 3/4 + 1/(4*sqrt(3/(19 - 304/(4103 + 273*sqrt(273))^(1/3) + 2*(4103 + 273*sqrt(273))^(1/3)))) + 1/2*sqrt(19/6 + 76/(3*(4103 + 273*sqrt(273))^(1/3)) - 1/6*(4103 + 273*sqrt(273))^(1/3) + 63/2*sqrt(3/(19 - 304/(4103 + 273*sqrt(273))^(1/3) + 2*(4103 + 273*sqrt(273))^(1/3)))) = 3.228704951094501729... is the root of the equation 5 - 24*d + 4*d^2 - 12*d^3 + 4*d^4 = 0 and c = 0.82499074317860885542266460957609663272... is the root of the equation -125 - 3376*c^2 - 22080*c^4 - 23296*c^6 + 93184*c^8 = 0. - Vaclav Kotesovec, added Sep 17 2013, updated Jan 04 2014
G.f.: 1/(9*(3-3*x+x^2))*(x^2+27- x^2*(2*x+3)^3*(x-6)^3/(9*(3-3*x+x^2)^3*S(0) - x^2*(2*x+3)^2*(x-6)^2 )), where S(k) = 4*k+3 - x^2*(2*x^2-9*x-18)^2*(3*k+4)*(6*k+5)/( 18*(4*k+5)*(3-3*x+x^2)^3 - x^2*(2*x^2-9*x-18)^2*(3*k+5)*(6*k+7)/S(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 26 2013

A054515 Number of ways to place non-intersecting diagonals in convex (n+2)-gon so as to create no quadrilaterals.

Original entry on oeis.org

1, 1, 2, 6, 21, 78, 301, 1198, 4888, 20340, 85986, 368239, 1594183, 6965380, 30675399, 136026759, 606848034, 2721783023, 12265670909, 55511013680, 252193872912, 1149742659556, 5258257323304, 24117924005616, 110915268468358, 511334146237807, 2362650323603539
Offset: 0

Views

Author

Len Smiley, Apr 08 2000

Keywords

Comments

Number of tree interval posets of permutations with n+1 minimal elements. - Mathilde Bouvel, Oct 21 2021

Examples

			a(3) = 6 because the pentagon allows null placement and five ways to place two diagonals.
		

Crossrefs

Cf. A046736, A049124, A003168, A054514, A348479 (free interv. posets not necess. trees).

Programs

  • Maple
    read("transforms") :
    taylor( (1-2*y+y^2-y^3)/(1-y),y=0,50) ;
    gfun[seriestolist](%) ;
    REVERT(%) ; # R. J. Mathar, Nov 04 2021
  • Mathematica
    InverseSeries[Series[(y-2*y^2+y^3-y^4)/(1-y), {y, 0, 24}], x] (* then A(x)=[y(x)-x]/x *)
  • PARI
    my(N=28, x='x+O('x^N)); Vec(serreverse((x-2*x^2+x^3-x^4)/(1-x))) \\ Hugo Pfoertner, Jan 26 2024

Formula

REVERT transform of (1-2*x+x^2-x^3)/(1-x) [Smiley].
a(n-1) = (1/n) * [binomial(2n-2,n-1) + Sum_{i=1..(n-3)} Sum_{k=1..Min(i,(n-i-1)/2)} binomial(n+i-1,i)*binomial(i,k)*binomial(n-i-k-2,k-1) ] if n>1. Proved in M. Bouvel, L. Cioni, B. Izart (Theorem 21) with offset 1. - Mathilde Bouvel, Oct 21 2021
G.f. A(z) = Sum_{n>=0} a(n)*z^n satisfies A(z) = 1 + z*A^2 + z^3*A^4/(1-z*A). Proved in M. Bouvel, L. Cioni, B. Izart (Equation (6) page 17 with offset 1). - Mathilde Bouvel, Oct 21 2021
Asymptotic behavior of a(n-1) is c*n^(-3/2)*r^n with c approximately 0.0792 and r approximately 4.8920. Proved in M. Bouvel, L. Cioni, B. Izart (Theorem 22). - Mathilde Bouvel, Oct 21 2021
D-finite with recurrence 23 *n *(n-1) *(12869043*n-33144451) *(n+1) *a(n) -n *(n-1) *(1989552043*n^2-6117767430*n+2643232213) * a(n-1) +(n-1) *(3359030609*n^3-15361701516*n^2+20123332181*n-6949961920) *a(n-2) +(-3560897749*n^4+25182507306*n^3-62054513365*n^2 +60006265908*n-16495478980) *a(n-3) +3*(146027817*n^4-1247820696*n^3+3378236999*n^2-2363753280*n-1468123920)*a(n-4) -3*(335627*n+695280) *(3*n-13) *(3*n-11) *(n-4) *a(n-5)=0. - R. J. Mathar, Oct 28 2021
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(n+k,k) * binomial(2*n-k,n-3*k). - Seiichi Manyama, Jan 26 2024

Extensions

a(0) = 1 prefixed by R. J. Mathar, Nov 04 2021

A307413 G.f. A(x) satisfies: A(x) = 1 + x*A(x)/(1 - x*A(x) - 2*x^2*A(x)^2).

Original entry on oeis.org

1, 1, 2, 7, 26, 102, 420, 1787, 7794, 34666, 156636, 716982, 3317700, 15494156, 72935624, 345701843, 1648489762, 7902956738, 38067806892, 184152092450, 894259126540, 4357738501844, 21302682030328, 104439435098718, 513390992000340, 2529846489669412, 12494572784556440
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 07 2019

Keywords

Comments

Unsigned version of A326564. - Paul D. Hanna, Aug 28 2019

Crossrefs

Programs

  • Mathematica
    terms = 26; A[] = 0; Do[A[x] = 1 + x A[x]/(1 - x A[x] - 2 x^2 A[x]^2) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]
    terms = 27; A[] = 0; Do[A[x] = 1 + Sum[(1/3) (2^k - (-1)^k) x^k A[x]^k, {k, 1, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]
    terms = 27; CoefficientList[1/x InverseSeries[Series[x (1 + x) (1 - 2 x)/(1 - 2 x^2), {x, 0, terms}], x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 + Sum_{k>=1} Jacobsthal(k)*x^k*A(x)^k, where Jacobsthal = A001045.
G.f.: A(x) = (1/x)*Series_Reversion(x*(1 + x)*(1 - 2*x)/(1 - 2*x^2)).
From Paul D. Hanna, Aug 29 2019: (Start)
G.f. A(x) satisfies: 0 = Sum_{n>=1} (1-(-1)^n - 2*A(x))^n * x^n / n.
G.f. A(x) satisfies: log(1 - 4*x^2*A(x)^2)/2 = arctanh(2*x - 2*x*A(x)). (End)

A307411 G.f. A(x) satisfies: A(x) = 1 + x*A(x)*(1 + 2*x*A(x))/(1 - x*A(x) - x^2*A(x)^2).

Original entry on oeis.org

1, 1, 4, 14, 60, 267, 1254, 6071, 30156, 152714, 785682, 4094752, 21573258, 114709363, 614777462, 3317589966, 18011350796, 98307220409, 539121535194, 2969177051678, 16415395615190, 91070109305056, 506843759000184, 2828968117483929, 15831944500607010, 88818114923080102
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 07 2019

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 25; A[] = 0; Do[A[x] = 1 + x A[x] (1 + 2 x A[x])/(1 - x A[x] - x^2 A[x]^2) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]
    terms = 26; A[] = 0; Do[A[x] = 1 + Sum[LucasL[k] x^k A[x]^k, {k, 1, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]
    terms = 26; CoefficientList[1/x InverseSeries[Series[x (1 - x - x^2)/(1 + x^2), {x, 0, terms}], x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 + Sum_{k>=1} Lucas(k)*x^k*A(x)^k, where Lucas = A000204.
G.f.: A(x) = (1/x)*Series_Reversion(x*(1 - x - x^2)/(1 + x^2)).

A307412 G.f. A(x) satisfies: A(x) = 1 + x*A(x)/(1 - 2*x*A(x) - x^2*A(x)^2).

Original entry on oeis.org

1, 1, 3, 12, 53, 250, 1234, 6295, 32925, 175616, 951596, 5223658, 28987546, 162349759, 916502869, 5209630108, 29792226533, 171284524184, 989460348216, 5740230703588, 33429379234924, 195361236443008, 1145312096390408, 6733896357727242, 39697441350016170, 234596104853541967
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 07 2019

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 25; A[] = 0; Do[A[x] = 1 + x A[x]/(1 - 2 x A[x] - x^2 A[x]^2) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]
    terms = 26; A[] = 0; Do[A[x] = 1 + Sum[Fibonacci[k, 2] x^k A[x]^k, {k, 1, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]
    terms = 26; CoefficientList[1/x InverseSeries[Series[x (1 - 2 x - x^2)/(1 - x - x^2), {x, 0, terms}], x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 + Sum_{k>=1} Pell(k)*x^k*A(x)^k, where Pell = A000129.
G.f.: A(x) = (1/x)*Series_Reversion(x*(1 - 2*x - x^2)/(1 - x - x^2)).
a(n) ~ sqrt((1 + 2^(1/3))*(4 + 7*2^(1/3))) * (2 + 3/2^(2/3) + 3/2^(1/3))^n / (3 * sqrt(Pi) * (2*n)^(3/2)). - Vaclav Kotesovec, Nov 05 2021

A269228 Number of nondirected diagonally convex polyominoes with perimeter 2n + 2.

Original entry on oeis.org

1, 2, 7, 28, 122, 556, 2618, 12634, 62128, 310212, 1568495, 8014742, 41323641, 214719610, 1123244757, 5910863420, 31268459118, 166185855552, 886961294034, 4751819567488, 25545030878475, 137756210983218, 745003421378887, 4039670554117446, 21957581725458521
Offset: 1

Views

Author

Svjetlan Feretic, Jul 11 2016

Keywords

Comments

The generating function satisfies an algebraic equation of degree eight. I computed that generating function using the "turbo Temperley" method.
The formula for the generating function is given in the enclosed Maple worksheet.
The most practical version of the "turbo Temperley" method was given in Bousquet-Mélou's paper cited below.
The first five terms are the same as in the sequence A005435.
A005435(n) is the number of column-convex polyominoes with perimeter 2n + 2.
A049124(n) is the number of directed diagonally convex polyominoes with perimeter 2n.

Examples

			a(7) = 2618, so there are 2618 nondirected diagonally convex polyominoes with perimeter 2*7 + 2 = 16.
		

Crossrefs

A290816 Number of dissections of an n-gon into polygons with odd number of sides counted up to rotations and reflections.

Original entry on oeis.org

1, 1, 2, 4, 8, 23, 65, 223, 757, 2824, 10559, 40994, 160734, 641420, 2584587, 10528305, 43237978, 178974779, 745814185, 3127246179, 13185588894, 55878618492, 237905685582, 1017225981255, 4366536472758, 18812074137141, 81320795918871, 352638701880227
Offset: 3

Views

Author

Evgeniy Krasko, Sep 03 2017

Keywords

Examples

			For a(5) = 2 the dissections of a pentagon are: a dissection into 3 triangles; a dissection into one pentagon.
		

Crossrefs

Cf. A049124 (counted distinctly).

Programs

Extensions

Terms a(16) and beyond from Andrew Howroyd, Nov 22 2017

A364833 G.f. satisfies A(x) = 1 + x*A(x)^2/(1 - x^3*A(x)^3).

Original entry on oeis.org

1, 1, 2, 5, 15, 49, 168, 595, 2160, 7997, 30083, 114660, 441840, 1718531, 6737820, 26600784, 105659970, 421949492, 1693120779, 6823018035, 27602090087, 112053680381, 456343848121, 1863893501065, 7633232165286, 31337360839387, 128944120202510
Offset: 0

Views

Author

Seiichi Manyama, Aug 28 2023

Keywords

Crossrefs

Programs

  • Maple
    A364833 := proc(n)
        add( binomial(n-2*k-1,k)*binomial(2*n-3*k+1,n-3*k)/ (2*n-3*k+1),k=0..floor(n/3)) ;
    end proc:
    seq(A364833(n),n=0..80); # R. J. Mathar, Aug 29 2023
  • PARI
    a(n) = sum(k=0, n\3, binomial(n-2*k-1, k)*binomial(2*n-3*k+1, n-3*k)/(2*n-3*k+1));

Formula

a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k-1,k) * binomial(2*n-3*k+1,n-3*k)/(2*n-3*k+1).
D-finite with recurrence 31*n*(626109182191*n-1858292669035) *(n-1)*(n+1) *a(n) -n*(n-1) *(244150473843619*n^2 -1454194662255591*n +2175006457069082) *a(n-1) +3*(n-1) *(292927551362415*n^3 -2593205532882651*n^2 +7084566217454162*n -5823331737745632)*a(n-2) +(-843955616916167*n^4 +9932491073296715*n^3 -42016891739306929*n^2 +76184884157722453*n -50166914106142776) *a(n-3) +18*(1509721335071*n^4 -40413442328880*n^3 +330301781039401*n^2 -1078322794857576*n +1231650372542192) *a(n-4) +18*(39673125909769*n^4 -598320530478001*n^3 +3228489073613917*n^2 -7321259523567459*n +5788776339353646) *a(n-5) +27*(n-5) *(3102413205331*n^3 -35996479327373*n^2 +114122791959960*n -64735736097804) *a(n-6) -243*(n-6) *(n-7)*(475638134099*n^2 -2399948859181*n +2877042451214) *a(n-7) -243*(45857481910*n -35520400961) *(n-5) *(n-7) *(n-8)*a(n-8)=0. - R. J. Mathar, Aug 29 2023
G.f.: (1/x) * Series_Reversion( x*(1 - x / (1 - x^3)) ). - Seiichi Manyama, Sep 28 2024
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