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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A002293 Number of dissections of a polygon: binomial(4*n, n)/(3*n + 1).

Original entry on oeis.org

1, 1, 4, 22, 140, 969, 7084, 53820, 420732, 3362260, 27343888, 225568798, 1882933364, 15875338990, 134993766600, 1156393243320, 9969937491420, 86445222719724, 753310723010608, 6594154339031800, 57956002331347120, 511238042454541545
Offset: 0

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Author

Keywords

Comments

The number of rooted loopless n-edge maps in the plane (planar with a distinguished outside face). - Valery A. Liskovets, Mar 17 2005
Number of lattice paths from (1,0) to (3*n+1,n) which, starting from (1,0), only utilize the steps +(1,0) and +(0,1) and additionally, the paths lie completely below the line y = (1/3)*x (i.e., if (a,b) is in the path, then b < a/3). - Joseph Cooper (jecooper(AT)mit.edu), Feb 07 2006
Number of length-n restricted growth strings (RGS) [s(0), s(1), ..., s(n-1)] where s(0) = 0 and s(k) <= s(k-1) + 3, see fxtbook link below. - Joerg Arndt, Apr 08 2011
From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates quartic trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m = 4. See the Graham et al. reference, p. 347. eq. 7.66. (Second edition, p. 361, eq. 7.67.) See also the Pólya-Szegő reference.
Also 4-Raney sequence. See the Graham et al. reference, pp. 346-347.
(End)
Bacher: "We describe the statistics of checkerboard triangulations obtained by coloring black every other triangle in triangulations of convex polygons." The current sequence (A002293) occurs on p. 12 as one of two "extremal sequences" of an array of coefficients of polynomials, whose generating functions are given in terms of hypergeometric functions. - Jonathan Vos Post, Oct 05 2007
A generating function in terms of a (labyrinthine) solution to a depressed quartic equation is given in the Copeland link for signed A005810. With D(z,t) that g.f., a g.f. for signed A002293 is {[-1+1/D(z,t)]/(4t)}^(1/3). - Tom Copeland, Oct 10 2012
For a relation to the inviscid Burgers's equation, see A001764. - Tom Copeland, Feb 15 2014
For relations to compositional inversion, the Legendre transform, and convex geometry, see the Copeland, the Schuetz and Whieldon, and the Gross (p. 58) links. - Tom Copeland, Feb 21 2017 (See also Gross et al. in A062994. - Tom Copeland, Dec 24 2019)
This is the number of A'Campo bicolored forests of degree n and co-dimension 0. This can be shown using generating functions or a combinatorial approach. See Combe and Jugé link below. - Noemie Combe, Feb 28 2017
Conjecturally, a(n) is the number of 3-uniform words over the alphabet [n] that avoid the patterns 231 and 221 (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
The compositional inverse o.g.f. pair in Copeland's comment above are related to a pair of quantum fields in Balduf's thesis by Theorem 4.2 on p. 92. Cf. A001764. - Tom Copeland, Dec 13 2019
a(n) is the total number of down steps before the first up step in all 3_1-Dyck paths of length 4*n. A 3_1-Dyck path is a lattice path with steps (1, 3), (1, -1) that starts and ends at y = 0 and stays above the line y = -1. - Sarah Selkirk, May 10 2020
a(n) is the number of pairs (A<=B) of noncrossing partitions of [2n] such that every block of A has exactly two elements. In fact, it is proved that a(n) is the number of planar tied arc diagrams with n arcs (see Aicardi link below). A planar diagram with n arcs represents a noncrossing partition A of [2n] with n blocks, each block containing the endpoints of one arc; each tie connects two arcs, so that the ties define a partition B >= A: the endpoints of two arcs connected by a tie belong to the same block of B. Ties do not cross arcs nor other ties iff B has a planar diagram, i.e., B is a noncrossing partition. - Francesca Aicardi, Nov 07 2022
Dropping the initial 1 (starting 1, 4, 22 with offset 1) yields the REVERT transformation 1, -4 ,10, -20, 35.. essentially A000292 without leading 0. - R. J. Mathar, Aug 17 2023
Number of rooted polyominoes composed of n pentagonal cells of the hyperbolic regular tiling with Schläfli symbol {5,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {5,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 4 of the generalized Catalan family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment of A130564. - _Wolfdieter Lang, Feb 05 2024
a(n) is the cardinality of the planar ramified Jones monoid PR(J_n). - Diego Arcis, Nov 21 2024

Examples

			There are a(2) = 4 quartic trees (vertex degree <= 4 and 4 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these four trees yields 4*4 + 6 = 22 = a(3) such trees.
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 23.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347.
  • Peter Hilton and Jean Pedersen, Catalan numbers, their generalization, and their uses, Math. Intelligencer 13 (1991), no. 2, 64-75.
  • V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, Heidelberg, New York, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
  • 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

Column k=3 of triangle A062993 and A070914.
Cf. A000260, A002295, A002296, A027836, A062994, A346646 (binomial transform), A346664 (inverse binomial transform).
Polyominoes: A005038 (oriented), A005040 (unoriented), A369471 (chiral), A369472 (achiral), A001764 {4,oo}, A002294 {6,oo}.
Cf. A130564 (for generalized Catalan C(k, n), for = 4).

Programs

  • GAP
    List([0..22],n->Binomial(4*n,n)/(3*n+1)); # Muniru A Asiru, Nov 01 2018
  • Magma
    [ Binomial(4*n,n)/(3*n+1): n in [0..50]]; // Vincenzo Librandi, Apr 19 2011
    
  • Maple
    series(RootOf(g = 1+x*g^4, g),x=0,20); # Mark van Hoeij, Nov 10 2011
    seq(binomial(4*n, n)/(3*n+1),n=0..20); # Robert FERREOL, Apr 02 2015
    # Using the integral representation above:
    Digits:=6;
    R:=proc(x)((I + sqrt(3))*(4*sqrt(256 - 27*x) - 12*I*sqrt(3)*sqrt(x))^(1/3))/16 - ((I - sqrt(3))*(4*sqrt(256 - 27*x) + 12*I*sqrt(3)*sqrt(x))^(1/3))/16;end;
    W:=proc(x) x^(-3/4) * sqrt(4*R(x) - 3^(3/4)*x^(1/4)/sqrt(R(x)))/(2*3^(1/4)*Pi);end;
    # Attention: W(x) is singular at x = 0. Integration is done from  a very small positive x to x = 256/27.
    # For a(8):  ... gives 420732
    evalf(int(x^8*W(x),x=0.000001..256/27));
    # Karol A. Penson, Jul 05 2024
  • Mathematica
    CoefficientList[InverseSeries[ Series[ y - y^4, {y, 0, 60}], x], x][[Range[2, 60, 3]]]
    Table[Binomial[4n,n]/(3n+1),{n,0,25}] (* Harvey P. Dale, Apr 18 2011 *)
    CoefficientList[1 + InverseSeries[Series[x/(1 + x)^4, {x, 0, 60}]], x] (* Gheorghe Coserea, Aug 12 2015 *)
    terms = 22; A[] = 0; Do[A[x] = 1 + x*A[x]^4 + O[x]^terms, terms];
    CoefficientList[A[x], x] (* Jean-François Alcover, Jan 13 2018 *)
  • PARI
    a(n)=binomial(4*n,n)/(3*n+1) /* Charles R Greathouse IV, Jun 16 2011 */
    
  • PARI
    my(x='x+O('x^33)); Vec(1 + serreverse(x/(1+x)^4)) \\ Gheorghe Coserea, Aug 12 2015
    
  • Python
    A002293_list, x = [1], 1
    for n in range(100):
        x = x*4*(4*n+3)*(4*n+2)*(4*n+1)//((3*n+2)*(3*n+3)*(3*n+4))
        A002293_list.append(x) # Chai Wah Wu, Feb 19 2016
    

Formula

O.g.f. satisfies: A(x) = 1 + x*A(x)^4 = 1/(1 - x*A(x)^3).
a(n) = binomial(4*n,n-1)/n, n >= 1, a(0) = 1. From the Lagrange series of the o.g.f. A(x) with its above given implicit equation.
From Karol A. Penson, Apr 02 2010: (Start)
Integral representation as n-th Hausdorff power moment of a positive function on the interval [0, 256/27]:
a(n) = Integral_{x=0..256/27}(x^n((3/256) * sqrt(2) * sqrt(3) * ((2/27) * 3^(3/4) * 27^(1/4) * 256^(/4) * hypergeom([-1/12, 1/4, 7/12], [1/2, 3/4], (27/256)*x)/(sqrt(Pi) * x^(3/4)) - (2/27) * sqrt(2) * sqrt(27) * sqrt(256) * hypergeom([1/6, 1/2, 5/6], [3/4, 5/4], (27/256)*x)/ (sqrt(Pi) * sqrt(x)) - (1/81) * 3^(1/4) * 27^(3/4) * 256^(1/4) * hypergeom([5/12, 3/4, 13/12], [5/4, 3/2], (27/256)*x/(sqrt(Pi)*x^(1/4)))/sqrt(Pi))).
This representation is unique as it represents the solution of the Hausdorff moment problem.
O.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 4/3], (256/27)*x);
E.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 1, 4/3], (256/27)*x). (End)
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 1
6, 6, 3, 1
...
(where 1, 3, 6, 10, ...) is the triangular series. - Gary W. Adamson, Jul 08 2011
O.g.f. satisfies g = 1+x*g^4. If h is the series reversion of x*g, so h(x*g)=x, then (x-h(x))/x^2 is the o.g.f. of A006013. - Mark van Hoeij, Nov 10 2011
a(n) = binomial(4*n+1, n)/(4*n+1) = A062993(n+2,2). - Robert FERREOL, Apr 02 2015
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1-i} Sum_{k=0..n-1-i-j} a(i)*a(j)*a(k)*a(n-1-i-j-k) for n>=1; and a(0) = 1. - Robert FERREOL, Apr 02 2015
a(n) ~ 2^(8*n+1/2) / (sqrt(Pi) * n^(3/2) * 3^(3*n+3/2)). - Vaclav Kotesovec, Jun 03 2015
From Peter Bala, Oct 16 2015: (Start)
A(x)^2 is o.g.f. for A069271; A(x)^3 is o.g.f. for A006632;
A(x)^5 is o.g.f. for A196678; A(x)^6 is o.g.f. for A006633;
A(x)^7 is o.g.f. for A233658; A(x)^8 is o.g.f. for A233666;
A(x)^9 is o.g.f. for A006634; A(x)^10 is o.g.f. for A233667. (End)
D-finite with recurrence: a(n+1) = a(n)*4*(4*n + 3)*(4*n + 2)*(4*n + 1)/((3*n + 2)*(3*n + 3)*(3*n + 4)). - Chai Wah Wu, Feb 19 2016
E.g.f.: F([1/4, 1/2, 3/4], [2/3, 1, 4/3], 256*x/27), where F is the generalized hypergeometric function. - Stefano Spezia, Dec 27 2019
x*A'(x)/A(x) = (A(x) - 1)/(- 3*A(x) + 4) = x + 7*x^2 + 55*x^3 + 455*x^4 + ... is the o.g.f. of A224274. Cf. A001764 and A002294 - A002296. - Peter Bala, Feb 04 2022
a(n) = hypergeom([1 - n, -3*n], [2], 1). Row sums of A173020. - Peter Bala, Aug 31 2023
G.f.: t*exp(4*t*hypergeom([1, 1, 5/4, 3/2, 7/4], [4/3, 5/3, 2, 2], (256*t)/27))+1. - Karol A. Penson, Dec 20 2023
From Karol A. Penson, Jul 03 2024: (Start)
a(n) = Integral_{x=0..256/27} x^(n)*W(x)dx, n>=0, where W(x) = x^(-3/4) * sqrt(4*R(x) - 3^(3/4)*x^(1/4)/sqrt(R(x)))/(2*3^(1/4)*Pi), with R(x) = ((i + sqrt(3))*(4*sqrt(256 - 27*x) -12*i*sqrt(3*x))^(1/3))/16 - ((i - sqrt(3))*(4*sqrt(256 - 27*x) + 12*i* sqrt(3*x))^(1/3))/16, where i is the imaginary unit.
The elementary function W(x) is positive on the interval x = (0, 256/27) and is equal to the combination of hypergeometric functions in my formula from 2010; see above.
(Pi*W(x))^6 satisfies an algebraic equation of order 6, with integer polynomials as coefficients. (End)
G.f.: (Sum_{n >= 0} binomial(4*n+1, n)*x^n) / (Sum_{n >= 0} binomial(4*n, n)*x^n). - Peter Bala, Dec 14 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7). - Seiichi Manyama, Jun 16 2025

A349331 G.f. A(x) satisfies A(x) = 1 + x * A(x)^4 / (1 - x).

Original entry on oeis.org

1, 1, 5, 31, 219, 1678, 13570, 114014, 985542, 8708099, 78298727, 714105907, 6590200215, 61427125994, 577456943614, 5468604044500, 52122539760992, 499613409224137, 4813105582181533, 46576519080852235, 452545041339982871, 4413071971740021275, 43177663974461532959
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 15 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> coeff(series(RootOf(1+x*A^4/(1-x)-A, A), x, n+1), x, n):
    seq(a(n), n=0..22);  # Alois P. Heinz, Nov 15 2021
  • Mathematica
    nmax = 22; A[] = 0; Do[A[x] = 1 + x A[x]^4/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    Table[Sum[Binomial[n - 1, k - 1] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 22}]
  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0);
    A[#A] = 1 + sum(k=1, m-1, (polcoeff(Ser(A)^4, k)) )); A[n+1]}
    for(n=0, 30, print1(a(n), ", ")) \\ Vaclav Kotesovec, Nov 23 2024, after Paul D. Hanna

Formula

a(n) = Sum_{k=0..n} binomial(n-1,k-1) * binomial(4*k,k) / (3*k+1).
a(n) ~ 283^(n + 1/2) / (2^(7/2) * sqrt(Pi) * n^(3/2) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Nov 15 2021

A349310 G.f. A(x) satisfies: A(x) = (1 + x * A(x)^4) / (1 - x).

Original entry on oeis.org

1, 2, 10, 74, 642, 6082, 60970, 635818, 6826690, 74958914, 837833482, 9500939978, 109037364930, 1264049402754, 14780619799722, 174121322204074, 2064572904600706, 24620095821589378, 295087003429677322, 3552841638851183690, 42950428996378731010
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 14 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = (1 + x A[x]^4)/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    Table[Sum[Binomial[n + 3 k, 4 k] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 20}]

Formula

a(n) = Sum_{k=0..n} binomial(n+3*k,4*k) * binomial(4*k,k) / (3*k+1).
a(n) = F([(1+n)/3, (2+n)/3, (3+n)/3, -n], [2/3, 1, 4/3], -1), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 14 2021
a(n) ~ sqrt(1 + 3*r) / (2^(13/6) * sqrt(3*Pi) * (1-r)^(1/6) * n^(3/2) * r^(n + 1/3)), where r = 0.0766602099042102089064087954661556186872273232742446843... is the smallest real root of the equation 3^3 * (1-r)^4 = 4^4 * r. - Vaclav Kotesovec, Nov 15 2021

A346647 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(5*k,k) / (4*k + 1).

Original entry on oeis.org

1, 2, 8, 54, 460, 4361, 43988, 462580, 5014252, 55624944, 628432101, 7205500484, 83632219892, 980710882430, 11601345881748, 138278231052451, 1659037424218780, 20020306637339944, 242835190201382648, 2958961154058610552, 36203518795424475661
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 26 2021

Keywords

Comments

Binomial transform of A002294.

Crossrefs

Programs

  • Maple
    A346647 := proc(n)
        hypergeom([-n,1/5,2/5,3/5,4/5],[1/2,3/4,1,5/4],-3125/256) ;
        simplify(%) ;
    end proc:
    seq(A346647(n),n=0..40) ; # R. J. Mathar, Jan 10 2023
  • Mathematica
    Table[Sum[Binomial[n, k] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 20}]
    nmax = 20; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^3 A[x]^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 20; CoefficientList[Series[Sum[(Binomial[5 k, k]/(4 k + 1)) x^k/(1 - x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5, -n}, {1/2, 3/4, 1, 5/4}, -3125/256], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(5*k,k)/(4*k + 1)); \\ Michel Marcus, Jul 26 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^3 * A(x)^5.
G.f.: Sum_{k>=0} ( binomial(5*k,k) / (4*k + 1) ) * x^k / (1 - x)^(k+1).
a(n) ~ 3381^(n + 3/2) / (78125 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)). - Vaclav Kotesovec, Jul 30 2021
D-finite with recurrence +8*n*(4*n+1) *(2*n-1)*(4*n-1)*a(n) +(-4405*n^4 +9322*n^3 -7655*n^2 +2978*n -480)*a(n-1) +12*(n-1) *(1255*n^3 -3829*n^2 +4204*n -1640) *a(n-2) -2*(n-1) *(n-2) *(10655*n^2 -32221*n +26076) *a(n-3) +4*(n-1) *(n-2) *(n-3)*(3445*n -6922) *a(n-4) -3381*(n-1)*(n-2) *(n-3)*(n-4)*a(n-5)=0. - R. J. Mathar, Aug 17 2023

A349289 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * A(x)^3)).

Original entry on oeis.org

1, 2, 9, 61, 493, 4371, 41065, 401563, 4044097, 41658044, 436862457, 4648331765, 50057856881, 544557984498, 5975422922413, 66059269445451, 735064865871889, 8226310738656892, 92531697191189777, 1045551973586825023, 11862334695799444993
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 13 2021

Keywords

Crossrefs

Programs

  • Maple
    A349289 := proc(n)
        add( binomial(n+2*k,3*k)*binomial(4*k,k)/(3*k+1),k=0..n)  ;
    end proc:
    seq(A349289(n),n=0..50) ; # R. J. Mathar, Feb 10 2024
  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = 1/((1 - x) (1 - x A[x]^3)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    Table[Sum[Binomial[n + 2 k, 3 k] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 20}]

Formula

a(n) = Sum_{k=0..n} binomial(n+2*k,3*k) * binomial(4*k,k) / (3*k+1).
a(n) = F([1/4, 1/2, 3/4, (1+n)/2, (2+n)/2, -n], [1/3, 2/3, 2/3, 1, 4/3], -2^10/3^6) where F is the generalized hypergeometric function. - Stefano Spezia, Nov 13 2021
a(n) ~ sqrt(1 + 2*r) / (4 * 2^(1/6) * sqrt(3*Pi*(1-r)) * n^(3/2) * r^(n + 1/3)), where r = 0.0816785448577670972635343365300887975661075663022821172271... is the root of the equation 4^4 * r = 3^3 * (1-r)^3. - Vaclav Kotesovec, Nov 14 2021
D-finite with recurrence 81*n*(3*n-1)*(3*n+1)*a(n) +3*(243*n^3-7101*n^2+9986*n-3560)*a(n-1) +(-115027*n^3+514908*n^2-699869*n+269580)*a(n-2) +(-85543*n^3+1604715*n^2-6291692*n+6995280)*a(n-3) +(580211*n^3-6643158*n^2+23063299*n-23830944)*a(n-4) +(-33473*n^3-2231073*n^2+26352470*n-70945392)*a(n-5) +(-872129*n^3+17812344*n^2-119542699*n+264170868)*a(n-6) +(667171*n^3-14196243*n^2+100393472*n-236010000)*a(n-7) -6*(3*n-23)*(9948*n^2-147805*n+548868)*a(n-8) +4044*(3*n-26)*(n-8)*(3*n-22)*a(n-9)=0. - R. J. Mathar, Feb 10 2024
a(n) = 1 + Sum_{i, j, k, l>=0 and i+j+k+l=n-1} a(i) * a(j) * a(k) * a(l). - Seiichi Manyama, Jul 10 2025

A346648 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(6*k,k) / (5*k + 1).

Original entry on oeis.org

1, 2, 9, 73, 751, 8587, 104425, 1323952, 17303503, 231455104, 3153167249, 43597546197, 610232050453, 8629733401556, 123114479858631, 1769728635257503, 25607523627970183, 372688563309335806, 5451995469296025115, 80122698147986922194, 1182341393088427774071
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 26 2021

Keywords

Comments

Binomial transform of A002295.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 20}]
    nmax = 20; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^4 A[x]^6 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 20; CoefficientList[Series[Sum[(Binomial[6 k, k]/(5 k + 1)) x^k/(1 - x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[HypergeometricPFQ[{1/6, 1/3, 1/2, 2/3, 5/6, -n}, {2/5, 3/5, 4/5, 1, 6/5}, -46656/3125], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(6*k,k)/(5*k + 1)); \\ Michel Marcus, Jul 26 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^4 * A(x)^6.
G.f.: Sum_{k>=0} ( binomial(6*k,k) / (5*k + 1) ) * x^k / (1 - x)^(k+1).
a(n) ~ 49781^(n + 3/2) / (3359232 * sqrt(3*Pi) * n^(3/2) * 5^(5*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A346649 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(7*k,k) / (6*k + 1).

Original entry on oeis.org

1, 2, 10, 95, 1146, 15343, 218407, 3241316, 49588850, 776483636, 12383420161, 200444399493, 3284531747403, 54378741581471, 908238222519566, 15284835717461020, 258933935458506210, 4412025177612412048, 75564998345532498844, 1300158755391113561288
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 26 2021

Keywords

Comments

Binomial transform of A002296.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] Binomial[7 k, k]/(6 k + 1), {k, 0, n}], {n, 0, 19}]
    nmax = 19; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^5 A[x]^7 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 19; CoefficientList[Series[Sum[(Binomial[7 k, k]/(6 k + 1)) x^k/(1 - x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[HypergeometricPFQ[{1/7, 2/7, 3/7, 4/7, 5/7, 6/7, -n}, {1/3, 1/2, 2/3, 5/6, 1, 7/6}, -823543/46656], {n, 0, 19}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(7*k,k)/(6*k + 1)); \\ Michel Marcus, Jul 26 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^5 * A(x)^7.
G.f.: Sum_{k>=0} ( binomial(7*k,k) / (6*k + 1) ) * x^k / (1 - x)^(k+1).
a(n) ~ 870199^(n + 3/2) / (282475249 * sqrt(Pi) * n^(3/2) * 2^(6*n + 2) * 3^(6*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A346650 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(8*k,k) / (7*k + 1).

Original entry on oeis.org

1, 2, 11, 120, 1661, 25484, 415619, 7066670, 123865313, 2222178999, 40604688117, 753051711707, 14138552326609, 268204210248763, 5132686807360949, 98973130183436759, 1921142366704203305, 37508070639707177792, 736080632477073862271, 14511777729474947626918
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 26 2021

Keywords

Comments

Binomial transform of A007556.
In general, for m > 1, Sum_{k=0..n} binomial(n,k) * binomial(m*k,k) / ((m-1)*k + 1) ~ (m^m + (m-1)^(m-1))^(n + 3/2) / (sqrt(2*Pi) * m^((3*m-1)/2) * n^(3/2) * (m-1)^((m-1)*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] Binomial[8 k, k]/(7 k + 1), {k, 0, n}], {n, 0, 19}]
    nmax = 19; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^6 A[x]^8 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 19; CoefficientList[Series[Sum[(Binomial[8 k, k]/(7 k + 1)) x^k/(1 - x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[HypergeometricPFQ[{1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, -n}, {2/7, 3/7, 4/7, 5/7, 6/7, 1, 8/7}, -16777216/823543], {n, 0, 19}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(8*k,k)/(7*k + 1)); \\ Michel Marcus, Jul 26 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^6 * A(x)^8.
G.f.: Sum_{k>=0} ( binomial(8*k,k) / (7*k + 1) ) * x^k / (1 - x)^(k+1).
a(n) ~ 17600759^(n + 3/2) / (34359738368 * sqrt(Pi) * n^(3/2) * 7^(7*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A346664 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(4*k,k) / (3*k + 1).

Original entry on oeis.org

1, 0, 3, 12, 73, 453, 2985, 20373, 142933, 1024302, 7466211, 55182240, 412586977, 3115105321, 23717115513, 181884676827, 1403719428485, 10894049061956, 84967420574247, 665643698649684, 5235570329071893, 41328838600501830, 327315349579739619, 2600034901186102182
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 27 2021

Keywords

Comments

Inverse binomial transform of A002293.

Crossrefs

Programs

  • Maple
    A346664 := proc(n)
        add( (-1)^(n-k)*binomial(n,k)*binomial(4*k,k)/(3*k+1),k=0..n) ;
    end proc:
    seq(A346664(n),n=0..80); # R. J. Mathar, Aug 17 2023
  • Mathematica
    Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 23}]
    nmax = 23; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x)^2 A[x]^4 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    nmax = 23; CoefficientList[Series[Sum[(Binomial[4 k, k]/(3 k + 1)) x^k/(1 + x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
    Table[(-1)^n HypergeometricPFQ[{1/4, 1/2, 3/4, -n}, {2/3, 1, 4/3}, 256/27], {n, 0, 23}]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n,k)*binomial(4*k,k)/(3*k+1)); \\ Michel Marcus, Jul 28 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^2 * A(x)^4.
G.f.: Sum_{k>=0} ( binomial(4*k,k) / (3*k + 1) ) * x^k / (1 + x)^(k+1).
a(n) ~ 229^(n + 3/2) / (2048 * sqrt(2*Pi) * n^(3/2) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021
D-finite with recurrence +3*n*(3*n-1)*(3*n+1)*a(n) -74*n*(2*n-1) *(n-1)*a(n-1) -6*(n-1) *(101*n^2 -202*n +105)*a(n-2) -330*(n-1) *(n-2)*(2*n-3) *a(n-3) -229*(n-1)*(n-2) *(n-3)*a(n-4)=0. - R. J. Mathar, Aug 17 2023

A345367 a(n) = Sum_{k=0..n} binomial(4*k,k) / (3*k + 1).

Original entry on oeis.org

1, 2, 6, 28, 168, 1137, 8221, 62041, 482773, 3845033, 31188921, 256757719, 2139691083, 18015030073, 153008796673, 1309402039993, 11279339531413, 97724562251137, 851035285261745, 7445189624293545, 65401191955640665, 576639234410182210, 5101317352349364430
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 28 2021

Keywords

Comments

Partial sums of A002293.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 22}]
    nmax = 22; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^3 A[x]^4 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
  • PARI
    a(n) = sum(k=0, n, binomial(4*k, k)/(3*k+1)); \\ Michel Marcus, Jul 28 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^3 * A(x)^4.
a(n) ~ 2^(8*n + 17/2) / (229 * sqrt(Pi) * n^(3/2) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Jul 28 2021
D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*a(n) +(-283*n^3+384*n^2-173*n+24)*a(n-1) +8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-2)=0. - R. J. Mathar, Aug 05 2021
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