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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A251575 E.g.f.: exp(5*x*G(x)^4) / G(x)^4 where G(x) = 1 + x*G(x)^5 is the g.f. of A002294.

Original entry on oeis.org

1, 1, 5, 65, 1505, 51505, 2354725, 135258625, 9373203425, 761486105825, 71001537157925, 7475144493546625, 877222642396170625, 113551974107296500625, 16073867927431440597125, 2470217878902686107522625, 409596824402404827033730625, 72890993386914239524503090625, 13857243751694786173837746653125
Offset: 0

Views

Author

Paul D. Hanna, Dec 06 2014

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 65*x^3/3! + 1505*x^4/4! + 51505*x^5/5! +...
such that A(x) = exp(5*x*G(x)^4) / G(x)^4
where G(x) = 1 + x*G(x)^5 is the g.f. of A002294:
G(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + 23751*x^6 +...
Note that
A'(x) = exp(5*x*G(x)^4) = 1 + 5*x + 65*x^2/2! + 1505*x^3/3! + 51505*x^4/4! +...
LOGARITHMIC DERIVATIVE.
The logarithm of the e.g.f. begins:
log(A(x)) = x + 4*x^2/2 + 26*x^3/3 + 204*x^4/4 + 1771*x^5/5 +...
and so A'(x)/A(x) = G(x)^4.
TABLE OF POWERS OF E.G.F.
Form a table of coefficients of x^k/k! in A(x)^n as follows.
n=1: [1, 1,  5,   65,  1505,   51505,  2354725,  135258625, ...];
n=2: [1, 2, 12,  160,  3680,  124560,  5637760,  321147200, ...];
n=3: [1, 3, 21,  291,  6705,  225315, 10112805,  571694355, ...];
n=4: [1, 4, 32,  464, 10784,  361120, 16101760,  904145920, ...];
n=5: [1, 5, 45,  685, 16145,  540645, 23993725, 1339552925, ...];
n=6: [1, 6, 60,  960, 23040,  774000, 34254720, 1903435200, ...];
n=7: [1, 7, 77, 1295, 31745, 1072855, 47438125, 2626525615, ...];
n=8: [1, 8, 96, 1696, 42560, 1450560, 64195840, 3545600000, ...]; ...
in which the main diagonal begins (see A251585):
[1, 2, 21, 464, 16145, 774000, 47438125, 3545600000, ...]
and is given by the formula:
[x^n/n!] A(x)^(n+1) = 5^(n-3) * (n+1)^(n-4) * (16*n^3 + 87*n^2 + 172*n + 125) for n>=0.

Crossrefs

Cf. A251585, A251665, A002294, A118971.
Cf. Variants: A243953, A251573, A251574, A251576, A251577, A251578, A251579, A251580.

Programs

  • Mathematica
    Flatten[{1,1,Table[Sum[5^k * n!/k! * Binomial[5*n-k-5, n-k] * (k-1)/(n-1),{k,0,n}],{n,2,20}]}] (* Vaclav Kotesovec, Dec 07 2014 *)
  • PARI
    {a(n) = local(G=1);for(i=1,n,G=1+x*G^5 +x*O(x^n)); n!*polcoeff(exp(5*x*G^4)/G^4, n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    {a(n) = if(n==0||n==1, 1, sum(k=0, n, 5^k * n!/k! * binomial(5*n-k-5,n-k) * (k-1)/(n-1) ))}
for(n=0, 20, print1(a(n), ", "))

Formula

Let G(x) = 1 + x*G(x)^5 be the g.f. of A002294, then the e.g.f. A(x) of this sequence satisfies:
(1) A'(x)/A(x) = G(x)^4.
(2) A'(x) = exp(5*x*G(x)^4).
(3) A(x) = exp( Integral G(x)^4 dx ).
(4) A(x) = exp( Sum_{n>=1} A118971(n-1)*x^n/n ), where A118971(n-1) = binomial(5*n-2,n)/(4*n-1).
(5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251585.
(6) A(x) = Sum_{n>=0} A251585(n)*(x/A(x))^n/n! and
(7) [x^n/n!] A(x)^(n+1) = (n+1)*A251585(n)
where A251585(n) = 5^(n-3) * (n+1)^(n-5) * (16*n^3 + 87*n^2 + 172*n + 125).
a(n) = Sum_{k=0..n} 5^k * n!/k! * binomial(5*n-k-5, n-k) * (k-1)/(n-1) for n>1.
Recurrence: 8*(2*n-3)*(4*n-7)*(4*n-5)*(25*n^3 - 210*n^2 + 598*n - 581)*a(n) = 5*(15625*n^7 - 240625*n^6 + 1592500*n^5 - 5883125*n^4 + 13135350*n^3 - 17781015*n^2 + 13566657*n - 4523904)*a(n-1) - 3125*(25*n^3 - 135*n^2 + 253*n - 168)*a(n-2). - Vaclav Kotesovec, Dec 07 2014
a(n) ~ 5^(5*n-11/2) * n^(n-2) / (2^(8*n-9) * exp(n-1)). - Vaclav Kotesovec, Dec 07 2014

A251665 E.g.f.: exp(5*x*G(x)^4) / G(x) where G(x) = 1 + x*G(x)^5 is the g.f. of A002294.

Original entry on oeis.org

1, 4, 47, 1034, 34349, 1540480, 87311275, 5991370390, 483100288985, 44778459212540, 4691799973171175, 548418557098305250, 70754785462138421125, 9987462340422594014200, 1531136319790275407365475, 253347224928445454055920750, 45001449932636667231257800625, 8541130421294458307989700672500
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2014

Keywords

Examples

			E.g.f.: A(x) = 1 + 4*x + 47*x^2/2! + 1034*x^3/3! + 34349*x^4/4! + 1540480*x^5/5! +...
such that A(x) = exp(5*x*G(x)^4) / G(x)
where G(x) = 1 + x*G(x)^5 is the g.f. of A002294:
G(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + 23751*x^6 +...

Crossrefs

Cf. A251575, A251695, A002294.
Cf. Variants: A243953, A251663, A251664, A251666, A251667, A251668, A251669, A251670.

Programs

  • Mathematica
    Table[Sum[5^k * n!/k! * Binomial[5*n-k-2,n-k] * (4*k-1)/(4*n-1),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 07 2014 *)
  • PARI
    {a(n)=local(G=1); for(i=0, n, G=1+x*G^5 +x*O(x^n)); n!*polcoeff(exp(5*x*G^4)/G, n)}
for(n=0, 20, print1(a(n), ", "))
  • PARI
    {a(n) = sum(k=0, n, 5^k * n!/k! * binomial(5*n-k-2,n-k) * (4*k-1)/(4*n-1) )}
for(n=0, 20, print1(a(n), ", "))

Formula

Let G(x) = 1 + x*G(x)^5 be the g.f. of A002294, then the e.g.f. A(x) of this sequence satisfies:
(1) A'(x)/A(x) = G(x)^4 + 3*G'(x)/G(x).
(2) A(x) = F(x/A(x)^4) where F(x) is the e.g.f. of A251695.
(3) A(x) = Sum_{n>=0} A251695(n)*(x/A(x)^4)^n/n! where A251695(n) = (3*n+1) * (4*n+1)^(n-2) * 5^n .
(4) [x^n/n!] A(x)^(4*n+1) = (3*n+1) * (4*n+1)^(n-1) * 5^n.
a(n) = Sum_{k=0..n} 5^k * n!/k! * binomial(5*n-k-2,n-k) * (4*k-1)/(4*n-1) for n>=0.
Recurrence: 8*(2*n-1)*(4*n-3)*(4*n-1)*(1875*n^4 - 13375*n^3 + 35700*n^2 - 41905*n + 17681)*a(n) = 5*(1171875*n^8 - 11875000*n^7 + 51765625*n^6 - 126596875*n^5 + 189126875*n^4 - 174442875*n^3 + 93137550*n^2 - 22362645*n - 233856)*a(n-1) - 3125*(1875*n^4 - 5875*n^3 + 6825*n^2 - 3130*n - 24)*a(n-2). - Vaclav Kotesovec, Dec 07 2014
a(n) ~ 3 * 5^(5*n-3/2) / 2^(8*n-1) * n^(n-1) / exp(n-1). - Vaclav Kotesovec, Dec 07 2014

A382089 Expansion of e.g.f. (1/x) * Series_Reversion( x * exp(-x * B(x)^4) ), where B(x) = 1 + x*B(x)^5 is the g.f. of A002294.

Original entry on oeis.org

1, 1, 11, 268, 10301, 543576, 36542527, 2987431168, 287751180537, 31916479461760, 4006558784401811, 561568192339405824, 86932015931716588789, 14730649112418719484928, 2711977587454133506904775, 539042371050858695696121856, 115046065096051639979478349553
Offset: 0

Views

Author

Seiichi Manyama, Mar 15 2025

Keywords

Crossrefs

Cf. A382086, A382087, A382088.
Cf. A002294.

Programs

  • PARI
    a(n) = if(n==0, 1, (n-1)!*sum(k=0, n-1, (n+1)^(n-k-1)*binomial(4*n+k-1, k)/(n-k-1)!));

Formula

E.g.f. A(x) satisfies A(x) = exp(x*A(x) * B(x*A(x))^4).
a(n) = (n-1)! * Sum_{k=0..n-1} (n+1)^(n-k-1) * binomial(4*n+k-1,k)/(n-k-1)! for n > 0.
E.g.f.: exp( Series_Reversion( x * (1-x)^4 * exp(-x) ) ).

A186182 Expansion of 1/(1-x*A002294(x)).

Original entry on oeis.org

1, 1, 2, 8, 50, 388, 3363, 31132, 301156, 3007000, 30753169, 320492869, 3391067666, 36331532588, 393353506931, 4296895624750, 47300050998991, 524168531729460, 5842914510975105, 65470405191871331, 737008925038212059, 8331166456981245521
Offset: 0

Views

Author

Vladimir Kruchinin, Feb 14 2011

Keywords

Links

Crossrefs

Cf. A002294.

Programs

  • Mathematica
    Join[{1},Table[Sum[k/(4n-3k) Binomial[5n-4k-1,n-k],{k,n}],{n,30}]] (* Harvey P. Dale, Feb 05 2012 *)
  • PARI
    a(n)=max(1,sum(k=1,n, k/(4*n-3*k)*binomial(5*n-4*k-1,n-k)))

Formula

a(n) = Sum_{k=1..n} k/(4*n-3*k)*binomial(5*n-4*k-1,n-k), n>0; a(0) = 1.
From Vaclav Kotesovec, Sep 22 2024: (Start)
Recurrence: 8*(n-1)*(2*n-3)*(4*n-5)*(4*n-3)*(302869201*n^9 - 8459245881*n^8 + 104437088286*n^7 - 748013787426*n^6 + 3425181067929*n^5 - 10398368877609*n^4 + 20929298529704*n^3 - 26931712087164*n^2 + 20105202695760*n - 6634504195200)*a(n) = 5*(5*n-21)*(5*n-19)*(5*n-18)*(5*n-17)*(302869201*n^9 - 5733423072*n^8 + 47666412474*n^7 - 228372041208*n^6 + 694720947369*n^5 - 1391357951688*n^4 + 1834379283596*n^3 - 1535202635232*n^2 + 740132688960*n - 156635942400)*a(n-1) + 8*(n-1)*(2*n-3)*(4*n-5)*(4*n-3)*(302869201*n^9 - 8459245881*n^8 + 104437088286*n^7 - 748013787426*n^6 + 3425181067929*n^5 - 10398368877609*n^4 + 20929298529704*n^3 - 26931712087164*n^2 + 20105202695760*n - 6634504195200)*a(n-3) - 3*(341333589527*n^13 - 11542804387321*n^12 + 178153937603069*n^11 - 1661124543043265*n^10 + 10436085419810511*n^9 - 46636299048022863*n^8 + 152460394393134247*n^7 - 369062312013610715*n^6 + 661348648146586462*n^5 - 866258572340414716*n^4 + 805991085205781784*n^3 - 504373614185279520*n^2 + 190252933034572800*n - 32669988422400000)*a(n-4) + 5*(5*n-21)*(5*n-19)*(5*n-18)*(5*n-17)*(302869201*n^9 - 5733423072*n^8 + 47666412474*n^7 - 228372041208*n^6 + 694720947369*n^5 - 1391357951688*n^4 + 1834379283596*n^3 - 1535202635232*n^2 + 740132688960*n - 156635942400)*a(n-5).
a(n) ~ 5^(5*n + 7/2) / (314721 * sqrt(Pi) * n^(3/2) * 2^(8*n - 9/2)). (End)

A001764 a(n) = binomial(3*n,n)/(2*n+1) (enumerates ternary trees and also noncrossing trees).

Original entry on oeis.org

1, 1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, 1430715, 8414640, 50067108, 300830572, 1822766520, 11124755664, 68328754959, 422030545335, 2619631042665, 16332922290300, 102240109897695, 642312451217745, 4048514844039120, 25594403741131680, 162250238001816900
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Smallest number of straight line crossing-free spanning trees on n points in the plane.
Number of dissections of some convex polygon by nonintersecting diagonals into polygons with an odd number of sides and having a total number of 2n+1 edges (sides and diagonals). - Emeric Deutsch, Mar 06 2002
Number of lattice paths of n East steps and 2n North steps from (0,0) to (n,2n) and lying weakly below the line y=2x. - David Callan, Mar 14 2004
With interpolated zeros, this has g.f. 2*sqrt(3)*sin(arcsin(3*sqrt(3)*x/2)/3)/(3*x) and a(n) = C(n+floor(n/2),floor(n/2))*C(floor(n/2),n-floor(n/2))/(n+1). This is the first column of the inverse of the Riordan array (1-x^2,x(1-x^2)) (essentially reversion of y-y^3). - Paul Barry, Feb 02 2005
Number of 12312-avoiding matchings on [2n].
Number of complete ternary trees with n internal nodes, or 3n edges.
Number of rooted plane trees with 2n edges, where every vertex has even outdegree ("even trees").
a(n) is the number of noncrossing partitions of [2n] with all blocks of even size. E.g.: a(2)=3 counts 12-34, 14-23, 1234. - David Callan, Mar 30 2007
Pfaff-Fuss-Catalan sequence C^{m}_n for m=3, see the Graham et al. reference, p. 347. eq. 7.66.
Also 3-Raney sequence, see the Graham et al. reference, p. 346-7.
The number of lattice paths from (0,0) to (2n,0) using an Up-step=(1,1) and a Down-step=(0,-2) and staying above the x-axis. E.g., a(2) = 3; UUUUDD, UUUDUD, UUDUUD. - Charles Moore (chamoore(AT)howard.edu), Jan 09 2008
a(n) is (conjecturally) the number of permutations of [n+1] that avoid the patterns 4-2-3-1 and 4-2-5-1-3 and end with an ascent. For example, a(4)=55 counts all 60 permutations of [5] that end with an ascent except 42315, 52314, 52413, 53412, all of which contain a 4-2-3-1 pattern and 42513. - David Callan, Jul 22 2008
Central terms of pendular triangle A167763. - Philippe Deléham, Nov 12 2009
With B(x,t)=x+t*x^3, the comp. inverse in x about 0 is A(x,t) = Sum_{j>=0} a(j) (-t)^j x^(2j+1). Let U(x,t)=(x-A(x,t))/t. Then DU(x,t)/Dt=dU/dt+U*dU/dx=0 and U(x,0)=x^3, i.e., U is a solution of the inviscid Burgers's, or Hopf, equation. Also U(x,t)=U(x-t*U(x,t),0) and dB(x,t)/dt = U(B(x,t),t) = x^3 = U(x,0). The characteristics for the Hopf equation are x(t) = x(0) + t*U(x(t),t) = x(0) + t*U(x(0),0) = x(0) + t*x(0)^3 = B(x(0),t). These results apply to all the Fuss-Catalan sequences with 3 replaced by n>0 and 2 by n-1 (e.g., A000108 with n=2 and A002293 with n=4), see also A086810, which can be generalized to A133437, for associahedra. - Tom Copeland, Feb 15 2014
Number of intervals (i.e., ordered pairs (x,y) such that x<=y) in the Kreweras lattice (noncrossing partitions ordered by refinement) of size n, see the Bernardi & Bonichon (2009) and Kreweras (1972) references. - Noam Zeilberger, Jun 01 2016
Number of sum-indecomposable (4231,42513)-avoiding permutations. Conjecturally, number of sum-indecomposable (2431,45231)-avoiding permutations. - Alexander Burstein, Oct 19 2017
a(n) is the number of topologically distinct endstates for the game Planted Brussels Sprouts on n vertices, see Ji and Propp link. - Caleb Ji, May 14 2018
Number of complete quadrillages of 2n+2-gons. See Baryshnikov p. 12. See also Nov 10 2014 comments in A134264. - Tom Copeland, Jun 04 2018
a(n) is the number of 2-regular words on the alphabet [n] that avoid the patterns 231 and 221. Equivalently, this is the number of 2-regular tortoise-sortable words on the alphabet [n] (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
a(n) is the number of Motzkin paths of length 3n with n steps of each type, with the condition that (1, 0) and (1, 1) steps alternate (starting with (1, 0)). - Helmut Prodinger, Apr 08 2019
a(n) is the number of uniquely sorted permutations of length 2n+1 that avoid the patterns 312 and 1342. - Colin Defant, Jun 08 2019
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. - Tom Copeland, Dec 13 2019
The sequences of Fuss-Catalan numbers, of which this is the first after the Catalan numbers A000108 (the next is A002293), appear in articles on random matrices and quantum physics. See Banica et al., Collins et al., and Mlotkowski et al. Interpretations of these sequences in terms of the cardinality of specific sets of noncrossing partitions are provided by A134264. - Tom Copeland, Dec 21 2019
Call C(p, [alpha], g) the number of partitions of a cyclically ordered set with p elements, of cyclic type [alpha], and of genus g (the genus g Faa di Bruno coefficients of type [alpha]). This sequence counts the genus 0 partitions (non-crossing, or planar, partitions) of p = 3n into n parts of length 3: a(n) = C(3n, [3^n], 0). For genus 1 see A371250, for genus 2 see A371251. - Robert Coquereaux, Mar 16 2024
a(n) is the total number of down steps before the first up step in all 2_1-Dyck paths of length 3*n for n > 0. A 2_1-Dyck path is a lattice path with steps (1,2), (1,-1) that starts and ends at y = 0 and does not go below the line y = -1. - Sarah Selkirk, May 10 2020
a(n) is the number of pairs (A<=B) of noncrossing partitions of [n]. - Francesca Aicardi, May 28 2022
a(n) is the number of parking functions of size n avoiding the patterns 231 and 321. - Lara Pudwell, Apr 10 2023
Number of rooted polyominoes composed of n square cells of the hyperbolic regular tiling with Schläfli symbol {4,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {4,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 3 of the family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment in A130564. - _Wolfdieter Lang, Feb 05 2024
The number of Apollonian networks (planar 3-trees) with n+3 vertices with a given base triangle. - Allan Bickle, Feb 20 2024
Number of rooted polyominoes composed of n tetrahedral cells of the hyperbolic regular tiling with Schläfli symbol {3,3,oo}. A rooted polyomino has one external face identified, and chiral pairs are counted as two. a(n) = T(n) in the second Beineke and Pippert link. - Robert A. Russell, Mar 20 2024

Examples

			a(2) = 3 because the only dissections with 5 edges are given by a square dissected by any of the two diagonals and the pentagon with no dissecting diagonal.
G.f. = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 + 43263*x^8 + ...

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 23.
  • I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
  • I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347. See also the Pólya-Szegő reference.
  • W. Kuich, Languages and the enumeration of planted plane trees. Nederl. Akad. Wetensch. Proc. Ser. A 73 = Indag. Math. 32, (1970), 268-280.
  • T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, p. 98.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, New York, Heidelberg, Berlin, 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).

Links

Crossrefs

Cf. A001762, A001763, A002294 - A002296, A006013, A025174, A063548, A064017, A072247, A072248, A134264, A143603, A258708, A256311, A188687 (binomial transform), A346628 (inverse binomial transform).
A column of triangle A102537.
Bisection of A047749 and A047761.
Row sums of triangles A108410 and A108767.
Second column of triangle A062993.
Mod 3 = A113047.
2D Polyominoes: A005034 (oriented), A005036 (unoriented), A369315 (chiral), A047749 (achiral), A000108 {3,oo}, A002293 {5,oo}.
3D Polyominoes: A007173 (oriented), A027610 (unoriented), A371350 (chiral), A371351 (achiral).
Cf. A130564 (for C(k, n) cases).

Programs

  • GAP
    List([0..25],n->Binomial(3*n,n)/(2*n+1)); # Muniru A Asiru, Oct 31 2018
  • Haskell
    a001764 n = a001764_list !! n
a001764_list = 1 : [a258708 (2 * n) n | n <- [1..]]
-- Reinhard Zumkeller, Jun 23 2015
  • Magma
    [Binomial(3*n,n)/(2*n+1): n in [0..30]]; // Vincenzo Librandi, Sep 04 2014
  • Maple
    A001764 := n->binomial(3*n,n)/(2*n+1): seq(A001764(n), n=0..25);
with(combstruct): BB:=[T,{T=Prod(Z,F),F=Sequence(B),B=Prod(F,Z,F)}, unlabeled]:seq(count(BB,size=i),i=0..22); # Zerinvary Lajos, Apr 22 2007
with(combstruct):BB:=[S, {B = Prod(S,S,Z), S = Sequence(B)}, labelled]: seq(count(BB, size=n)/n!, n=0..21); # Zerinvary Lajos, Apr 25 2008
n:=30:G:=series(RootOf(g = 1+x*g^3, g),x=0,n+1):seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 03 2015
alias(PS=ListTools:-PartialSums): A001764List := proc(m) local A, P, n;
A := [1,1]; P := [1]; for n from 1 to m - 2 do P := PS(PS([op(P), P[-1]]));
A := [op(A), P[-1]] od; A end: A001764List(25); # Peter Luschny, Mar 26 2022
  • Mathematica
    InverseSeries[Series[y-y^3, {y, 0, 24}], x] (* then a(n)=y(2n+1)=ways to place non-crossing diagonals in convex (2n+4)-gon so as to create only quadrilateral tiles *) (* Len Smiley, Apr 08 2000 *)
Table[Binomial[3n,n]/(2n+1),{n,0,25}] (* Harvey P. Dale, Jul 24 2011 *)
  • PARI
    {a(n) = if( n<0, 0, (3*n)! / n! / (2*n + 1)!)};
  • PARI
    {a(n) = if( n<0, 0, polcoeff( serreverse( x - x^3 + O(x^(2*n + 2))), 2*n + 1))};
  • PARI
    {a(n) = my(A); if( n<0, 0, A = 1 + O(x); for( m=1, n, A = 1 + x * A^3); polcoeff(A, n))};
  • PARI
    b=vector(22);b[1]=1;for(n=2,22,for(i=1,n-1,for(j=1,n-1,for(k=1,n-1,if((i-1)+(j-1)+(k-1)-(n-2),NULL,b[n]=b[n]+b[i]*b[j]*b[k])))));a(n)=b[n+1]; print1(a(0));for(n=1,21,print1(", ",a(n))) \\ Gerald McGarvey, Oct 08 2008
  • PARI
    Vec(1 + serreverse(x / (1+x)^3 + O(x^30))) \\ Gheorghe Coserea, Aug 05 2015
  • Python
    from math import comb
def A001764(n): return comb(3*n,n)//(2*n+1) # Chai Wah Wu, Nov 10 2022
  • Sage
    def A001764_list(n) :
    D = [0]*(n+1); D[1] = 1
    R = []; b = false; h = 1
    for i in range(2*n) :
        for k in (1..h) : D[k] += D[k-1]
        if not b : R.append(D[h])
        else : h += 1
        b = not b
    return R
A001764_list(22) # Peter Luschny, May 03 2012

Formula

From Karol A. Penson, Nov 08 2001: (Start)
G.f.: (2/sqrt(3*x))*sin((1/3)*arcsin(sqrt(27*x/4))).
E.g.f.: hypergeom([1/3, 2/3], [1, 3/2], 27/4*x).
Integral representation as n-th moment of a positive function on [0, 27/4]: a(n) = Integral_{x=0..27/4} (x^n*((1/12) * 3^(1/2) * 2^(1/3) * (2^(1/3)*(27 + 3 * sqrt(81 - 12*x))^(2/3) - 6 * x^(1/3))/(Pi * x^(2/3)*(27 + 3 * sqrt(81 - 12*x))^(1/3)))), n >= 0. This representation is unique. (End)
G.f. A(x) satisfies A(x) = 1+x*A(x)^3 = 1/(1-x*A(x)^2) [Cyvin (1998)]. - Ralf Stephan, Jun 30 2003
a(n) = n-th coefficient in expansion of power series P(n), where P(0) = 1, P(k+1) = 1/(1 - x*P(k)^2).
G.f. Rev(x/c(x))/x, where c(x) is the g.f. of A000108 (Rev=reversion of). - Paul Barry, Mar 26 2010
From Gary W. Adamson, Jul 07 2011: (Start)
Let M = the production matrix:
1, 1
2, 2, 1
3, 3, 2, 1
4, 4, 3, 2, 1
5, 5, 4, 3, 2, 1
...
a(n) = upper left term in M^n. Top row terms of M^n = (n+1)-th row of triangle A143603, with top row sums generating A006013: (1, 2, 7, 30, 143, 728, ...). (End)
Recurrence: a(0)=1; a(n) = Sum_{i=0..n-1, j=0..n-1-i} a(i)a(j)a(n-1-i-j) for n >= 1 (counts ternary trees by subtrees of the root). - David Callan, Nov 21 2011
G.f.: 1 + 6*x/(Q(0) - 6*x); Q(k) = 3*x*(3*k + 1)*(3*k + 2) + 2*(2*(k^2) + 5*k +3) - 6*x*(2*(k^2) + 5*k + 3)*(3*k + 4)*(3*k + 5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 27 2011
D-finite with recurrence: 2*n*(2n+1)*a(n) - 3*(3n-1)*(3n-2)*a(n-1) = 0. - R. J. Mathar, Dec 14 2011
REVERT transform of A115140. BINOMIAL transform is A188687. SUMADJ transform of A188678. HANKEL transform is A051255. INVERT transform of A023053. INVERT transform is A098746. - Michael Somos, Apr 07 2012
(n + 1) * a(n) = A174687(n).
G.f.: F([2/3,4/3], [3/2], 27/4*x) / F([2/3,1/3], [1/2], (27/4)*x) where F() is the hypergeometric function. - Joerg Arndt, Sep 01 2012
a(n) = binomial(3*n+1, n)/(3*n+1) = A062993(n+1,1). - Robert FERREOL, Apr 03 2015
a(n) = A258708(2*n,n) for n > 0. - Reinhard Zumkeller, Jun 23 2015
0 = a(n)*(-3188646*a(n+2) + 20312856*a(n+3) - 11379609*a(n+4) + 1437501*a(n+5)) + a(n+1)*(177147*a(n+2) - 2247831*a(n+3) + 1638648*a(n+4) - 238604*a(n+5)) + a(n+2)*(243*a(n+2) + 31497*a(n+3) - 43732*a(n+4) + 8288*a(n+5)) for all integer n. - Michael Somos, Jun 03 2016
a(n) ~ 3^(3*n + 1/2)/(sqrt(Pi)*4^(n+1)*n^(3/2)). - Ilya Gutkovskiy, Nov 21 2016
Given g.f. A(x), then A(1/8) = -1 + sqrt(5), A(2/27) = (-1 + sqrt(3))*3/2, A(4/27) = 3/2, A(3/64) = -2 + 2*sqrt(7/3), A(5/64) = (-1 + sqrt(5))*2/sqrt(5), etc. A(n^2/(n+1)^3) = (n+1)/n if n > 1. - Michael Somos, Jul 17 2018
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/3)*binomial(3*n,n)*x^n/n ).
The sequence defined by b(n) := [x^n] A(x)^n = A224274(n) for n >= 1 and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. Cf. A060941. (End)
G.f.: 1/sqrt(B(x)+(1-6*x)/(9*B(x))+1/3), with B(x):=((27*x^2-18*x+2)/54-(x*sqrt((-(4-27*x))*x))/(2*3^(3/2)))^(1/3). - Vladimir Kruchinin, Sep 28 2021
x*A'(x)/A(x) = (A(x) - 1)/(- 2*A(x) + 3) = x + 5*x^2 + 28*x^3 + 165*x^4 + ... is the o.g.f. of A025174. Cf. A002293 - A002296. - Peter Bala, Feb 04 2022
a(n) = hypergeom([1 - n, -2*n], [2], 1). Row sums of A108767. - Peter Bala, Aug 30 2023
G.f.: z*exp(3*z*hypergeom([1, 1, 4/3, 5/3], [3/2, 2, 2], (27*z)/4)) + 1.
- Karol A. Penson, Dec 19 2023
G.f.: hypergeometric([1/3, 2/3], [3/2], (3^3/2^2)*x). See the e.g.f. above. - Wolfdieter Lang, Feb 04 2024
a(n) = (3*n)! / (n!*(2*n+1)!). - Allan Bickle, Feb 20 2024
Sum_{n >= 0} a(n)*x^n/(1 + x)^(3*n+1) = 1. See A316371 and A346627. - Peter Bala, Jun 02 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^5). - Seiichi Manyama, Jun 16 2025

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

Views

Author

N. J. A. Sloane

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).

Links

Crossrefs

Column k=3 of triangle A062993 and A070914.
Cf. A000260, A002295, A002296, A027836, A062994, A346646 (binomial transform), A346664 (inverse binomial transform).
Cf. A006632, A006633, A006634, A025174, A069271, A196678, A224274, A233658, A233666, A233667, A277877, A283049, A283101, A283102, A283103.
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

A003463 a(n) = (5^n - 1)/4.

Original entry on oeis.org

0, 1, 6, 31, 156, 781, 3906, 19531, 97656, 488281, 2441406, 12207031, 61035156, 305175781, 1525878906, 7629394531, 38146972656, 190734863281, 953674316406, 4768371582031, 23841857910156, 119209289550781, 596046447753906, 2980232238769531
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

5^a(n) is the highest power of 5 dividing (5^n)!. - Benoit Cloitre, Feb 04 2002
n such that A002294(n) is not divisible by 5. - Benoit Cloitre, Jan 14 2003
Without leading zero, i.e., sequence {a(n+1) = (5*5^n-1)/4}, this is the binomial transform of A003947. - Paul Barry, May 19 2003 [Edited by M. F. Hasler, Oct 31 2014]
Numbers n such that a(n) is prime are listed in A004061(n) = {3, 7, 11, 13, 47, 127, 149, 181, 619, 929, ...}. Corresponding primes a(n) are listed in A086122(n) = {31, 19531, 12207031, 305175781, 177635683940025046467781066894531, ...}. 3^(m+1) divides a(2*3^m*k). 31 divides a(3k). 13 divides a(4k). 11 divides a(5k). 71 divides a(5k). 7 divides a(6k). 19531 divides a(7k). 313 divides a(8k). 19 divides a(9k). 829 divides a(9k). 71 divides a(10k). 521 divides a(10k). 17 divides a(16k). p divides a(p-1) for all prime p except p = {2,5}. p^(m+1) divides a(p^m*(p-1)) for all prime p except p = {2,5}. p divides a((p-1)/2) for prime p = {11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, ...} = A045468, Primes congruent to {1, 4} mod 5. p divides a((p-1)/3) for prime p = {13, 67, 127, 163, 181, 199, 211, 241, 313, 337, 367, 379, 409, 457, ...}. p divides a((p-1)/4) for prime p = {101, 109, 149, 181, 269, 389, 401, 409, 449, 461, 521, 541, ...} = A107219, Primes of the form x^2+100y^2. p divides a((p-1)/5) for prime p = {31, 191, 251, 271, 601, 641, 761, 1091, 1861, ...}. p divides a((p-1)/6) for prime p = {181, 199, 211, 241, 379, 409, 631, 691, 739, 769, 1039, ...}. - Alexander Adamchuk, Jan 23 2007
Starting with 1 = convolution square of A026375: (1, 3, 11, 45, 195, 873, ...). - Gary W. Adamson, May 17 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 27 2010
This is the sequence A(0,1;4,5;2) = A(0,1;6,-5;0) of the family of sequences [a,b:c,d:k] considered by Gary Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
It is the Lucas sequence U(6,5). - Felix P. Muga II, Mar 21 2014
a(2*n+1) is the sum of the numerators and denominators of the reduced fractions 0 < b/5^n < 1 plus 1, with b < 5^n. - J. M. Bergot, Jul 24 2015
The sequence multiplied by 10 (0, 10, 60, 310, 1560, ...) is the maximum number of coins which can be decided by n weighings on 2 balances in the counterfeit coin problem with undecided under/overweight. [Halbeisen and Hungerbuhler, Disc. Math. 147 (1995) 139 Theorem 1]. - R. J. Mathar, Sep 10 2015
Order of the rank-n projective geometry PG(n-1,5) over the finite field GF(5). - Anthony Hernandez, Oct 05 2016
Number of zeros in the substitution system {0 -> 11100, 1 -> 11110} at step n from initial string "1" (1 -> 11110 -> 1111011110111101111011100 -> ...). - Ilya Gutkovskiy, Apr 10 2017
a(n) is the numerator of Sum_{k=1..n} 1/5^k, which approaches a limit of 1/4. The denominators are 5^n. In general, Sum_{k=1..n} 1/x^k approaches a limit of 1/(x-1). It is of interest to note that as x increases, so does the rate of convergence. See Crossrefs for numerators for other values of x which have the general form (x^n-1)/(x-1). - Gary Detlefs, Aug 31 2021

Examples

			Base 5...........decimal
0......................0
1......................1
11.....................6
111...................31
1111.................156
11111................781
111111..............3906
1111111............19531
11111111...........97656
111111111.........488281
1111111111.......2441406
etc. ...............etc.
- _Zerinvary Lajos_, Jan 14 2007

References

  • Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 282.

Links

Crossrefs

Cf. A004061, A026375, A045468, A060458, A074479, A086122, A107219.
Cf. A003462, A002450, A003464, A023000, A023001, A002452, A002275.

Programs

Formula

Second binomial transform of A015518; binomial transform of A000302 (preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=1..n} binomial(n,k)*4^(k-1). - Paul Barry, Mar 28 2003
a(n) = (-1)^n times the (i, j)-th element of M^n (for all i and j such that i is not equal to j), where M = ((1, -1, 1, -2), (-1, 1, -2, 1), (1, -2, 1, -1), (-2, 1, -1, 1)). - Simone Severini, Nov 25 2004
a(n) = A125118(n,4) for n>3. - Reinhard Zumkeller, Nov 21 2006
a(n) = ((3+sqrt(4))^n - (3-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = 6*a(n-1) - 5*a(n-2) n>1, a(0)=0, a(1)=1. - Philippe Deléham, Jan 01 2009
From Wolfdieter Lang, Oct 18 2010: (Start)
O.g.f.: x/((1-5*x)*(1-x)).
a(n) = 4*a(n-1) + 5*a(n-2) + 2, a(0)=0, a(1)=1.
a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3), a(0)=0, a(1)=1, a(2)=6. Observation by G. Detlefs. See the W. Lang comment and link. (End)
a(n) = 5*a(n-1) + 1 with n>0, a(0)=0. - Vincenzo Librandi, Nov 17 2010
a(n) = a(n-1) + A000351(n-1) n>0, a(0)=0. - Felix P. Muga II, Mar 19 2014
a(n) = a(n-1) + 20*a(n-2) + 5 for n > 1, a(0)=0, a(1)=1. - Felix P. Muga II, Mar 19 2014
a(n) = A060458(n)/2^(n+2), for n > 0. - R. J. Cano, Sep 25 2014
From Ilya Gutkovskiy, Oct 05 2016: (Start)
E.g.f.: (exp(4*x) - 1)*exp(x)/4.
Convolution of A000351 and A057427. (End)

A002295 Number of dissections of a polygon: binomial(6n,n)/(5n+1).

Original entry on oeis.org

1, 1, 6, 51, 506, 5481, 62832, 749398, 9203634, 115607310, 1478314266, 19180049928, 251857119696, 3340843549855, 44700485049720, 602574657427116, 8175951659117794, 111572030260242090, 1530312970340384580, 21085148778264281865, 291705220704719165526
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates sextic (6-ary) trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m=6. See the Graham et al. reference, p. 347. eq. 7.66. See also the Pólya-Szegő reference.
Also 6-Raney sequence. See the Graham et al. reference, p. 346-7. (End)
This is instance k = 6 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

Examples

			There are a(2)=6 sextic trees (vertex degree <= 6 and 6 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these 6 trees yields 6*6 + binomial(6,2) = 51 = 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.
  • 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.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nürnberg, Jul 27 1994
  • 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).
  • Editor's note: "Über die Bestimmung der Anzahl der verschiedenen Arten, auf welche sich ein n-Eck durch Diagonalen in lauter m-Ecke zerlegen laesst, mit Bezug auf einige Abhandlungen der Herren Lamé, Rodrigues, Binet, Catalan und Duhamel in dem Journal de Mathématiques pures et appliquées, publié par Joseph Liouville. T. III. IV.", Archiv der Mathematik u. Physik, 1 (1841), pp. 193ff; see especially p. 198.

Links

Crossrefs

Cf. A001764, A002293, A002294, A002296, A130564.
Fifth column of triangle A062993.

Programs

Formula

O.g.f.: A(x) = 1 + x*A(x)^6 = 1/(1-x*A(x)^5).
a(n) = binomial(6*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.
a(n) = upper left term in M^n, M = the production matrix:
1, 1
5, 5, 1
15, 15, 5, 1
35, 35, 15, 5, 1
...
where (1, 5, 15, 35, ...) = A000332 starting with 1. - Gary W. Adamson, Jul 08 2011
a(n) are special values of Jacobi polynomials, in Maple notation:
a(n) = JacobiP(n-1, 5*n+1, -n, 1)/n, n=1, 2, ... . - Karol A. Penson, Mar 17 2015
a(n) = binomial(6*n+1, n)/(6*n+1) = A062993(n+4,4). - Robert FERREOL, Apr 03 2015
a(0) = 1; a(n) = Sum_{i1+i2+...+i6=n-1} a(i1)*a(i2)*...*a(i6) for n>=1. - Robert FERREOL, Apr 03 2015
D-finite with recurrence: 5*n*(5*n+1)*(5*n-3)*(5*n-2)*(5*n-1)*a(n) - 72*(6*n-5)*(6*n-1)*(3*n-1)*(2*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Sep 06 2016
From Ilya Gutkovskiy, Jan 15 2017: (Start)
O.g.f.: 5F4(1/6,1/3,1/2,2/3,5/6; 2/5,3/5,4/5,6/5; 46656*x/3125).
E.g.f.: 5F5(1/6,1/3,1/2,2/3,5/6; 2/5,3/5,4/5,1,6/5; 46656*x/3125).
a(n) ~ 3^(6*n+1/2)*64^n/(sqrt(Pi)*5^(5*n+3/2)*n^(3/2)). (End)
x*A'(x)/A(x) = (A(x) - 1)/(- 5*A(x) + 6) = x + 11*x^2 + 136*x^3 + 1771*x^4 + ... = (1/6)*Sum_{n >= 1} binomial(6*n,n)*x^n. Cf. A001764 and A002293 - A002296. - Peter Bala, Feb 04 2022
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^11). - Seiichi Manyama, Jun 16 2025

Extensions

More terms from Stefan Steinerberger, Apr 06 2006
Edited by M. F. Hasler, Apr 08 2015

A002296 Number of dissections of a polygon: binomial(7n,n)/(6n+1).

Original entry on oeis.org

1, 1, 7, 70, 819, 10472, 141778, 1997688, 28989675, 430321633, 6503352856, 99726673130, 1547847846090, 24269405074740, 383846168712104, 6116574500860880, 98106248306858715, 1582638261961640247, 25661404527790252375, 417980115131315136400
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n), n>=1, enumerates heptic (7-ary) trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m=7. See the Graham et al. reference, p. 347. eq. 7.66. See also the Pólya-Szegő reference.
Also 7-Raney sequence. See the Graham et al. reference, pp. 346-347.
a(n) = A258708(3*n,2*n) for n > 0. - Reinhard Zumkeller, Jun 23 2015
This is instance k = 7 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

Examples

			There are a(2)=7 heptic trees (vertex degree <= 7 and 7 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these 7 trees yields 7*7 + binomial(7,2) = 70 = a(3) such trees.

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347.
  • 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.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nürnberg, Jul 27 1994.
  • 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).

Links

Crossrefs

Cf. A001764, A002293, A002294, A002295.
Sixth column of triangle A062993.
Cf. A235535: binomial(9n,3n)/(6n+1); A235536: binomial(8n,2n)/(6n+1).
Cf. A258708.
Cf. A130564.

Programs

  • Haskell
    a002296 n = a002296_list !! n
a002296_list = [a258708 (4 * n) (3 * n) | n <- [1..]]
-- Reinhard Zumkeller, Jun 23 2015
  • Maple
    seq(binomial(7*n+1, n)/(7*n+1), n=0..30); # Robert FERREOL, Apr 02 2015
n:=30: G:=series(RootOf(g = 1+x*g^7, g), x=0, n+1): seq(coeff(G, x, k), k=0..n); # Robert FERREOL, Apr 02 2015
  • Mathematica
    Table[Binomial[7n,n]/(6n+1),{n,0,20}] (* Harvey P. Dale, Nov 21 2011 *)
  • PARI
    a(n)=binomial(7*n,n)/(6*n+1) \\ Charles R Greathouse IV, Feb 06 2012

Formula

O.g.f. A(x) = 1 + x*A(x)^7 = 1/(1-x*A(x)^6).
a(n) = binomial(7*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.
D-finite with recurrence: 72*n*(6*n-1)*(3*n-1)*(2*n-1)*(3*n-2)*(6*n+1)*a(n) - 7*(7*n-3)*(7*n-6)*(7*n-2)*(7*n-5)*(7*n-1)*(7*n-4)*a(n-1) = 0. - R. J. Mathar, Nov 16 2012
a(n) are special values of Jacobi polynomials, in Maple notation:
a(n) = JacobiP(n-1, 6*n+1, -n, 1)/n, n = 1, 2, ... . - Karol A. Penson, Mar 16 2015
a(n) = binomial(7*n+1, n)/(7*n+1) = A062993(n+5,5). - Robert FERREOL, Apr 02 2015
a(0) = 1; a(n) = Sum_{i1+i2+...+i7=n-1} a(i1)*a(i2)*...*a(i7) for n>=1. - Robert FERREOL, Apr 02 2015
From Ilya Gutkovskiy, Jan 16 2017: (Start)
O.g.f.: 6F5(1/7,2/7,3/7,4/7,5/7,6/7; 1/3,1/2,2/3,5/6,7/6; 823543*x/46656).
E.g.f.: 6F6(1/7,2/7,3/7,4/7,5/7,6/7; 1/3,1/2,2/3,5/6,1,7/6; 823543*x/46656).
a(n) ~ 7^(7*n+1/2)/(sqrt(Pi)*3^(6*n+3/2)*4^(3*n+1)*n^(3/2)). (End)
x*A'(x)/A(x) = (A(x) - 1)/(- 6*A(x) + 7) = x + 13*x^2 + 190*x^3 + 2925*x^4 + ... = (1/7)*Sum_{n >= 1} binomial(7*n,n)*x^n. Cf. A001764 and A002293, A002294, A002295. - Peter Bala, Feb 04 2022
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^13). - Seiichi Manyama, Jun 16 2025

Extensions

Pfaff-Fuss-Catalan, Raney, o.g.f. and 7-ary tree comments from Wolfdieter Lang, Sep 14 2007

A130564 Member k=5 of a family of generalized Catalan numbers.

Original entry on oeis.org

1, 5, 40, 385, 4095, 46376, 548340, 6690585, 83615350, 1064887395, 13770292256, 180320238280, 2386316821325, 31864803599700, 428798445360120, 5809228810425801, 79168272296871450, 1084567603590147950
Offset: 1

Views

Author

Wolfdieter Lang, Jul 13 2007

Keywords

Comments

The generalized Catalan numbers C(k,n):= binomial(k*n+1,n)/(k*n+1) become for negative k=-|k|, with |k|>=2, ((-1)^(n-1))*binomial((|k|+1)*n-2,n)/(|k|*n-1), n>=0.
The family c(k,n):=binomial((k+1)*n-2,n)/(k*n-1), n>=1, has the members A000108, A006013, A006632, A118971 for k=1,2,3,4, respectively (but the offset there is 0).
The members of the C(k,n) family for positive k are: A000012 (powers of 1), A000108, A001764, A002293, A002294, A002295, A002296, A007556, A062994, for k=1..9.
The ordinary generating functions for the k-family {c(k, n+1)}{n>=0}, k >= 1, are G(k, x) = hypergeometric(Aseq(k+1), Bseq(k), ((k+1)^(k+1)/k^k)*x), with Aseq(k+1) = [a(k)_1,..., a(k){k+1}], where a(k)j = (2*k - (j-1))/(k+1), and Bseq(k) = [b(k)_1, ..., b(k)_k], where b(k)_j = (2*k - (j-1))/k. The e.g.f. has an extra 1 in the B-section, which leads to a cancellation with the A-section 1 term. Thanks to Dixon J. Jones for asking for the general formulas. - _Wolfdieter Lang, Feb 04 2024
The o.g.f. of {C(k, n)}{n>=0} is in the Graham-Knuth-Patashnik book denoted as B_k(z) on pp. 200, 349 (2nd ed. 1994, pp. 200, 363). - _Wolfdieter Lang, Mar 07 2024

References

  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1994, pp. 200, 363.

Links

Crossrefs

Cf. A000012, A000108, A001764, A002293, A002294, A002295, A002296, A006013, A062994, A006632, A007556, A118971, A130565, A234466, A234513, A234573, A235340.

Programs

  • Mathematica
    Rest@ CoefficientList[InverseSeries[Series[y (1 - y)^5, {y, 0, 18}], x], x] (* Michael De Vlieger, Oct 13 2019 *)

Formula

a(n) = binomial((k+1)*n-2,n)/(k*n-1), with k=5.
G.f.: inverse series of y*(1-y)^5.
a(n) = (5/6)*binomial(6*n,n)/(6*n-1). [Bruno Berselli, Jan 17 2014]
From Wolfdieter Lang, Feb 06 2020: (Start)
G.f.: (5/6)*(1 - hypergeom([-1, 1, 2, 3, 4]/6, [1, 2, 3, 4]/5,(6^6/5^5)*x)).
E.g.f.: (5/6)*(1 - hypergeom([-1, 1, 2, 3, 4]/6, [1, 2, 3, 4, 5]/5,(6^6/5^5)*x)). (End)
D-finite with recurrence 5*n*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n) -72*(6*n-7)*(3*n-1)*(2*n-1)*(3*n-2)*(6*n-5)*a(n-1)=0. - R. J. Mathar, May 07 2021
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