A066357 -id:A066357 - cp's OEIS frontend

A005900 Octahedral numbers: a(n) = n(2n^2 + 1)/3.

0, 1, 6, 19, 44, 85, 146, 231, 344, 489, 670, 891, 1156, 1469, 1834, 2255, 2736, 3281, 3894, 4579, 5340, 6181, 7106, 8119, 9224, 10425, 11726, 13131, 14644, 16269, 18010, 19871, 21856, 23969, 26214, 28595, 31116, 33781, 36594, 39559, 42680

Offset: 0

N. J. A. Sloane

Series reversion of g.f.: A(x) is Sum_{n>0} - A066357(n)(-x)^n.
Partial sums of centered square numbers A001844. - Paul Barry, Jun 26 2003
Also as a(n) = (1/6)*(4n^3 + 2n), n>0: structured tetragonal diamond numbers (vertex structure 5) (cf. A000447 - structured diamonds); and structured trigonal anti-prism numbers (vertex structure 5) (cf. A100185 - structured anti-prisms). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {3,4}.
If X is an n-set and Y and Z are disjoint 2-subsets of X then a(n-4) is equal to the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Starting with 1 = binomial transform of [1, 5, 8, 4, 0, 0, 0, ...] where (1, 5, 8, 4) = row 3 of the Chebyshev triangle A081277. - Gary W. Adamson, Jul 19 2008
a(n) = largest coefficient of (1 + ... + x^(n-1))^4. - R. H. Hardin, Jul 23 2009
Convolution square root of (1 + 6x + 19x^3 + ...) = (1 + 3x + 5x^2 + 7x^3 + ...) = A005408(x). - Gary W. Adamson, Jul 27 2009
Starting with offset 1 = the triangular series convolved with [1, 3, 4, 4, 4, ...]. - Gary W. Adamson, Jul 28 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral, and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Let b be any product of four different primes. Then the divisor lattice of b^n is of width a(n+1). - Jean Drabbe, Oct 13 2010
Arises in Bezdek's proof on contact numbers for congruent sphere packings (see preprint). - Jonathan Vos Post, Feb 08 2011
Euler transform of length 2 sequence [6, -2]. - Michael Somos, Mar 27 2011
a(n+1) is the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = 2n. - Clark Kimberling, Mar 19 2012
a(n) is the number of semistandard Young tableaux over all partitions of 3 with maximal element <= n. - Alois P. Heinz, Mar 22 2012
Self convolution of the odd numbers. - Reinhard Zumkeller, Apr 04 2012
a(n) is the number of (w,x,y,z) with all terms in {1,...,n} and w+x=y+z; also the number of (w,x,y,z) with all terms in {0,...,n} and |w-x|<=y. - Clark Kimberling, Jun 02 2012
The sequence is the third partial sum of (0, 1, 3, 4, 4, 4, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the number of join-irreducible elements in the Weyl group of type B_n with respect to the strong Bruhat order. - Rafael Mrden, Aug 26 2020
Number of unit octahedra contained in an n-scale octahedron composed of a tetrahedral-octahedral honeycomb. The number of unit tetrahedra in it is 8*A000292(n-1) = 4*(n^3 - n)/3. Also, the number of unit tetrahedra and unit octahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb is respectively A006527(n) = (n^3 + 2*n)/3 and A000292(n-1) = (n^3 - n)/6. - Jianing Song, Feb 24 2025

			G.f. = x + 6*x^2 + 19*x^3 + 44*x^4 + 85*x^5 + 146*x^6 + 231*x^7 + ...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
X. Acloque, Polynexus Numbers and other mathematical wonders [broken link]
Karoly Bezdek, Contact numbers for congruent sphere packings, arXiv:1102.1198 [math.MG], 2011.
Matteo Cavaleri and Alfredo Donno, Some degree and distance-based invariants of wreath products of graphs, arXiv:1805.08989 [math.CO], 2018.
Y-h. Guo, Some n-Color Compositions, J. Int. Seq. 15 (2012) 12.1.2, eq (5), m=2.
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Muhammad Fatih Killik and Bünyamin Şahi̇n, Further Results on Level Matrix, Int'l J. Math. Combin. (2024) Vol 4, 68-74. See p. 71.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Hankyung Ko, Volodymyr Mazorchuk and Rafael Mrđen, Join operation for the Bruhat order and Verma modules, arXiv:2109.01067 [math.RT], 2021. See Remark 5.10 p. 19.
A. Lascoux and M.-P. Schützenberger, Treillis et bases des groupes de Coxeter, Electron. J. Combin. 3 (1996), #R27.
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
J. K. Merikoski, R. Kumar and R. A. Rajput, Upper bounds for the largest eigenvalue of a bipartite graph, Electronic Journal of Linear Algebra ISSN 1081-3810, A publication of the International Linear Algebra Society, Volume 26, pp. 168-176, April 2013.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Octahedral Number.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Sums of 2 consecutive terms give A001845. Cf. A001844.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A022521.
Cf. A081277.
Row n=3 of A210391. - Alois P. Heinz, Mar 22 2012
Cf. A005408.
Cf. A053676, A053677, A053678.
Cf. A002061.
Cf. A000292 (tetrahedral numbers), A000578 (cubes), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Similar sequence: A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193(m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

a005900 n = sum $ zipWith (*) odds $ reverse odds
            where odds = take n a005408_list
a005900_list = scanl (+) 0 a001844_list
-- Reinhard Zumkeller, Jun 16 2013, Apr 04 2012

[n*(2*n^2+1)/3: n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015

I:=[0,1,6,19]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Sep 12 2015

al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(3,n), n=0..100)];
A005900:=(z+1)**2/(z-1)**4; # Simon Plouffe in his 1992 dissertation
with(combinat): seq(fibonacci(4,2*n)/12, n=0..40); # Zerinvary Lajos, Apr 21 2008

Table[(2n^3+n)/3, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,6,19},50] (* Harvey P. Dale, Oct 10 2013 *)
CoefficientList[Series[x (1 + x)^2/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)

makelist(n*(2*n^2+1)/3, n, 0, 20); /* Martin Ettl, Jan 07 2013 */

PARI
```
{a(n) = n*(2*n^2+1)/3};
```

concat([0],Vec(x*(1 + x)^2/(1 - x)^4 + O(x^50))) \\ Indranil Ghosh, Mar 16 2017

def a(n): return n*(2*n*n + 1)//3
print([a(n) for n in range(41)]) # Michael S. Branicky, Sep 03 2021

a(n) = 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2. - Amarnath Murthy, May 28 2001
G.f.: x * (1 + x)^2 / (1 - x)^4. a(n) = -a(-n) = (2*n^3 + n) / 3.
a(n) = ( ((n+1)^5-n^5) - (n^5-(n-1)^5) )/30. - Xavier Acloque, Oct 17 2003
a(n) is the sum of the products pq, where p and q are both positive and odd and p + q = 2n, e.g., a(4) = 7*1 + 5*3 + 3*5 + 1*7 = 44. - Jon Perry, May 17 2005
a(n) = 4*binomial(n,3) + 4*binomial(n,2) + binomial(n,1). - Mitch Harris, Jul 06 2006
a(n) = binomial(n+2,3) + 2*binomial(n+1,3) + binomial(n,3), (this pair generalizes; see A014820, the 4-cross polytope numbers).
Sum_{n>=1} 1/a(n) = 3*gamma + 3*Psi((I*(1/2))*sqrt(2)) - (1/2)*(3*I)*Pi*coth((1/2)*Pi*sqrt(2)) - (1/2)*(3*I)*sqrt(2) = A175577, where I=sqrt(-1). - Stephen Crowley, Jul 14 2009
a(n) = A035597(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = A000578(n) - 2*A000292(n-1) for n>0. - J. M. Bergot, Apr 05 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Wesley Ivan Hurt, Sep 11 2015
E.g.f.: (1/3)*x*(3 + 6*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Mar 16 2017
a(n) = (A002061(A002061(n+1)) - A002061(A002061(n)))/6. - Daniel Poveda Parrilla, Jun 10 2017
a(n) = 6*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 6*log(2) - 4 = 1/(6 + 2/(6 + 6/(6 + ... + n*(n-1)/(6 + ...)))). See A142983. - Peter Bala, Mar 06 2024

A001448 a(n) = binomial(4n,2n) or (4n)!/((2n)!(2n)!).

Original entry on oeis.org

1, 6, 70, 924, 12870, 184756, 2704156, 40116600, 601080390, 9075135300, 137846528820, 2104098963720, 32247603683100, 495918532948104, 7648690600760440, 118264581564861424, 1832624140942590534, 28453041475240576740, 442512540276836779204, 6892620648693261354600

Offset: 0

N. J. A. Sloane

Corollary 8 in Chapman et alia says: "For n>=1, there are binomial(4n,2n) binary sequences of length 4n+1 with the property that for all j, the j-th occurrence of 10 appears in positions 4j+1 and 4j+2 or later (if it exists at all)." - Peter Luschny, Nov 21 2011

Sequence terms are given by [x^n] ( (1 + x)^(k+2)/(1 - x)^k )^n for k = 2. See the cross references for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

			a(n) = (1/Pi)*Integral_{x=0..4} x^(2n)/sqrt(4-(x-2)^2) dx. - _Paul Barry_, Sep 17 2010
G.f. = 1 + 6*x + 70*x^2 + 924*x^3 + 12870*x^4 + 184756*x^5 + 2704156*x^6 + ...

T. D. Noe, Table of n, a(n) for n = 0..100
R. J. Chapman, T. Y. Chowa, A. Khetana, D. P. Moulton and R. J. Waters, Simple formulas for lattice paths avoiding certain periodic staircase boundaries, Journal of Combinatorial Theory, Series A 116 (2009) 205-214.
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
Quy Nhan, A finite sum of binomial coefficients, MathStackExchange, 2025-06-21.
Tito Piezas III, Fermat primes and Binomial sums, tpiezas, A Collection of Algebraic IdentitiesJune 3, 2012.
K. H. Pilehrood and T. H. Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7.
Ricardo A. Podestá, New identities for binary Krawtchouk polynomials, binomial coefficients and Catalan numbers, arXiv:1603.09156 [math.CO], 2016.
Renzo Sprugnoli, Sums of reciprocals of the central binomial coefficients, INTEGERS 6 (2006) #A27.

Bisection of A000984. Cf. A002458, A066357, A000984 (k = 0), A091527 (k = 1), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6).

Cf. A370100, A370101.

[Factorial(4*n)/(Factorial(2*n)*Factorial(2*n)): n in [0..20]]; // Vincenzo Librandi, Sep 13 2011

Maple
```
A001448 := n-> binomial(4*n,2*n) ;
```

Table[Binomial[4n,2n],{n,0,20}] (* Harvey P. Dale, Apr 26 2014 *)
a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-2 n, -2 n}, {1}, 1]]; (* Michael Somos, Oct 22 2014 *)

a(n)=binomial(4*n,2*n) \\ Charles R Greathouse IV, Sep 13 2011

from math import comb
def A001448(n): return comb(n<<2,n<<1) # Chai Wah Wu, Aug 10 2023

a(n) = A000984(2*n).
Using Stirling's formula in sequence A000142 it is easy to get the asymptotic expression a(n) ~ 16^n / sqrt(2 * Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
From Wolfdieter Lang, Dec 13 2001: (Start)
a(n) = 2*A001700(2*n-1) = (2*n+1)*C(2*n), n >= 1, C(n) := A000108(n) (Catalan).
G.f.: (1-y*((1+4*y)*c(y)-(1-4*y)*c(-y)))/(1-(4*y)^2) with y^2=x, c(y) = g.f. for A000108 (Catalan). (End)
a(n) ~ 2^(-1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - (1/16)*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
a(n) = (1/Pi)*Integral_{x=-2..2} (2+x)^(2*n)/sqrt((2-x)*(2+x)) dx. Peter Luschny, Sep 12 2011
G.f.: (1/2) * (1/sqrt(1+4*sqrt(x)) + 1/sqrt(1-4*sqrt(x))). - Mark van Hoeij, Oct 25 2011
Sum_{n>=1} 1/a(n) = 16/15 + Pi*sqrt(3)/27 - 2*sqrt(5)*log(phi)/25, [T. Trif, Fib Quart 38 (2000) 79] with phi=A001622. - R. J. Mathar, Jul 18 2012
D-finite with recurrence n*(2*n-1)*a(n) -2*(4*n-1)*(4*n-3)*a(n-1)=0. - R. J. Mathar, Dec 02 2012
G.f.: sqrt((1 + sqrt(1-16*x))/(2*(1-16*x))) = 1 + 6*x/(G(0)-6*x), where G(k) = 2*x*(4*k+3)*(4*k+1) + (2*k+1)*(k+1) - 2*x*(k+1)*(2*k+1)*(4*k+5)*(4*k+7)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jun 30 2013
a(n) = hypergeom([1-2*n,-2*n],[2],1)*(2*n+1). - Peter Luschny, Sep 22 2014
From Michael Somos, Oct 22 2014: (Start)
0 = a(n)*(+65536*a(n+2) - 16896*a(n+3) + 858*a(n+4)) + a(n+1)*(-3584*a(n+2) + 1176*a(n+3) - 66*a(n+4)) + a(n+2)*(+14*a(n+2) - 14*a(n+3) + a(n+4)) for all n in Z.
0 = a(n)^2*(+196608*a(n+1)^2 - 40960*a(n+1)*a(n+2) + 2100*a(n+2)^2) + a(n)*a(n+1)*(-12288*a(n+1)^2 + 2840*a(n+1)*a(n+2) - 160*a(n+2)^2) + a(n+1)^2*(+180*a(n+1)^2 - 48*a(n+1)*a(n+2) + 3*a(n+2)^2) for all n in Z. (End)
a(n) = [x^n] ( (1 + x)^4/(1 - x)^2 )^n; exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... = Sum_{n >= 0} A066357(n+1)*x^n. - Peter Bala, Jun 23 2015
a(n) = Sum_{i = 0..n} binomial(4*n,i) * binomial(3*n-i-1,n-i). - Peter Bala, Sep 29 2015
a(n) = A000984(n)*Product_{j=0..n} (2^j/(j!*(2*j-1)!!))*A068424(n, j)^2, with A068424 the falling factorial. See (5.4) in Podestá link. - Michel Marcus, Mar 31 2016
a(n) = GegenbauerC(2*n, -2*n, -1). - Peter Luschny, May 07 2016
a(n) = [x^n] 1/sqrt(1 - 4*x)^(2*n+1). - Ilya Gutkovskiy, Oct 10 2017
a(n) is the n-th moment of the positive weight function w(x) on (0,16), i.e. a(n) = Integral_{x=0..16} x^n*w(x) dx, n = 0,1,..., where w(x) = (1/(2*Pi))/(sqrt(4 - sqrt(x))*x^(3/4)). The function w(x) is the solution of the Hausdorff moment problem and is unique. - Karol A. Penson, Mar 06 2018
a(n) = (16^n*(Beta(2*n - 1/2, 1/2) - Beta(2*n - 1/2, 3/2)))/Pi. - Peter Luschny, Mar 06 2018
E.g.f.: hypergeom([1/4,3/4],[1/2,1],16*x). - Karol A. Penson, Mar 08 2018
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(2*n)] (1 + x + y)^(4*n). (End)
a(n) = (2^n/n!)*Product_{k = n..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = Sum_{k = 0..2*n} binomial(2*n+k-1, k). - Peter Bala, Nov 02 2024
Sum_{n>=0} (-1)^n/a(n) = 16/17 + 4*sqrt(34)*(sqrt(17)-2)*arctan(sqrt(2/(sqrt(17)-1)))/(289*sqrt(sqrt(17)-1)) + 2*sqrt(34)*(sqrt(17)+2)*log((sqrt(sqrt(17)+1)-sqrt(2))/(sqrt(sqrt(17)+1)+sqrt(2)))/(289*sqrt(sqrt(17)+1)) (Sprugnoli, 2006, Theorem 3.8, p. 11; Piezas, 2012). - Amiram Eldar, Nov 03 2024
For n >= 1, a(n) = Sum_{k=1}^n a(n-k) * A337350(n) = Sum_{k=1}^n a(n-k) * a(k) * (8k + 1) / (8k^2 + 2k - 1).  For proof, see the Quy Nhan link. - Lucas A. Brown, Jun 26 2025
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(2*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(2*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((4*n-1)/2,n).
a(n) = [x^n] (1+4*x)^((4*n-1)/2). (End)

A079489 a(n) = (24^nbinomial(2n, n) - binomial(4n + 1, 2*n)) / (n + 1).

Original entry on oeis.org

1, 3, 22, 211, 2306, 27230, 338444, 4362627, 57788170, 781825066, 10757497972, 150073096238, 2117778107732, 30176799215196, 433586825237912, 6274885068167651, 91383942213277530, 1338275570267001458, 19695358741104824036, 291137841642777382330, 4320734864185863437820

Offset: 0

N. J. A. Sloane, Jan 20 2003

a(n) is the number of ordered trees on 2n-1 edges in which every subtree of the root (including its rooting edge) has an even number of edges, except for the leftmost subtree which has an odd number of edges (including its rooting edge). - David Callan, Apr 10 2012
a(n) is the number of 2 X 2n Young tableaux with a wall between the first and second row in each even column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
  Example for a(1)=3:
   3 4  2 4  2 3
     -    -    -
   1 2, 1 3, 1 4. - Michael Wallner, Mar 09 2022

G. C. Greubel, Table of n, a(n) for n = 0..830
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
Feihu Liu and Guoce Xin, Simple Generating Functions for Certain Young Tableaux with Periodic Walls, arXiv:2401.14627 [math.CO], 2024.
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (Table A.1).
Alon Regev, Enumerating triangulations by parallel diagonals, arXiv:1208.3915 [math.CO], 2012. - _N. J. A. Sloane_, Dec 25 2012
Alon Regev, Enumerating Triangulations by Parallel Diagonals, Journal of Integer Sequences, Vol. 15 (2012), #12.8.5.

Final diagonal of triangle in A078990.

Cf. A000108, A006318, A066357, A060941, A066357, A300386 - A300389, A333563.

a := n -> (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1):
seq(a(n), n = 0..20);  # Peter Luschny, Aug 26 2024

((Sqrt[2] Sqrt[1 + Sqrt[1 - 16 x]] - Sqrt[1 - 16 x] - 1)/(4 x) + O[x]^20)[[3]] (* Vladimir Reshetnikov, Sep 25 2016 *)
CoefficientList[Series[-(1 - Sqrt[1 - 4*Sqrt[x]])*(1 - Sqrt[1 + 4*Sqrt[x]])/(4*x), {x,0,50}], x] (* G. C. Greubel, Apr 13 2017 *)

a(n)=if(n<0,0,polcoeff(serreverse(x*(1-x^2)/(1+x^2)^2+O(x^(2*n+3))),2*n+1))

{a(n)=polcoeff(exp(sum(m=1,n,binomial(4*m-1,2*m)*x^m/m)+x*O(x^n)),n)} \\ Paul D. Hanna, Dec 30 2010

Series reversion of x(1-x^2)/(1+x^2)^2 expanded in odd powers of x. [Previous name.]
If x = y*(1-y^2)/(1+y^2)^2 then y = x + 3*x^3 + 22*x^5 + 211*x^7 + 2306*x^9 + ...
G.f. A(x) satisfies x*A(x^2) = (C(x) - C(-x))/(C(x) + C(-x)) where C(x) is g.f. of the Catalan numbers A000108.
a(n) = Sum_{k=0..2n} (-1)^k * A000108(2*n-k) * A000108(k). - David Callan, Aug 16 2006
a(n) = ((2^(4n+2))/Gamma(1/2)) * ((Gamma(n+1/2)/(2*Gamma(n+2))) - Gamma(2n+3/2)/Gamma(2n+3)). [David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009]
G.f.: exp( Sum_{n>=1} C(4n-1,2n)*x^n/n ). - Paul D. Hanna, Dec 30 2010
G.f.: C(sqrt(x))*C(-sqrt(x)) where C(x) is the g.f. for the Catalan numbers A000108. - David Callan, Apr 10 2012
D-finite with recurrence n*(n+1)*(2*n+1)*a(n) -2*n*(32*n^2-32*n+11)*a(n-1) +16*(4*n-5)*(4*n-3)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 29 2012
a(n) ~ (2-sqrt(2))*16^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 20 2013
a(n) = 2^(2*n+1)*Catalan(n) - Catalan(2*n+1) (see Regev). It follows that the 2-adic valuations of a(n) and Catalan(n) are equal. In particular, a(n) is odd iff n is of the form 2^m - 1. - Peter Bala, Aug 02 2016
G.f.: (sqrt(2) * sqrt(1 + sqrt(1-16*x)) - sqrt(1-16*x) - 1)/(4*x). - Vladimir Reshetnikov, Sep 25 2016
G.f. A(x) satisfies A(x^2) = C(x)^2*r(-x*C(x)^2), where C(x) is g.f. of the Catalan numbers A000108, and r(x) is g.f. of the large Schröder numbers A006318. - Alexander Burstein, Nov 21 2019
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/2)*binomial(4*n,2*n)*x^n/n ).
1 + x*A(x) is the o.g.f. of A066357.
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, ...]  and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. See A333563. Cf. A060941. (End)
From Peter Bala, Oct 23 2024: (Start)
For integer r and positive integer s, define a sequence {u(n) : n >= 0} by setting u(n) = [x^(s*n)] A(x)^(r*n). We conjecture that the supercongruence u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k.
Let B(x) = 1/x * series_reversion(x*A(x)). Define a sequence {v(n) : n >= 0} by setting v(n) = [x^(s*n)] B(x)^(r*n). We conjecture that the supercongruence v(n*p^k) == v(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k. (End)

New name by Peter Luschny, Aug 26 2024

A079513 Triangular array (a Riordan array) related to tennis ball problem, read by rows.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 3, 2, 1, 6, 6, 6, 3, 1, 0, 22, 16, 10, 4, 1, 53, 53, 53, 31, 15, 5, 1, 0, 211, 158, 105, 52, 21, 6, 1, 554, 554, 554, 343, 185, 80, 28, 7, 1, 0, 2306, 1752, 1198, 644, 301, 116, 36, 8, 1, 6362, 6362, 6362, 4056, 2304, 1106, 462, 161, 45, 9, 1

Offset: 0

N. J. A. Sloane, Jan 22 2003

Riordan array (2/(2-x*c(x)+x*c(-x)), x*c(x)), with c(x) the g.f. of Catalan numbers (A000108). - Ralf Stephan, Dec 29 2013

			Triangle starts
     1;
     0,    1;
     1,    1,    1;
     0,    3,    2,    1;
     6,    6,    6,    3,    1;
     0,   22,   16,   10,    4,    1;
    53,   53,   53,   31,   15,    5,   1;
     0,  211,  158,  105,   52,   21,   6,   1;
   554,  554,  554,  343,  185,   80,  28,   7,  1;
     0, 2306, 1752, 1198,  644,  301, 116,  36,  8, 1;
  6362, 6362, 6362, 4056, 2304, 1106, 462, 161, 45, 9, 1;

G. C. Greubel, Rows n=0..100 of triangle, flattened
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (Table A.2).

First column is A066357 interspersed with 0's, 2nd column gives A079514.

Cf. A079514, A079515, A079516, A079517, A079518, A079519, A079520, A079521.

c[t_]:= (1-Sqrt[1-4*t])/(2*t); d[t_]:= (1-(1+2*t)*Sqrt[1-4*t] -(1-2*t)*Sqrt[1+4*t] +Sqrt[1-16*t^2])/(4*t^2); g[t_, r_]:= d[t]*(t*c[t])^r; Table[SeriesCoefficient[Series[g[t, k], {t, 0, n}], n], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Jan 16 2019 *)

Edited and more terms added by Ralf Stephan, Dec 29 2013

A365847 Expansion of (1/x) * Series_Reversion( x*(1-x)^4/(1+x)^4 ).

Original entry on oeis.org

1, 8, 96, 1368, 21440, 356968, 6197408, 110947768, 2033381760, 37963483592, 719495148768, 13806129179928, 267693334199616, 5236670783633960, 103227182363423008, 2048451544990578552, 40888361539777714944, 820400146864231266184

Offset: 0

Seiichi Manyama, Sep 20 2023

nonn

Cf. A066357, A365754, A365846, A365848.
Cf. A032349, A365622, A365843.
Cf. A260332.

a(n) = sum(k=0, n, binomial(4*n+k+3, k)*binomial(4*(n+1), n-k))/(n+1);

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

Conjecture: g.f.: B^4, where B is the g.f. of A260332.

A262737 O.g.f. exp( Sum_{n >= 1} A262732(n)*x^n/n ).

Original entry on oeis.org

1, 8, 95, 1336, 20642, 338640, 5791291, 102108760, 1842857390, 33879118384, 632210693270, 11944142806064, 228010741228740, 4391334026631072, 85221618348230355, 1664901954576830360, 32716286416687895862, 646228961799752926320, 12823701194384778672322

Offset: 0

Peter Bala, Sep 29 2015

O.g.f. is 1/x * the series reversion of x*(1 - x)^k/(1 + x)^(k+2) for k = 3. See the cross references for related sequences obtained from other values of k.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion

Cf. A000108 (k = 0), A007297 (k = 1), A066357 (k = 2), A262738 (k = 4), A262739 (k = 5), A262740 (k = 6), A262732.

A262737 := proc (n) option remember; if n = 0 then 1 else add(1/k!*(5*k)!/GAMMA(5*k/2 + 1)*GAMMA(3*k/2 + 1)/(3*k)!*A262737(n-k), k = 1 .. n)/n end if; end proc:
seq(A262737(n), n = 0 .. 20);

a(n) = sum(k=0, n, binomial(5*(n+1),k)*binomial(4*(n+1)-k-2,(n+1)-k-1))/(n+1); \\ Altug Alkan, Oct 03 2015

a(n-1) = 1/n * Sum_{i = 0..n-1} binomial(5*n,i)*binomial(4*n-i-2,n-i-1).
O.g.f.: A(x) = exp ( Sum_{n >= 1} 1/n! * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!*x^n/n ) = 1 + 8*x + 195*x^2 + 1336*x^3 + ....
1 + x*A'(x)/A(x) is the o.g.f. for A262732.
O.g.f. is the series reversion of x*(1 - x)^3/(1 + x)^5.
a(0) = 1 and for n >= 1, a(n) = 1/n * Sum {k = 1..n} 1/k!*(5*k)!/GAMMA(5*k/2+1)*GAMMA(3*k/2+1)/(3*k)! * a(n-k).

A262738 O.g.f. exp( Sum_{n >= 1} A211419(n)*x^n/n ).

Original entry on oeis.org

1, 10, 149, 2630, 51002, 1050132, 22539085, 498732014, 11296141454, 260613866380, 6103074997890, 144696786555580, 3466352150674324, 83776927644646952, 2040261954214847421, 50018542073019175806, 1233419779839067305350, 30572886836581693309020

Offset: 0

Peter Bala, Sep 29 2015

O.g.f. is 1/x * the series reversion of x*(1 - x)^k/(1 + x)^(k+2) at k = 4. See the cross references for related sequences obtained from other values of k.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion

Cf. A211419, A000108 (k = 0), A007297 (k = 1), A066357 (k = 2), A262737 (k = 3), A262739 (k = 5), A262740 (k = 6).

A262738 := proc(n) option remember; if n = 0 then 1 else add((6*k)!*(2*k)!/((4*k)!*(3*k)!*k!)*A262738(n-k), k = 1 .. n)/n end if; end proc:
seq(A262738(n), n = 0..20);

a(n) = sum(k=0, n, binomial(6*(n+1),k)*binomial(5*(n+1)-k-2,(n+1)-k-1))/(n+1); \\ Altug Alkan, Oct 03 2015

a(n-1) = 1/n * Sum_{i = 0..n-1} binomial(6*n,i)*binomial(5*n-i-2,n-i-1).
O.g.f.: A(x) = exp ( Sum_{n >= 1} (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!)*x^n/n ) = 1 + 10*x + 149*x^2 + 2630*x^3 + ....
1 + x*A'(x)/A(x) is the o.g.f. for A211419.
O.g.f. is the series reversion of x*(1 - x)^4/(1 + x)^6.
a(0) = 1 and for n >= 1, a(n) = 1/n * Sum {k = 1..n} (6*k)!*(2*k)!/((4*k)!*(3*k)!*k!)*a(n-k).

A262739 O.g.f. exp( Sum_{n >= 1} A262733(n)*x^n/n ).

Original entry on oeis.org

1, 12, 215, 4564, 106442, 2635704, 68031147, 1810302340, 49308457334, 1368019979976, 38525145673126, 1098380420669000, 31641932951483220, 919622628946689648, 26931762975278938035, 793967020231145502564, 23543663463050594677310, 701763102761640853890600, 21014048069544552257072530, 631868353403527700756671320, 19070677448561228207945931276

Offset: 0

Peter Bala, Sep 29 2015

O.g.f. is 1/x * the series reversion of x*(1 - x)^k/(1 + x)^(k+2) when k = 5. See the cross references for related sequences obtained from other values of k.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion

Cf. A262733, A211419, A000108 (k = 0), A007297 (k = 1), A066357 (k = 2), A262737 (k = 3), A262738 (k = 4), A262740 (k = 6).

A262739 := proc (n) option remember; if n = 0 then 1 else add(1/k!*(7*k)!/GAMMA(7*k/2 + 1)*GAMMA(5*k/2 + 1)/(5*k)!*A262739(n-k), k = 1 .. n)/n end if; end proc:
seq(A262739(n), n = 0..20);

a(n) = sum(k=0, n, binomial(7*(n+1),k)*binomial(6*(n+1)-k-2,(n+1)-k-1))/(n+1); \\ Altug Alkan, Oct 03 2015

a(n-1) = 1/n * Sum_{i = 0..n-1} binomial(7*n,i)*binomial(6*n-i-2,n-i-1).
O.g.f.: A(x) = exp ( Sum_{n >= 1} 1/n! * (7*n)!/(7*n/2)! * (5*n/2)!/(5*n)!*x^n/n ) = 1 + 12*x + 215*x^2 + 4564*x^3 + ....
1 + x*A'(x)/A(x) is the o.g.f. for A262733.
O.g.f. is the series reversion of x*(1 - x)^5/(1 + x)^7,
a(0) = 1 and for n >= 1, a(n) = 1/n * Sum {k = 1..n} 1/k! * (7*k)!/(7*k/2)! * (5*k/2)!/(5*k)!*a(n-k).

A262740 O.g.f. exp( Sum_{n >= 1} A211421(n)*x^n/n ).

Original entry on oeis.org

1, 14, 293, 7266, 197962, 5726364, 172662765, 5367187226, 170772853790, 5534640052292, 182070248073826, 6063785526898644, 204055962203476788, 6927718839334775608, 236994877398511998717, 8161492483543100398410, 282705062046649346154006, 9843330120848835962213940

Offset: 0

Peter Bala, Sep 29 2015

O.g.f. is 1/x * the series reversion of x*(1 - x)^k/(1 + x)^(k+2) at k = 6. See the cross references for related sequences obtained from other values of k.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion

Cf. A211421, A000108 (k = 0), A007297 (k = 1), A066357 (k = 2), A262737 (k = 3), A262738 (k = 4), A262739 (k = 5).

#A262740
A262740 := proc (n) option remember; if n = 0 then 1 else add(1/k!*(8*k)!/(4*k)!*(3*k)!/(6*k)!*A262740(n-k), k = 1 .. n)/n end if; end proc:
seq(A262740(n), n = 0..17);

a(n) = sum(k=0, n, binomial(8*(n+1),k)*binomial(7*(n+1)-k-2,(n+1)-k-1))/(n+1); \\ Altug Alkan, Oct 03 2015

a(n-1) = 1/n * Sum_{i = 0..n-1} binomial(8*n,i)*binomial(7*n-i-2,n-i-1).
O.g.f.: A(x) = exp ( Sum_{n >= 1} 1/n! * (8*n)!/(4*n)! * (3*n)!/(6*n)!*x^n/n ) = 1 + 14*x + 293*x^2 + 7266*x^3 + ....
1 + x*A'(x)/A(x) is the o.g.f. for A211421.
O.g.f. is the series reversion of x*(1 - x)^6/(1 + x)^8.
a(0) = 1 and for n >= 1, a(n) = 1/n * Sum {k = 1..n} 1/k! * (8*k)!/(4*k)! * (3*k)!/(6*k)!*a(n-k).

A104184 a(n) is the number of paths from (0,0) to (n,0) using steps of the form (1,2),(1,1),(1,0),(1,-1) or (1,-2) and staying above the x-axis. Also, a(n) is the number of possible combinations of balls on the lawn after n turns, using a Motzkin variation of the (4,2)-case of the tennis ball problem considered by D. Merlini, R. Sprugnoli and M. C. Verri.

Original entry on oeis.org

1, 1, 3, 9, 32, 120, 473, 1925, 8034, 34188, 147787, 647141, 2864508, 12796238, 57615322, 261197436, 1191268350, 5462080688, 25162978925, 116414836445, 540648963645, 2519574506595, 11779011525030, 55225888341334, 259612579655392, 1223396051745310

Offset: 0

Nicholas Biller (billern(AT)gmail.com), Mar 11 2005

nonn

The (4,2)-case of the Motzkin Tennis Ball Problem is a variation of the Tennis Ball Problem that generates a(n). On each turn, i, four balls labeled i are placed in the bucket and then any two are removed and placed on the lawn. We consider all possible combinations of balls on the lawn after n turns.

The number of ways to choose n numbers, ranging from 0 to 4, so that their sum is 2n and so that when you take k numbers from the left, the sum of these numbers is <= 2k (e.g. the combination of {141} is impossible, for 1+4 > 2k). Thus a(1) = {2}; a(2) = {04}, {13} and {22}; a(3) = {024}, {033}, {042}, {114}, {123}, {132}, {204}, {213} and {222}. - Joost Vermeij (joost_vermeij(AT)hotmail.com), Jun 12 2005

			a(3) = 9, since the possible combinations of balls on the lawn after 3 turns is 111122, 111123, 111133, 111222, 111223, 111233, 112222, 112223, 112233, if on each turn there are 4 identically labeled balls received and 2 selected.

Alois P. Heinz, Table of n, a(n) for n = 0..1437
C. Banderier, C. Krattenthaler, A. Krinik, D. Kruchinin, V. Kruchinin, D. Nguyen, and M. Wallner, Explicit formulas for enumeration of lattice paths: basketball and the kernel method, arXiv preprint arXiv:1609.06473 [math.CO], 2016.
AJ Bu and Doron Zeilberger, Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas, arXiv:2305.09030 [math.CO], 2023.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 512
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), pp. 307-344.

Cf. A066357, A001006, A005191.

a:= proc(n) option remember; `if`(n<5, [1, 1, 3, 9, 32][n+1],
      ((n+1)*(43*n^3-48*n^2-7*n+2)*a(n-1)
      -(124*n^4-370*n^3+255*n^2+15*n-14)*a(n-2)
      +5*(n-2)*(2*n^3-52*n^2+65*n-1)*a(n-3)
      +25*(n-2)*(n-3)*(8*n^2-8*n-1)*a(n-4)
      -125*n*(n-2)*(n-3)*(n-4)*a(n-5))/
       (2*(n-1)*(n+1)*(n+2)*(2*n+1)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Oct 11 2013

CoefficientList[Series[(1 + x + Sqrt[1 - 6*x + 5*x^2] - Sqrt[2]*Sqrt[1 + Sqrt[1 - 6*x + 5*x^2] + x*(-2 - 5*x + Sqrt[1 - 6*x + 5*x^2])])/(4*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 09 2014 *)

a(n):=((sum(binomial(n+1,l)*sum(binomial(i,2*l-1)*sum(binomial(j,n-j-i-1) *binomial(n-l+1,j),j,0,n-l+1),i,0,n-1),l,1,n+1))+sum(binomial(j,n-j) *binomial(n+1,j),j,0,n+1))/(n+1); /* Vladimir Kruchinin, Jun 26 2015 */

{a(n)=local(A=1);A=exp(sum(m=1,n+1,polcoeff(((1-x^5)/(1-x) +O(x^(2*m+1)))^m, 2*m)*x^m/m)+x*O(x^n));polcoeff(A,n)} /* Paul D. Hanna */

G.f. (for offset 1): (1/(4*x))*(1+x+sqrt((1-6*x+5*x^2)) - sqrt(2)*sqrt(1+sqrt((1-6*x+5*x^2)) + x*(-2-5*x+sqrt((1-6*x+5*x^2))))). - N-E. Fahssi, Jan 10 2008
Let M be the infinite pentadiagonal matrix with all 1's in the 1st and 2nd subdiagonals, the main diagonal, and the 1st and 2nd superdiagonals, and with the rest 0's. V = vector [1,0,0,0,...]. The sequence starting with offset 1 = iterates of M*V, leftmost column. - Gary W. Adamson, Jun 06 2011
From Paul D. Hanna, Oct 19 2011: (Start)
Logarithmic derivative yields the central pentanomial coefficients (A005191).
G.f.: exp( Sum_{n>=1} A005191(n)*x^n/n ).
G.f.: (1/x)*Series_Reversion(x*(1-x^5)*(1-x^2)*(1-x)/(1-x^10)).
G.f. satisfies: A(x) = (1-x^10*A(x)^10)/((1-x^5*A(x)^5)*(1-x^2*A(x)^2)*(1-x*A(x))). (See formula from Michael Somos in A005191.) (End)
a(n) ~ (3-sqrt(5)) * 5^(n+1) / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 09 2014
a(n) = ((Sum_{l=1..n+1} (C(n+1,l)*Sum_{i=0..n-1} (C(i,2*l-1) * Sum_{j=0..n-l+1} (C(j,n-j-i-1)*C(n-l+1,j))))) + Sum_{j=0..n+1} (C(j,n-j) * C(n+1,j)))/(n+1). - Vladimir Kruchinin, Jun 26 2015
-2*(n-1)*(2*n+1)*(n+2)*(n+1)*a(n) +(n+1)*(43*n^3-48*n^2-7*n+2)*a(n-1) +(-124*n^4+370*n^3-255*n^2-15*n+14)*a(n-2) +5*(n-2)*(2*n^3-52*n^2+65*n-1)*a(n-3) +25*(n-2)*(n-3)*(8*n^2-8*n-1)*a(n-4) -125*n*(n-2)*(n-3)*(n-4)*a(n-5)=0. - R. J. Mathar, Jul 23 2017

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A005900 Octahedral numbers: a(n) = n*(2*n^2 + 1)/3.

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A001448 a(n) = binomial(4n,2n) or (4*n)!/((2*n)!*(2*n)!).

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A079489 a(n) = (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1).

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A079513 Triangular array (a Riordan array) related to tennis ball problem, read by rows.

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A365847 Expansion of (1/x) * Series_Reversion( x*(1-x)^4/(1+x)^4 ).

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A262737 O.g.f. exp( Sum_{n >= 1} A262732(n)*x^n/n ).

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A005900 Octahedral numbers: a(n) = n(2n^2 + 1)/3.

A001448 a(n) = binomial(4n,2n) or (4n)!/((2n)!(2n)!).

A079489 a(n) = (24^nbinomial(2n, n) - binomial(4n + 1, 2*n)) / (n + 1).