A118968 -id:A118968 - cp's OEIS frontend

A118971 a(n) = binomial(5*n+3,n)/(n+1).

1, 4, 26, 204, 1771, 16380, 158224, 1577532, 16112057, 167710664, 1772645420, 18974357220, 205263418941, 2240623268512, 24648785802336, 272994644359580, 3041495503591365, 34064252968167180, 383302465665133014

Offset: 0

Paul Barry, May 07 2006

A quadrisection of A118968.
For n >= 1, a(n-1) is the number of lattice paths from (0,0) to (4n,n) using only the steps (1,0) and (0,1) and which stay strictly below the line y = x/4 except at the path's endpoints. -  Lucas A. Brown, Aug 21 2020
This is instance k = 4 of the family {c(k, n+1)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment in A130564. - _Wolfdieter Lang, Feb 04 2024

Michael De Vlieger, Table of n, a(n) for n = 0..924
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
J. Sawada, J. Sears, A. Trautrim, and A. Williams, Demystifying our Grandparent's De Bruijn Sequences with Concatenation Trees, arXiv:2308.12405 [math.CO], 2023.

Cf. A000108, A006013, A006632, A130564, A130565, A234466, A234513, A234573, A235340 (members of the same family).

Table[4*Binomial[5n+3,n]/(4n+4),{n,0,30}] (* Harvey P. Dale, Apr 09 2012 *)

G.f.: If the inverse series of y*(1-y)^4 is G(x) then A(x)=G(x)/x.
D-finite with recurrence 8*(4*n+1)*(2*n+1)*(4*n+3)*(n+1)*a(n) -5*(5*n+1)*(5*n+2)*(5*n+3)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Nov 26 2012
a(n) = (4/5)*binomial(5*(n+1),n+1)/(5*(n+1)-1). - Bruno Berselli, Jan 17 2014
E.g.f.: 4F4(4/5,6/5,7/5,8/5; 5/4,3/2,7/4,2; 3125*x/256). - Ilya Gutkovskiy, Jan 23 2018
G.f.: 5F4([4,5,6,7,8]/5, [5,6,7,8]/4; (5^5/4^4)*x) = (4/(5*x))*(1 - 4F3([-1,1,2,3]/5, [1,2,3]/4; (5^5/4^4)*x)). - Wolfdieter Lang, Feb 15 2024

A118970 a(n) = 3*binomial(5n+2,n)/(4n+3).

Original entry on oeis.org

1, 3, 18, 136, 1155, 10530, 100688, 996336, 10116873, 104819165, 1103722620, 11777187240, 127067830773, 1383914371728, 15194457001440, 167996704221280, 1868870731122405, 20903064321375315, 234927317665726686

Offset: 0

Paul Barry, May 07 2006

A quadrisection of A118968.

Convolved with A118969 (1, 2, 11, 80, 665, ...) = A002294: (1, 5, 35, 285, 2530, ...) - Gary W. Adamson, Nov 07 2011

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B33.

Michael De Vlieger, Table of n, a(n) for n = 0..924
Clemens Heuberger, Sarah J. Selkirk, and Stephan Wagner, Enumeration of Generalized Dyck Paths Based on the Height of Down-Steps Modulo k, arXiv:2204.14023 [math.CO], 2022.
Henri Muehle and Philippe Nadeau, A Poset Structure on the Alternating Group Generated by 3-Cycles, arXiv:1803.00540 [math.CO], 2018.

Cf. A118969, A002294, A000108, A163456.

ogf := series(RootOf(A = 1 + x * A^5,A)^3, x=0, 30); # Mark van Hoeij, Apr 22 2013

Array[3 Binomial[5 # + 2, #]/(4 # + 3) &, 19, 0] (* Michael De Vlieger, May 30 2018 *)
CoefficientList[Series[HypergeometricPFQ[{3/5,4/5,6/5,7/5},{1,5/4,3/2,7/4},(5^5/4^4)x],{x,0,18}],x]Range[0,18]! (* Stefano Spezia, Oct 01 2024 *)

a(n)=3*binomial(5*n+2,n)/(4*n+3); \\ Joerg Arndt, Apr 23 2013

G.f.: F^3 where F is the g.f. of A002294. - Mark van Hoeij, Apr 23 2013
8*n*(4*n+1)*(2*n+1)*(4*n+3)*a(n) -5*(5*n+1)*(5*n+2)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Dec 02 2014
From Peter Bala, Oct 08 2015: (Start)
O.g.f. A(x) = (1/x) * series reversion ( x/C(x)^3 ), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
(1/3)*x*A'(x)/A(x) = x + 9*x^2 + 91*x^3 + 969*x^4 + ... is the o.g.f. for A163456. (End)
E.g.f.: hypergeom([3/5, 4/5, 6/5, 7/5], [1, 5/4, 3/2, 7/4], (5^5/4^4)*x). - Stefano Spezia, Oct 01 2024

A118969 a(n) = 2binomial(5n+1,n)/(4*n+2).

Original entry on oeis.org

1, 2, 11, 80, 665, 5980, 56637, 556512, 5620485, 57985070, 608462470, 6474009360, 69682358811, 757366074080, 8300675584120, 91634565938880, 1018002755977245, 11372548404732930, 127677890035721025, 1439777493407492640

Offset: 0

Paul Barry, May 07 2006

A quadrisection of A118968.

If y = x + 2*x^3 + x^5, the series reversion is x = y - 2*y^3 + 11*y^5 - 80*y^7 + 665*y^9 - ... - R. J. Mathar, Sep 29 2012

			a(3) = 80 = sum of top row terms in M^n = (35 + 35 + 9 + 1).

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Clemens Heuberger, Sarah J. Selkirk, and Stephan Wagner, Enumeration of Generalized Dyck Paths Based on the Height of Down-Steps Modulo k, arXiv:2204.14023 [math.CO], 2022.
Karol A. Penson and Karol Zyczkowski, Product of Ginibre matrices: Fuss-Catalan and Raney distribution, arXiv:1103.3453 [math-ph], 2011.
Karol A. Penson and Karol Zyczkowski, Product of Ginibre matrices: Fuss-Catalan and Raney distribution, Phys. Rev. E 83, 061118 (2011).
Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 7.
Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. (in Mandarin)

Cf. A000292, A118968, A006013, A233832, A234505, A234868, A002294.

[2*Binomial(5*n+1,n)/(4*n+2): n in [0..20]]; // Vincenzo Librandi, Aug 12 2011

Table[2*Binomial[5n+1,n]/(4n+2),{n,0,20}] (* Harvey P. Dale, Aug 21 2011 *)

a(n)=2*binomial(5*n+1,n)/(4*n+2); \\ Joerg Arndt, Apr 20 2013

From Gary W. Adamson, Aug 11 2011: (Start)
a(n) is sum of top row terms in M^n, where M is an infinite square production matrix with the tetrahedral series in each column (A000292), as follows:
   1,  1,  0,  0,  0,  0, ...
   4,  1,  1,  0,  0,  0, ...
  10, 10,  4,  1,  0,  0, ...
  20, 20, 10,  4,  1,  0, ...
  35, 35, 20, 10,  4,  1, ...
  ... (End)
G.f.: hypergeom([1/5, 2/5, 3/5, 4/5],[1/2, 3/4, 5/4], 3125*x/256)^2. - Mark van Hoeij, Apr 19 2013
a(n) = 2*binomial(5n+1,n-1)/n for n>0, a(0)=1. - Bruno Berselli, Jan 19 2014
D-finite with recurrence 8*n*(4*n+1)*(2*n+1)*(4*n-1)*a(n) - 5*(5*n+1)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1) = 0. - R. J. Mathar, Oct 10 2014
G.f. A(x) satisfies: A(x) = 1 / (1 - x * A(x)^2)^2. - Ilya Gutkovskiy, Nov 13 2021

A124753 a(3n+k) = (k+1)*binomial(4n+k, n)/(3n+k+1), where k is n reduced mod 3.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 4, 9, 15, 22, 52, 91, 140, 340, 612, 969, 2394, 4389, 7084, 17710, 32890, 53820, 135720, 254475, 420732, 1068012, 2017356, 3362260, 8579560, 16301164, 27343888, 70068713, 133767543, 225568798, 580034052, 1111731933, 1882933364, 4855986044, 9338434700

Offset: 0

Paul Barry, Nov 06 2006

Row sums of Riordan array (1,x(1-x^3))^(-1). Also row sums of A124752.
a(n) is the number of ordered trees (A000108) with n vertices in which every non-leaf non-root vertex has exactly two children that are leaves. For example, a(4) counts the 2 trees
  \ /
   |   and  \|/ . - David Callan, Aug 22 2014

J.-B. Priez and A. Virmaux, Non-commutative Frobenius characteristic of generalized parking functions: Application to enumeration, arXiv:1411.4161 [math.CO], 2014-2015.

Cf. A084080, A002293, A069271 (trisection), A006632 (trisection).

Cf. A047749, A118968.

A124753 := proc(n)
    local k,np;
    k := modp(n,3) ;
    np := floor(n/3) ;
    (k+1)*binomial(np+n,np)/(n+1) ;
end proc:
seq(A124753(n),n=0..40) ; # R. J. Mathar, Oct 30 2014

a[n_] := Module[{q, k}, {q, k} = QuotientRemainder[n, 3]; (k+1)*Binomial[4q + k, q]/(3q + k + 1)];
Table[a[n], {n, 0, 34}] (* Jean-François Alcover, Nov 20 2017 *)

{a(n)=local(A=1+x); for(i=1,n,A=1+x*A*exp(sum(m=1,n\3,3*polcoeff(log(A+x*O(x^n)),3*m)*x^(3*m))+x*O(x^n))); polcoeff(A,n)} \\ Paul D. Hanna, Jun 04 2012

apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
a(n) = apr(n\3, 4, n%3+1); \\ Seiichi Manyama, Jul 20 2025

a(3n) = A002293(n), a(3n+1) = A069271(n), a(3n+2) = A006632(n+1).
a(n) = ((mod(n,3)+1)*C(4*floor(n/3)+mod(n,3), floor(n/3))/ (3*floor(n/3) + 1 + mod(n, 3))). - Paul Barry, Dec 14 2006
G.f. satisfies: A(x) = 1 + x*A(x)^2*A(w*x)*A(w^2*x), where w = exp(2*Pi*I/3). - Paul D. Hanna, Jun 04 2012
G.f. satisfies: A(x) = 1 + x*A(x)*G(x^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293. - Paul D. Hanna, Jun 04 2012
Conjecture: +8019*n*(n-1)*(n+1)*a(n) +17496*n*(n-1)*(n-3)*a(n-1) +2592*(3*n-5)*(n-1)*(3*n-16)*a(n-2) +216*(-224*n^3+48*n^2+3926*n-6331)*a(n-3) +576*(-288*n^3+2448*n^2-6558*n+5443)*a(n-4) +768*(-288*n^3+3600*n^2-14878*n+20375)*a(n-5) -8192*(4*n-23)*(2*n-11)*(4*n-21)*a(n-6)=0. - R. J. Mathar, Oct 30 2014
a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/3)} a(3*k) * a(n-1-3*k). - Seiichi Manyama, Jul 07 2025

A386380 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/6)} a(6k) a(n-1-6*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 15, 24, 34, 45, 57, 70, 154, 253, 368, 500, 650, 819, 1827, 3045, 4495, 6200, 8184, 10472, 23562, 39627, 59052, 82251, 109668, 141778, 320866, 543004, 814506, 1142295, 1533939, 1997688, 4540200, 7718340, 11633440, 16398200, 22137570

Offset: 0

Seiichi Manyama, Jul 20 2025

nonn

Wikipedia, Fuss-Catalan number

Cf. A002296, A233832, A233833, A386392, A233834, A130565.

Cf. A047749, A118968, A124753, A386379, A386396.

A386380 := proc(n)
    option remember ;
    if n = 0 then
        1;
    else
        add(procname(6*k)*procname(n-1-6*k),k=0..floor((n-1)/6)) ;
    end if;
end proc:
seq(A386380(n),n=0..80) ; # R. J. Mathar, Jul 30 2025

apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
a(n) = apr(n\6, 7, n%6+1);

For k=0..5, a(6*n+k) = (k+1) * binomial(7*n+k+1,n)/(7*n+k+1).

G.f. A(x) satisfies A(x) = 1/(1 - x * Product_{k=0..5} A(w^k*x)), where w = exp(Pi*i/3).

A386379 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/5)} a(5k) a(n-1-5*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 13, 21, 30, 40, 51, 114, 190, 280, 385, 506, 1150, 1950, 2925, 4095, 5481, 12586, 21576, 32736, 46376, 62832, 145299, 250971, 383838, 548340, 749398, 1741844, 3025308, 4654320, 6690585, 9203634, 21475146, 37456650, 57887550

Offset: 0

Seiichi Manyama, Jul 20 2025

nonn

Wikipedia, Fuss-Catalan number

Cf. A002295, A212071, A212072, A212073, A130564.

Cf. A047749, A118968, A124753, A386380.

apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
a(n) = apr(n\5, 6, n%5+1);

For k=0..4, a(5*n+k) = (k+1) * binomial(6*n+k+1,n)/(6*n+k+1).

G.f. A(x) satisfies A(x) = 1/(1 - x * Product_{k=0..4} A(w^k*x)), where w = exp(2*Pi*i/5).

A385725 E.g.f. A(x) satisfies A(x) = exp( x(A(x) + A(ix) + A(-x) + A(-i*x))/4 ), where i is the imaginary unit.

Original entry on oeis.org

1, 1, 1, 1, 1, 6, 31, 106, 281, 3160, 29701, 176056, 768241, 12702704, 173361371, 1466276176, 8937060081, 195180709248, 3494232292681, 38426220716416, 301057954180801, 8174141246647552, 181144607099402871, 2452803139819922176, 23494461553739152201, 762800754226165963776

Offset: 0

Seiichi Manyama, Jul 08 2025

nonn

Cf. A000272, A058014, A385691.

Cf. A118968.

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/4)} (4*k+1) * binomial(n-1,4*k) * a(4*k) * a(n-1-4*k).

A386203 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/4)} binomial(n-1,4k) a(4k) a(n-1-4*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 7, 22, 57, 184, 949, 4984, 21649, 99728, 659443, 4777648, 29500593, 179618176, 1441372201, 13153104256, 105727977601, 808208897792, 7631709900607, 83311277669632, 825548919414057, 7638849184574464, 83126488334117149, 1050853652511099904

Offset: 0

Seiichi Manyama, Jul 14 2025

nonn

Cf. A000111, A000142, A386202.

Cf. A118968, A385725.

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, (i-1)\4, binomial(i-1, 4*j)*v[4*j+1]*v[i-4*j])); v;

E.g.f. A(x) satisfies A'(x) = A(x) * (A(x) + A(i*x) + A(-x) + A(-i*x))/4, where i is the imaginary unit.

A386396 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/7)} a(7k) a(n-1-7*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 17, 27, 38, 50, 63, 77, 92, 200, 325, 468, 630, 812, 1015, 1240, 2728, 4488, 6545, 8925, 11655, 14763, 18278, 40508, 67158, 98728, 135751, 178794, 228459, 285384, 635628, 1059380, 1566040, 2165800, 2869685, 3689595, 4638348

Offset: 0

Seiichi Manyama, Jul 20 2025

nonn

Wikipedia, Fuss-Catalan number

Cf. A007556, A234461, A234462, A234463, A234464, A234465, A234466.

Cf. A047749, A118968, A124753, A386379, A386380.

apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
a(n) = apr(n\7, 8, n%7+1);

For k=0..6, a(7*n+k) = (k+1) * binomial(8*n+k+1,n)/(8*n+k+1).

G.f. A(x) satisfies A(x) = 1/(1 - x * Product_{k=0..6} A(w^k*x)), where w = exp(2*Pi*i/7).

A214374 G.f. satisfies: A(x) = 1 + xA(x)^3A(-x)A(Ix)A(-Ix).

Original entry on oeis.org

1, 1, 2, 5, 14, 77, 272, 954, 3390, 20942, 81886, 313216, 1200192, 7744531, 31604104, 125795060, 500225166, 3302458490, 13797942732, 56190743623, 228422293810, 1529418005565, 6486104136768, 26809045163856, 110586420610480, 747542548312112, 3203080302079898

Offset: 0

Paul D. Hanna, Jul 14 2012

nonn

			G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 77*x^5 + 272*x^6 + 954*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 5*x^2 + 14*x^3 + 42*x^4 + 202*x^5 + 779*x^6 + 2900*x^7 +...
A(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + 90*x^4 + 402*x^5 + 1631*x^6 + 6372*x^7 +...
A(x)*A(-x) = 1 + 3*x^2 + 22*x^4 + 421*x^6 + 5386*x^8 + 127595*x^10 +...
A(x)*A(-x)*A(I*x)*A(-I*x) = 1 + 35*x^4 + 8730*x^8 + 3122301*x^12 + 1308019950*x^16 + 599139933987*x^20 + 290661820794000*x^24 +...

Cf. A118968.

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

cp's OEIS Frontend

A118971 a(n) = binomial(5*n+3,n)/(n+1).

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A118970 a(n) = 3*binomial(5n+2,n)/(4n+3).

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A118969 a(n) = 2*binomial(5*n+1,n)/(4*n+2).

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A124753 a(3n+k) = (k+1)*binomial(4n+k, n)/(3n+k+1), where k is n reduced mod 3.

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A386380 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/6)} a(6*k) * a(n-1-6*k).

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A386379 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/5)} a(5*k) * a(n-1-5*k).

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A385725 E.g.f. A(x) satisfies A(x) = exp( x*(A(x) + A(i*x) + A(-x) + A(-i*x))/4 ), where i is the imaginary unit.

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A386203 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/4)} binomial(n-1,4*k) * a(4*k) * a(n-1-4*k).

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A118969 a(n) = 2binomial(5n+1,n)/(4*n+2).

A386380 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/6)} a(6k) a(n-1-6*k).

A386379 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/5)} a(5k) a(n-1-5*k).

A385725 E.g.f. A(x) satisfies A(x) = exp( x(A(x) + A(ix) + A(-x) + A(-i*x))/4 ), where i is the imaginary unit.

A386203 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/4)} binomial(n-1,4k) a(4k) a(n-1-4*k).

A386396 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/7)} a(7k) a(n-1-7*k).

A214374 G.f. satisfies: A(x) = 1 + xA(x)^3A(-x)A(Ix)A(-Ix).