A124428 -id:A124428 - cp's OEIS frontend

A124429 Antidiagonal sums of triangle A124428.

1, 1, 1, 2, 3, 5, 8, 13, 22, 36, 61, 102, 172, 292, 493, 841, 1429, 2439, 4169, 7124, 12216, 20930, 35940, 61749, 106155, 182749, 314638, 542338, 935195, 1613593, 2786037, 4811920, 8316435, 14378247, 24870062, 43036264, 74496224, 129008514

Offset: 0

Paul D. Hanna, Oct 31 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A124428, A124430.

[(&+[Binomial(Floor((n-k)/2), k)*Binomial(Floor((n-k+1)/2), k): k in [0..Floor(n/3)]]): n in [0..40]]; // G. C. Greubel, Feb 24 2019

Table[Sum[Binomial[Floor[(n-k)/2], k]*Binomial[Floor[(n-k+1)/2], k], {k, 0, Floor[n/3]}], {n, 0, 40}] (* G. C. Greubel, Feb 24 2019 *)

a(n)=sum(k=0,n\3,binomial((n-k)\2,k)*binomial((n-k+1)\2,k))

[sum(binomial(floor((n-k)/2),k)*binomial(floor((n-k+1)/2),k) for k in (0..floor(n/3))) for n in (0..40)] # G. C. Greubel, Feb 24 2019

a(n) = Sum_{k=0..[n/3]} C([(n-k)/2],k)*C([(n-k+1)/2],k).

A124430 Eigenvector of triangle A124428.

Original entry on oeis.org

1, 1, 2, 3, 7, 13, 31, 61, 144, 296, 714, 1534, 3761, 8303, 20495, 46115, 114461, 261445, 651114, 1503207, 3749017, 8726147, 21788311, 51072555, 127698665, 301244477, 754496298, 1790598079, 4494019431, 10726676701, 26983034009

Offset: 0

Paul D. Hanna, Oct 31 2006

nonn

			a(5) = 1*a(0) + 6*a(1) + 3*a(2) = 1*1 + 6*1 + 3*2 = 13;
a(6) = 1*a(0) + 9*a(1) + 9*a(2) + 1*a(3) = 1*1 + 9*1 + 9*2 + 1*3 = 31.
Triangle A124428(n,k) = C([n/2],k)*C([(n+1)/2],k) begins:
  1;
  1;
  1,  1;
  1,  2;
  1,  4,  1;
  1,  6,  3;
  1,  9,  9,  1;
  1, 12, 18,  4;
  1, 16, 36, 16,  1; ...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A124428, A124429.

a[n_]:= a[n] = If[n==0, 1, Sum[Binomial[Floor[n/2], k]*Binomial[Floor[(n + 1)/2], k]*a[k], {k,0,Floor[n/2]}]]; Table[a[n], {n, 0, 30}] (* G. C. Greubel, Feb 24 2019 *)

{a(n)=if(n==0,1,sum(k=0,n\2,a(k)*binomial(n\2,k)*binomial((n+1)\2,k)))}

a(n) = Sum_{k=0..[n/2]} a(k)*C([n/2],k)*C([(n+1)/2],k) for n>0, with a(0)=1 and [] means floor().

A006542 a(n) = binomial(n,3)*binomial(n-1,3)/4.

Original entry on oeis.org

1, 10, 50, 175, 490, 1176, 2520, 4950, 9075, 15730, 26026, 41405, 63700, 95200, 138720, 197676, 276165, 379050, 512050, 681835, 896126, 1163800, 1495000, 1901250, 2395575, 2992626, 3708810, 4562425, 5573800, 6765440, 8162176, 9791320, 11682825, 13869450

Offset: 4

N. J. A. Sloane

Number of permutations of n+4 that avoid the pattern 132 and have exactly 3 descents. - Mike Zabrocki, Aug 26 2004
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 20 2005
a(n) = number of Dyck n-paths with exactly 4 peaks. - David Callan, Jul 03 2006
Six-dimensional figurate numbers for a hyperpyramid with pentagonal base. This corresponds to the sum(sum(sum(sum(1+sum(5*n))))) interpretation, see the Munafo webpage. - Robert Munafo, Jun 18 2009

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 166, no. 1).
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 238.
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 = 4..200
Isaac Ahern and Sam Cook, Affine Symmetry Tensors in Minkowski Space, American Journal of Undergraduate Research, Volume 13, Issue 3, August 2016.
P. Aluffi, Degrees of projections of rank loci, arXiv:1408.1702 [math.AG], 2014. ["After compiling the results of many explicit computations, we noticed that many of the numbers d_{n,r,S} appear in the existing literature in contexts far removed from the enumerative geometry of rank conditions; we owe this surprising (to us) observation to perusal of [Slo14]."]
Brandy Amanda Barnette, Counting Convex Sets on Products of Totally Ordered Sets, Masters Theses & Specialist Projects, Paper 1484, 2015.
V. E. Hoggatt, Jr., Letter to N. J. A. Sloane, Apr 1977
G. Kreweras, Traitement simultané du "Problème de Young" et du "Problème de Simon Newcomb", Cahiers du Bureau Universitaire de Recherche Opérationnelle. Institut de Statistique, Université de Paris, 10 (1967), 23-31.
G. Kreweras, Traitement simultané du "Problème de Young" et du "Problème de Simon Newcomb", Cahiers du Bureau Universitaire de Recherche Opérationnelle. Institut de Statistique, Université de Paris, 10 (1967), 23-31. [Annotated scanned copy]
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See pp. 6, 25.
Robert Munafo, C(n,3)C(n-1,3)/4
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
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

The expression binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1) for the values m = 2 through 14 produces the sequences A000012, A000217, A002415, A006542, A006857, A108679, A134288, A134289, A134290, A134291, A140925, A140935, A169937.
Cf. A000332, A000579, A001263, A002378, A004068, A005585, A005891, A006322, A006414, A047819, A107891, A114242.
Fourth column of the table of Narayana numbers A001263.
Apart from a scale factor, a column of A124428.

List([4..40], n-> n*(n-1)^2*(n-2)^2*(n-3)/144); # G. C. Greubel, Feb 24 2019

[ n*((n-1)*(n-2))^2*(n-3)/144 : n in [4..40] ]; // Wesley Ivan Hurt, Jun 17 2014

A006542:=-(1+3*z+z**2)/(z-1)**7; # conjectured by Simon Plouffe in his 1992 dissertation
A006542:=n->n*((n-1)*(n-2))^2*(n-3)/144; seq(A006542(n), n=4..40); # Wesley Ivan Hurt, Jun 17 2014

Table[Binomial[n, 3]*Binomial[n-1, 3]/4, {n, 4, 40}]

PARI
```
a(n)=n*((n-1)*(n-2))^2*(n-3)/144
```

[n*(n-1)^2*(n-2)^2*(n-3)/144 for n in (4..40)] # G. C. Greubel, Feb 24 2019

a(n) = C(n, 3)*C(n-1, 3)/4 = n*(n-1)^2*(n-2)^2*(n-3)/144.
a(n) = A000292(n-3)*A000292(n-2)/4.
E.g.f.: x^4*(6 + 6*x + x^2)*exp(x)/144. - Vladeta Jovovic, Jan 29 2003
a(n) = Sum(Sum(Sum(Sum(1 + Sum(5*n))))) = Sum (A006414). - Xavier Acloque, Oct 08 2003
a(n) = C(n, 6) + 3*C(n+1, 6) + C(n+2, 6). - Mike Zabrocki, Aug 26 2004
G.f.: x^4*(1 + 3*x + x^2)/(1-x)^7. - Emeric Deutsch, Jun 20 2005
a(n) = C(n-2, n-4)*C(n-1, n-3)*C(n, n-2)/18. - Zerinvary Lajos, Jul 29 2005
a(n) = C(n,4)*C(n,3)/n. - Mitch Harris, Jul 06 2006
a(n+2) = (1/4)*Sum_{1 <= x_1, x_2 <= n} x_1*x_2*(det V(x_1,x_2))^2 = (1/4)*Sum_{1 <= i,j <= n} i*j*(i-j)^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
a(n) = C(n-1,3)^2 - C(n-1,2)*C(n-1,4). - Gary Detlefs, Dec 05 2011
a(n) = A000292(A000217(n-1)) - A000217(A000292(n-1)). - Ivan N. Ianakiev, Jun 17 2014
a(n) = Product_{i=1..3} A002378(n-4+i)/A002378(i). - Bruno Berselli, Nov 12 2014 (Rewritten, Sep 01 2016.)
Sum_{n>=4} 1/a(n) = 238 - 24*Pi^2. - Jaume Oliver Lafont, Jul 10 2017
Sum_{n>=4} (-1)^n/a(n) = 134 - 192*log(2). - Amiram Eldar, Oct 19 2020
a(n) = A000332(n) + 5*A000579(n+1). - Yasser Arath Chavez Reyes, Aug 18 2024

Zabroki and Lajos formulas offset corrected by Gary Detlefs, Dec 05 2011

A006011 a(n) = n^2*(n^2 - 1)/4.

Original entry on oeis.org

0, 0, 3, 18, 60, 150, 315, 588, 1008, 1620, 2475, 3630, 5148, 7098, 9555, 12600, 16320, 20808, 26163, 32490, 39900, 48510, 58443, 69828, 82800, 97500, 114075, 132678, 153468, 176610, 202275, 230640, 261888, 296208, 333795, 374850, 419580, 468198

Offset: 0

N. J. A. Sloane, Simon Plouffe

Products of two consecutive triangular numbers (A000217).
a(n) is the number of Lyndon words of length 4 on an n-letter alphabet. A Lyndon word is a primitive word that is lexicographically earliest in its cyclic rotation class. For example, a(2)=3 counts 1112, 1122, 1222. - David Callan, Nov 29 2007
For n >= 2 this is the second rightmost column of A163932. - Johannes W. Meijer, Oct 16 2009
Partial sums of A059270. - J. M. Bergot, Jun 27 2013
Using the integers, triangular numbers, and squares plot the points (A001477(n),A001477(n+1)), (A000217(n), A000217(n+1)), and (A000290(n),A000290(n+1)) to create the vertices of a triangle. One-half the area of this triangle = a(n). - J. M. Bergot, Aug 01 2013
a(n) is the Wiener index of the triangular graph T(n+1). - Emeric Deutsch, Aug 26 2013

			From _Bruno Berselli_, Aug 29 2014: (Start)
After the zeros, the sequence is provided by the row sums of the triangle:
   3;
   4, 14;
   5, 16, 39;
   6, 18, 42,  84;
   7, 20, 45,  88, 155;
   8, 22, 48,  92, 160, 258;
   9, 24, 51,  96, 165, 264, 399;
  10, 26, 54, 100, 170, 270, 406, 584;
  11, 28, 57, 104, 175, 276, 413, 592, 819;
  12, 30, 60, 108, 180, 282, 420, 600, 828, 1110; etc.,
where T(r,c) = c*(c^2+r+1), with r = row index, c = column index, r >= c > 0. (End)

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

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Miguel Azaola and Francisco Santos, The number of triangulations of the cyclic polytope C(n,n-4), Discrete Comput. Geom., Vol. 27 (2002), pp. 29-48 (see Prop. 4.2(a)).
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926. (Annotated scanned copy)
Eric Weisstein's World of Mathematics, Triangular Graph.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Thrice A002415. Row 4 of A074650.
Cf. A000217, A000290, A000537, A001477, A002415, A008911, A047928, A059270, A083374, A126274, A163932, A228317
A column of A124428.

[n^2*(n^2-1)/4: n in [0..40]]; // Vincenzo Librandi, Sep 14 2011

A006011 := proc(n)
    n^2*(n^2-1)/4 ;
end proc: # R. J. Mathar, Nov 29 2015

Table[n^2 (n^2 - 1)/4, {n, 0, 38}]
Binomial[Range[20]^2, 2]/2 (* Eric W. Weisstein, Sep 08 2017 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 3, 18, 60, 150}, 20] (* Eric W. Weisstein, Sep 08 2017 *)
CoefficientList[Series[-3 x (1 + x)/(-1 + x)^5, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 08 2017 *)
Join[{0},Times@@@Partition[Accumulate[Range[0,40]],2,1]] (* Harvey P. Dale, Aug 08 2025 *)

a(n)=binomial(n^2,2)/2 \\ Charles R Greathouse IV, Jun 27 2013

G.f.: 3*(1 + x) / (1 - x)^5.
a(n) = (n-1)*n/2 * n*(n+1)/2 = A000217(n-1)*A000217(n) = 1/2*(n^2-1)*n^2/2 = 1/2*A000217(n^2-1). - Alexander Adamchuk, Apr 13 2006
a(n) = 3*A002415(n) = A047928(n-1)/4 = A083374(n-1)/2 = A008911(n)*3/2. - Zerinvary Lajos, May 09 2007
a(n) = (A126274(n) - A000537(n+1))/2. - Enrique Pérez Herrero, Mar 11 2013
Ceiling(sqrt(a(n)) + sqrt(a(n-1)))/2 = A000217(n). - Richard R. Forberg, Aug 14 2013
a(n) = Sum_{i=1..n-1} i*(i^2+n) for n > 1 (see Example section). - Bruno Berselli, Aug 29 2014
Sum_{n>=2} 1/a(n) = 7 - 2*Pi^2/3 = 0.42026373260709425411... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n^2+n) - A000217(n)*A000217(n+1). - Charlie Marion, Feb 15 2020
Sum_{n>=2} (-1)^n/a(n) = Pi^2/3 - 3. - Amiram Eldar, Nov 02 2021
E.g.f.: exp(x)*x^2*(6 + 6*x + x^2)/4. - Stefano Spezia, Mar 12 2024

A102038 a(n+1) = n*a(n) + a(n-1), a(1)=1, a(2)=2.

Original entry on oeis.org

1, 2, 5, 17, 73, 382, 2365, 16937, 137861, 1257686, 12714721, 141119617, 1706150125, 22321071242, 314201147513, 4735338283937, 76079613690505, 1298088771022522, 23441677492095901, 446689961120844641

Offset: 1

Eric Angelini, Feb 12 2005

a(n) = x(n) + y(n) where x(n)/y(n) is the continued fraction [1,2,3,4,...,n].

Using a(n) = x(n) - y(n) instead of a(n) = x(n) + y(n) would give A058307.

G. C. Greubel, Table of n, a(n) for n = 1..449
Juan S. Auli and Sergi Elizalde, Wilf equivalences between vincular patterns in inversion sequences, arXiv:2003.11533 [math.CO], 2020.
Zhicong Lin and Sherry H. F. Yan, Vincular patterns in inversion sequences, Applied Mathematics and Computation (2020), Vol. 364, 124672.
Shuzhen Lv and Philip B. Zhang, Joint equidistributions of mesh patterns 123 and 321 with symmetric and antipodal shadings, arXiv:2501.00357 [math.CO], 2024. See p. 17.

Numerators are in A001040 and denominators in A001053.

Cf. A124428.

a:=[1,2];; for n in [3..30] do a[n]:=(n-1)*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Feb 23 2019

I:=[1,2]; [n le 2 select I[n] else (n-1)*Self(n-1) +Self(n-2): n in [1..30]]; // G. C. Greubel, Feb 23 2019

a[1]= 1; a[2]= 2; a[n_]:= a[n] = (n-1)*a[n-1]+a[n-2]; Table[a[n], {n,20}] (* Robert G. Wilson v, Feb 14 2005 *)
RecurrenceTable[{a[1]==1,a[2]==2,a[n+1]==n*a[n]+a[n-1]},a,{n,20}] (* Harvey P. Dale, Sep 04 2018 *)

a(n)=sum(k=0,n,k!*binomial((n+k)\2,k)*binomial((n+k+1)\2,k)) \\ Paul D. Hanna, Oct 31 2006

[sum(factorial(k)*binomial(floor((n+k-1)/2), k)*binomial(floor((n+k)/2), k) for k in (0..n)) for n in (1..30)] # G. C. Greubel, Feb 23 2019

a(n) = A001040(n) + A001053(n) for n>1.
From Paul D. Hanna, Oct 31 2006: (Start)
a(n+1) = Sum_{k=0..n} k!*C(floor((n+k)/2),k)*C(floor((n+k+1)/2),k).
a(n+1) = Sum_{k=0..n} k!*A124428(n+k,k). (End)

Edited and extended by Robert G. Wilson v, Feb 14 2005

A091964 Number of left factors of peakless Motzkin paths of length n.

Original entry on oeis.org

1, 2, 4, 9, 21, 50, 121, 296, 730, 1812, 4521, 11328, 28485, 71844, 181674, 460443, 1169283, 2974574, 7578937, 19337489, 49401526, 126350742, 323495259, 829033334, 2126454271, 5458711430, 14023219126, 36049991901, 92734505565

Offset: 0

Emeric Deutsch, Mar 13 2004

nonn

Number of paths from (0,0) to the line x=n, consisting of steps u=(1,1), h=(1,0), d=(1,-1), that never go below the x-axis and a u step is never followed by a d step.
a(n) is also the number of peakless Motzkin paths of length n in which the (1,0)-steps at level 0 come in 2 colors. Example: a(4)=21 because, denoting u=(1,1), h=(1,0), and d=(1,-1), we have 2^4 = 16 paths of shape hhhh, 2 paths of shape huhd, 2 paths of shape uhdh, and 1 path of shape uhhd. - Emeric Deutsch, May 03 2011
Equals diagonal sums of triangle A124428. - Paul D. Hanna, Oct 31 2006

			a(2)=4 because we have hh, hu, uh and uu.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Andrei Asinowski, Axel Bacher, Cyril Banderier, Bernhard Gittenberger, Analytic Combinatorics of Lattice Paths with Forbidden Patterns: Enumerative Aspects, in International Conference on Language and Automata Theory and Applications, S. Klein, C. Martín-Vide, D. Shapira (eds), Springer, Cham, pp 195-206, 2018.
Andrei Asinowski, Axel Bacher, Cyril Banderier, Bernhard Gittenberger, Analytic combinatorics of lattice paths with forbidden patterns, the vectorial kernel method, and generating functions for pushdown automata, Laboratoire d'Informatique de Paris Nord (LIPN 2019).
Ivo L. Hofacker, Christian M. Reidys, and Peter F. Stadler, Symmetric circular matchings and RNA folding. Discr. Math., 312:100-112, 2012. See Prop. 5, C_2^{1}(z).
Asamoah Nkwanta, Lattice paths and RNA secondary structures, DIMACS Series in Discrete Math. and Theoretical Computer Science, 34, 1997, 137-147.
Helmut Prodinger, Cornerless, peakless, valleyless Motzkin paths (regular and skew) and applications to bargraphs, arXiv:2501.13645 [math.CO], 2025. See p. 8.

Cf. A004148, A104559, A124428.

[(&+[Binomial(Floor((n+k)/2),k)*Binomial(Floor((n+k+1)/2),k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Feb 26 2019

CoefficientList[Series[2/(1-3*x+x^2+Sqrt[1-2*x-x^2-2*x^3+x^4]), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

a(n)=sum(k=0,n,binomial(n-k\2,(k+1)\2)*binomial(n-(k+1)\2,k\2)) \\ Paul D. Hanna, Mar 24 2005

a(n)=sum(k=0,n,binomial((n+k)\2,k)*binomial((n+k+1)\2,k)) \\ Paul D. Hanna, Oct 31 2006

[sum(binomial(floor((n+k)/2),k)*binomial(floor((n+k+1)/2),k) for k in (0..n)) for n in (0..30)] # G. C. Greubel, Feb 26 2019

G.f.: 2/(1 - 3*z + z^2 + sqrt(1 - 2*z - z^2 - 2*z^3 + z^4)).
a(n) = Sum_{k=0..n} C(n-floor(k/2), floor((k+1)/2)) * C(n-floor((k+1)/2), floor(k/2)). - Paul D. Hanna, Mar 24 2005
a(n) = Sum_{k=0..n} C(floor((n+k)/2),k)*C(floor((n+k+1)/2),k). - Paul D. Hanna, Oct 31 2006
G.f.: 1/(1-x-x/(1-x^2/(1-x/(1-x^2/(1-x/(1-x^2/(1-... (continued fraction). - Paul Barry, Jun 30 2009
D-finite with recurrence (n+1)*a(n) + 2*(-n-1)*a(n-1) + (-n+1)*a(n-2) + 2*(-n+3)*a(n-3) + (n-3)*a(n-4) = 0. - R. J. Mathar, Nov 24 2012
a(n) ~ (3+sqrt(5))^n / (sqrt(7*sqrt(5)-15) * sqrt(Pi*n) * 2^(n-1/2)). - Vaclav Kotesovec, Feb 12 2014
Equivalently, a(n) ~ phi^(2*n + 2) / (5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021

A056953 Denominators of continued fraction for alternating factorial.

Original entry on oeis.org

1, 1, 2, 3, 7, 13, 34, 73, 209, 501, 1546, 4051, 13327, 37633, 130922, 394353, 1441729, 4596553, 17572114, 58941091, 234662231, 824073141, 3405357682, 12470162233, 53334454417, 202976401213, 896324308634, 3535017524403, 16083557845279, 65573803186921

Offset: 0

Aleksandar Petojevic, Sep 05 2000

Starting (1, 2, 3, ...) with offset 0 = eigensequence of an infinite lower triangular matrix with 1's in the main diagonal and the natural numbers repeated in the subdiagonal. - Gary W. Adamson, Feb 14 2011

a(n) is the number of involutions of [n] such that every 2-cycle contains one odd and one even element; a(4) = 7: 1234, 1243, 1324, 2134, 2143, 4231, 4321. - Alois P. Heinz, Feb 14 2013

Alois P. Heinz, Table of n, a(n) for n = 0..400
Francesca Aicardi, Diego Arcis, and Jesús Juyumaya, Ramified inverse and planar monoids, arXiv:2210.17461 [math.RT], 2022.
Index entries for sequences related to Laguerre polynomials

Bisections are A000262 and A002720.

Cf. A124428, diagonals of A088699.

[(&+[Factorial(k)*Binomial(Floor(n/2),k)*Binomial(Floor((n+1)/2),k): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, May 16 2018

a:= proc(n) option remember; `if`(n<4, [1, 1, 2, 3][n+1],
      ((4*n-2)*a(n-2) +2*a(n-3) -(n-2)*(n-3)*a(n-4)) /4)
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Feb 14 2013

Table[Sum[k!*Binomial[Floor[n/2], k]*Binomial[Floor[(n+1)/2], k] , {k,0,Floor[n/2]}], {n,0,30}] (* G. C. Greubel, May 16 2018 *)

a(n)=sum(k=0,n\2,k!*binomial(n\2,k)*binomial((n+1)\2,k)) \\ Paul D. Hanna, Oct 31 2006

a(0)=1; a(1)=1; a(n) = a(n-1) + n*a(n-2)/2.
a(n) = Sum_{k=0..[n/2]} k!*C([n/2],k)*C([(n+1)/2],k). - Paul D. Hanna, Oct 31 2006
a(n) ~ n^(n/2 + 1/4) / (2^(n/2 + 3/4) * exp(n/2 - sqrt(2*n) + 1/2)) * (1 + (25 + 6*(-1)^n)/(24*sqrt(2*n)) + (397 + 156*(-1)^n)/(2304*n)). - Vaclav Kotesovec, Feb 22 2019

A191521 Triangle read by rows: T(n,k) is the number of left factors of Dyck paths of length n that have k valleys (i.e., a (1,-1)-step followed by a (1,1)-step).

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 3, 3, 6, 1, 4, 12, 4, 4, 18, 12, 1, 5, 30, 30, 5, 5, 40, 60, 20, 1, 6, 60, 120, 60, 6, 6, 75, 200, 150, 30, 1, 7, 105, 350, 350, 105, 7, 7, 126, 525, 700, 315, 42, 1, 8, 168, 840, 1400, 840, 168, 8, 8, 196, 1176, 2450, 1960, 588, 56, 1, 9, 252, 1764, 4410, 4410, 1764, 252, 9

Offset: 0

Emeric Deutsch, Jun 05 2011

Row n>=1 contains ceiling(n/2) entries.
Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
Sum_{k>=0} k*T(n,k) = A191522(n).

			T(4,1)=3 because we have U(DU)D, U(DU)U, and UU(DU), where U=(1,1) and D=(1,-1) (the valleys are shown between parentheses).
Triangle starts:
  1;
  1;
  2;
  2,  1;
  3,  3;
  3,  6,  1;
  4, 12,  4;
  4, 18, 12,  1;
  ...

Alois P. Heinz, Rows n = 0..300, flattened

Cf. A001405, A124428, A191522.

Q := sqrt(((1-z)^2-t*z^2)*((1+z)^2-t*z^2)): G := (1+t*z^2-z^2-Q)/(t*z*(t*z^2-1+2*z-z^2+Q)): Gser := simplify(series(G, z = 0, 19)): for n from 0 to 16 do P[n] := sort(coeff(Gser, z, n)) end do: 1; for n to 16 do seq(coeff(P[n], t, k), k = 0 .. ceil((1/2)*n)-1) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(x, y, t) option remember; expand(`if`(x=0, 1,
     `if`(y>0, b(x-1, y-1, z), 0)+b(x-1, y+1, 1)*t))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0, 1)):
seq(T(n), n=0..30);  # Alois P. Heinz, Mar 29 2017

T[n_, m_] := If [n == 0 && m == 0, 1, If[n == 0, 0, If[OddQ[n-1], (2* Binomial[n/2, m]*Binomial[n/2, m+1]*(n/2 + 1))/n, Binomial[(n+1)/2, m+1]*Sum[(-1)^(k-1)*Binomial[(n+1)/2, m-k+1], {k, 1, (n+1)/2}]]]];
Table[T[n, m], {n, 0, 16}, {m, 0, If[n <= 2, 0, Quotient[n-1, 2]]}] // Flatten (* Jean-François Alcover, Feb 16 2021, after Vladimir Kruchinin *)

T(n,m):=if n=0 and m=0 then 1 else if n=0 then 0 else if oddp(n-1) then (2*binomial(n/2,m)*binomial(n/2,m+1)*(n/2+1))/n else binomial((n+1)/2,m+1)*sum((-1)^(k-1)*binomial((n+1)/2,m-k+1),k,1,(n+1)/2);
/* Vladimir Kruchinin, Jul 24 2019 */

G.f.: G(t,z) = (1+t*z^2-z^2-Q)/(t*z*(t*z^2-1+2*z-z^2+Q)), where Q = sqrt(((1-z)^2-t*z^2)*((1+z)^2-t*z^2)).

T(n,k) = 2*C(n/2,k)*C(n/2,k+1)*(n/2+1)/n, for even n, C((n+1)/2,k+1)*Sum_{j=1..(n+1)/2} (-1)^(j-1)*C((n+1)/2,k-j+1), for odd n, T(0,0)=1. - Vladimir Kruchinin, Jul 24 2019

A124431 a(n) = Sum_{k=0..n} 2^kC([(n+k)/2],k)C([(n+k+1)/2],k) where [x]=floor(x).

Original entry on oeis.org

1, 3, 9, 29, 97, 331, 1145, 4001, 14089, 49915, 177713, 635293, 2278841, 8198227, 29567729, 106872961, 387038993, 1404052659, 5101219929, 18559193245, 67605310097, 246541193883, 899999057385, 3288522934433, 12026324883865

Offset: 0

Paul D. Hanna, Oct 31 2006

nonn

This is the inverse Motzkin transform of A026378 assuming offset 1 here. - R. J. Mathar, Jul 07 2009
Hankel transform is Somos-4 variant A162547. - Paul Barry, Jan 09 2011
a(n) is the number of peakless Motzkin paths of length n in which the (1,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=29 because, denoting  U=(1,1), H=(1,0), and D=(1,-1), we have 3^3 = 27 paths of shape HHH and 2 paths of shape UHD. - Emeric Deutsch, May 03 2011
Conjecture: (n+1)*a(n) -2*(2*n+1)*a(n-1) +2*(n-1)*a(n-2) +2*(5-2*n)*a(n-3) +(n-3)*a(n-4) =0. - R. J. Mathar, Aug 09 2012

			G.f. = 1 + 3*x + 9*x^2 + 29*x^3 + 97*x^4 + 331*x^5 + 1145*x^6 + 4001*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A124428.

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (Sqrt((1+x^2)/(1-4*x+x^2)) -1)/(2*x) )); // G. C. Greubel, Feb 26 2019

Table[Sum[2^k Binomial[Floor[(n+k)/2],k]Binomial[Floor[(n+k+1)/2],k],{k,0,n}],{n,0,30}] (* Harvey P. Dale, May 20 2012 *)
CoefficientList[Series[(Sqrt[(1+x^2)/(1-4*x+x^2)] -1)/(2*x), {x,0,30}],x] (* G. C. Greubel, Feb 26 2019 *)

a(n)=sum(k=0,n,2^k*binomial((n+k)\2,k)*binomial((n+k+1)\2,k))

my(x='x+O('x^30)); Vec((sqrt((1+x^2)/(1-4*x+x^2)) -1)/(2*x)) \\ G. C. Greubel, Feb 26 2019

((sqrt((1+x^2)/(1-4*x+x^2)) -1)/(2*x)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019

a(n) = Sum_{k=0..n} 2^k*A124428(n+k,k).
G.f.: (((x^2+1)*(1-4*x+x^2))^(1/2) - (1-4*x+x^2))/(2*x*(1-4*x+x^2)). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009
G.f.: (1/(1-4*x+x^2))*c(-x/(1-4*x+x^2)), c(x) the g.f. of A000108. - Paul Barry, Jan 09 2011
G.f.: G(0)/(2*x) - 1/(2*x), where G(k)= 1 + 4*x*(4*k+1)/( (1+x^2)*(4*k+2) - x*(1+x^2)*(4*k+2)*(4*k+3)/(x*(4*k+3) + (1+x^2)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 26 2013
a(n) ~ sqrt(14*sqrt(3)-24) * (2+sqrt(3))^(n+2) / (2*sqrt(3*Pi*n)). - Vaclav Kotesovec, Feb 03 2014
0 = a(n)*(+a(n+1) - 6*a(n+2) + 6*a(n+3) - 18*a(n+4) + 5*a(n+5)) + a(n+1)*(-2*a(n+1) + 14*a(n+2) - 10*a(n+3) + 61*a(n+4) - 18*a(n+5)) + a(n+2)*(+4*a(n+2) - 28*a(n+3) - 10*a(n+4) + 6*a(n+5)) + a(n+3)*(+4*a(n+3) + 14*a(n+4) - 6*a(n+5)) + a(n+4)*(-2*a(n+4) + a(n+5)) if n>=0. - Michael Somos, Aug 06 2014
Conjecture: +(n+1)*a(n) +2*(-2*n-1)*a(n-1) +2*(n-1)*a(n-2) +2*(-2*n+5)*a(n-3) +(n-3)*a(n-4)=0. - R. J. Mathar, Jun 17 2016

A191579 Triangular array related to continued fractions of square root of (N^2 - 1) for N>1, apparently containing A004148 and summing to A091964.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 4, 6, 6, 4, 1, 8, 13, 13, 10, 5, 1, 17, 28, 30, 24, 15, 6, 1, 37, 62, 69, 59, 40, 21, 7, 1, 82, 140, 160, 144, 105, 62, 28, 8, 1, 185, 320, 375, 350, 271, 174, 91, 36, 9, 1, 423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1

Offset: 1

Kenneth J Ramsey, Jun 07 2011

The row sums of this triangle seems to be A091964 (verified to 12 terms), cf. diagonal sums of the triangle A124428. The 1st column seems to be A004148.  The 2nd and 3rd column seems to be A089735, and A098075 (verified to 10 terms).
As each of these sequence is related to enumeration of RNA molecule structures, but was generated independently by reference to square array A192062 (re continued fractions for square roots of n^2-1 for n>1, see comments in the example below), it could be interesting to check this further for a relationship. As Mathar noted, this triangle appears identical to A097724. - edited by Kenneth J Ramsey, Oct 25 2012
Is this (apart from offsets) the same as A097724? - R. J. Mathar, Aug 01 2011

			The triangle begins
1;
1, 1;
1, 2, 1;
2, 3, 3, 1;
4, 6, 6, 4, 1;
8, 13, 13, 10, 5, 1;
17, 28, 30, 24, 15, 6, 1;
37, 62, 69, 59, 40, 21, 7, 1;
82, 140, 160, 144, 105, 62, 28, 8, 1;
185, 320, 375, 350, 271, 174, 91, 36, 9, 1;
423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1;
...
The 4th row is 2,3,3,1 because the 2nd,4th,6th and 8th terms of columns j = 1-5 of square array T(i,j) A192062  form the 4*5 matrix {{1,3,8,21},{1,4,15,56},{1,5,24,115},{1,6,35,204},{1,7,48,329}}. Solving the resulting system of linear equations results in the identities:
2*1 + 3*3 + 3*8 + 1*21 = 56 = T(8,2) of A192062
2*1 + 3*4 + 3*15+ 1*56 = 115 = T(8,3) of A192062
2*1 + 3*5 + 3*24 + 1*115 = 204 = T(8,4) of A192062
2*1 + 3*6 + 3*35 + 1*204 = 329 = T(8,5) of A192062

Cf. A091964, A124428, A004148, A089735, A098075, A192062.

The only way I know to generate this triangle is by reference to the square array A192062. The columns of that array, T(i,j) are such that for any given i>0, each term T(i,2*n) equals the sum as k = 1 to n, T(i-1,2*k)*C_k where C_k is the k th term of the n th row of this triangle. So solving the system of linear equations for each n > 0 gives the n th row of this triangle.

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A124429 Antidiagonal sums of triangle A124428.

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A124430 Eigenvector of triangle A124428.

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A006542 a(n) = binomial(n,3)*binomial(n-1,3)/4.

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A006011 a(n) = n^2*(n^2 - 1)/4.

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A102038 a(n+1) = n*a(n) + a(n-1), a(1)=1, a(2)=2.

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A091964 Number of left factors of peakless Motzkin paths of length n.

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A124431 a(n) = Sum_{k=0..n} 2^kC([(n+k)/2],k)C([(n+k+1)/2],k) where [x]=floor(x).