A132894 -id:A132894 - cp's OEIS frontend

A005430 Apéry numbers: nC(2n,n).

0, 2, 12, 60, 280, 1260, 5544, 24024, 102960, 437580, 1847560, 7759752, 32449872, 135207800, 561632400, 2326762800, 9617286240, 39671305740, 163352435400, 671560012200, 2756930576400, 11303415363240, 46290177201840, 189368906734800, 773942488394400

Offset: 0

Simon Plouffe

Appears as diagonal in A003506. - Zerinvary Lajos, Apr 12 2006
The aerated sequence 1,0,2,0,12,0,60,0,... has e.g.f. 1+x*Bessel_I(1,2x). - Paul Barry, Mar 29 2010
Conjecture: the terms of the inverse binomial transform are 2*A132894(n). - R. J. Mathar, Oct 21 2012

Frank Harary and Edgar M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 78, (3.5.25).
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..200
Kunle Adegoke, Robert Frontczak, and Taras Goy, Fibonacci-Catalan Series, Integers: Electronic Journal of Combinatorial Number Theory, Vol. 22 (2022), #A110.
Laurent Alonso and Edward M. Reingold, Analysis of Boyer and Moore's MJRTY Algorithm, 2012.
T. Amdeberhan and Henri Cohen, Bernoulli sum meets golden number, MathOverflow, version of 2017-06-15.
Libor Caha and Daniel Nagaj, The pair-flip model: a very entangled translationally invariant spin chain, arXiv:1805.07168 [quant-ph], 2018.
Benjamin Ruoyu Kan, Polynomial Approximations for Quantum Hamiltonian Complexity, Bachelor's thesis, Harvard Univ., 2023.
Hans J. H. Tuenter, Walking into an absolute sum, arXiv:math/0606080 [math.NT], 2006. Published version on Walking into an absolute sum, The Fibonacci Quarterly, 40(2):175-180, May 2002.
Alfred J. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3), Math. Intelligencer 1 (1978/1979), 195-203.
I. J. Zucker, On the series Sum(k>=1) C(2k,k)^(-1)*k^(-n) and related sums, J. Number Theory, Vol. 20, No. 1 (1985), 92-102.
Wadim Zudilin, An elementary proof of Apery's theorem, arXiv:math/0202159 [math.NT], 2002.

Cf. A002011, A002457, A002736, A005258, A005259, A005429, 1/beta(n, n+1) in A061928.

Cf. A001803, A003506.

List([0..30], n-> n*Binomial(2*n,n)); # G. C. Greubel, Dec 09 2018

[n*Binomial(2*n,n): n in [0..30]]; // G. C. Greubel, Dec 09 2018

Maple
```
A005430 := n -> n*binomial(2*n, n);
```

Table[n*Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, May 29 2015 *)

a(n)=-(-1)^n*real(polcoeff(serlaplace(x^2*besselh1(1,2*x)),2*n)) \\ Ralf Stephan

[n*binomial(2*n,n) for n in range(30)] # G. C. Greubel, Dec 09 2018

a(n) = A002011(n-1)/2 = 2 * A002457(n-1).
Sum_{n >= 1} 1/a(n) = Pi*sqrt(3)/9. - Benoit Cloitre, Apr 07 2002
G.f.:  2*x/sqrt((1-4*x)^3). - Marco A. Cisneros Guevara, Jul 25 2011
E.g.f.: a(n) = n!* [x^n] exp(2*x)*2*x*(BesselI(0, 2*x)+BesselI(1, 2*x)). - Peter Luschny, Aug 25 2012
D-finite with recurrence (-n+1)*a(n) + 2*(2*n-1)*a(n-1) = 0. - R. J. Mathar, Dec 03 2012
G.f.: 2*x*(1-4*x)^(-3/2) = -G(0)/2 where G(k) =  1 - (2*k+1)/(1 - 2*x/(2*x - (k+1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
a(n-1) = Sum_{k=0..floor(n/2)} k*C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 4^n*sqrt(n)/sqrt(Pi).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(phi)/sqrt(5) = A086466, where phi is the golden ratio. (End)
1/a(n) = (-1)^n*Sum_{j=0..n-1} binomial(n-1,j)*Bernoulli(j+n)/(j+n) for n >= 1. See the Amdeberhan & Cohen link. - Peter Luschny, Jun 20 2017
1/a(n) = Sum_{k=0..n} (-1)^(k+1)*binomial(n,k)*HarmonicNumber(n+k) for n >= 1. - Peter Luschny, Aug 15 2017
Sum_{n>=1} x^n/a(n) = 2*sqrt(x/(4-x))*arcsin(sqrt(x)/2), for abs(x) < 4 (Adegoke et al., 2022, section 6, p. 11). - Amiram Eldar, Dec 07 2024

More terms from James Sellers, May 01 2000

A328347 Number T(n,k) of n-step walks on cubic lattice starting at (0,0,0), ending at (0,k,n-k) and using steps (0,0,1), (0,1,0), (1,0,0), (-1,1,1), (1,-1,1), and (1,1,-1); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 1, 1, 3, 4, 3, 7, 15, 15, 7, 19, 52, 72, 52, 19, 51, 175, 300, 300, 175, 51, 141, 576, 1185, 1480, 1185, 576, 141, 393, 1869, 4473, 6685, 6685, 4473, 1869, 393, 1107, 6000, 16380, 28392, 33880, 28392, 16380, 6000, 1107, 3139, 19107, 58572, 115332, 159264, 159264, 115332, 58572, 19107, 3139

Offset: 0

Alois P. Heinz, Oct 13 2019

These walks are not restricted to the first (nonnegative) octant.

			Triangle T(n,k) begins:
     1;
     1,    1;
     3,    4,     3;
     7,   15,    15,     7;
    19,   52,    72,    52,    19;
    51,  175,   300,   300,   175,    51;
   141,  576,  1185,  1480,  1185,   576,   141;
   393, 1869,  4473,  6685,  6685,  4473,  1869,  393;
  1107, 6000, 16380, 28392, 33880, 28392, 16380, 6000, 1107;
  ...

Alois P. Heinz, Rows n = 0..200, flattened
Wikipedia, Lattice path
Wikipedia, Self-avoiding walk

Columns k=0-1 give: A002426, A132894 = n*A005773(n).
Row sums give A084609.
T(2n,n) gives A328426.
Cf. A007318, A328300, A328345.

b:= proc(l) option remember; `if`(l[-1]=0, 1, (r-> add(add(
      add(`if`(i+j+k=1, (h-> `if`(add(t, t=h)<0, 0, b(h)))(
      sort(l-[i, j, k])), 0), k=r), j=r), i=r))([$-1..1]))
    end:
T:= (n, k)-> b(sort([0, k, n-k])):
seq(seq(T(n, k), k=0..n), n=0..12);

b[l_List] := b[l] = If[l[[-1]] == 0, 1, Sum[If[i + j + k == 1, Function[h, If[Total[h] < 0, 0, b[h]]][Sort[l - {i, j, k}]], 0], {i, {-1, 0, 1}}, {j, {-1, 0, 1}}, {k, {-1, 0, 1}}]];
T[n_, k_] := b[Sort[{0, k, n - k}]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 30 2020, after Alois P. Heinz *)

T(n,k) = T(n,n-k).

A107230 A number triangle of inverse Chebyshev transforms.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 6, 3, 1, 6, 12, 12, 4, 1, 10, 30, 30, 20, 5, 1, 20, 60, 90, 60, 30, 6, 1, 35, 140, 210, 210, 105, 42, 7, 1, 70, 280, 560, 560, 420, 168, 56, 8, 1, 126, 630, 1260, 1680, 1260, 756, 252, 72, 9, 1, 252, 1260, 3150, 4200, 4200, 2520, 1260, 360, 90, 10, 1

Offset: 0

Paul Barry, May 13 2005

First column is A001405, second column is A100071, third column is A107231. Row sums are A005773(n+1), diagonal sums are A026003. The inverse Chebyshev transform concerned takes a g.f. g(x)->(1/sqrt(1-4x^2))g(xc(x^2)) where c(x) is the g.f. of A000108. It transforms a(n) to b(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*a(n-2k). Then a(n) = Sum_{k=0..floor(n/2)} (n/(n-k))*(-1)^k*binomial(n-k,k) *b(n-2k).

Triangle read by rows: T(n,k) is the number of paths of length n with steps U=(1,1), D=(1,-1) and H=(1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having k H steps. Example: T(3,1)=6 because we have HUD. HUU, UDH, UHD, UHU and UUH. Sum_{k=0..n} k*T(n,k) = A132894(n). - Emeric Deutsch, Oct 07 2007

			Triangle begins
   1;
   1,  1;
   2,  2,  1;
   3,  6,  3,  1;
   6, 12, 12,  4,  1;
  10, 30, 30, 20,  5,  1;

Jinyuan Wang, Rows n=0..200 of triangle, flattened
Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.

Cf. A132894.

[[Binomial(n, k)*Binomial(n-k, Floor((n-k)/2)): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 11 2019

T:=proc(n,k) options operator, arrow: binomial(n, k)*binomial(n-k, floor((1/2)*n-(1/2)*k)) end proc: for n from 0 to 11 do seq(T(n,k),k=0..n) end do; # yields sequence in triangular form - Emeric Deutsch, Oct 07 2007

Table[Binomial[n, k]*Binomial[n-k, Floor[(n-k)/2]], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 11 2019 *)

T(n, k) = binomial(n, k)*binomial(n-k, (n-k)\2); \\ Michel Marcus, Feb 10 2019

[[binomial(n, k)*binomial(n-k, floor((n-k)/2)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 11 2019

T(n,k) = binomial(n,k)*binomial(n-k, floor((n-k)/2)).
G.f.: G=G(t,z) satisfies z*(1-2*z-t*z)*G^2+(1-2*z-t*z)*G-1=0. - Emeric Deutsch, Oct 07 2007
E.g.f.: exp(x*y)*(BesselI(0,2*x)+BesselI(1,2*x)). - Vladeta Jovovic, Dec 02 2008

A383948 Expansion of 1/sqrt((1-3x)^3 (1-7*x)).

Original entry on oeis.org

1, 8, 51, 308, 1855, 11340, 70665, 448320, 2887155, 18815240, 123759097, 819969276, 5464090177, 36580917716, 245837438055, 1657396783440, 11204207037315, 75918595916520, 515462211835305, 3506072423912940, 23885410548196701, 162951783575205108, 1113110415733083531

Offset: 0

Seiichi Manyama, Aug 19 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..500

Cf. A098409, A360318.
Cf. A383949, A383950.
Cf. A132894.

R := PowerSeriesRing(Rationals(), 34); f := 1/Sqrt((1- 3*x)^3 * (1-7*x)); coeffs := [ Coefficient(f, n) : n in [0..33] ]; coeffs; // Vincenzo Librandi, Aug 27 2025

CoefficientList[Series[ 1/Sqrt[(1-3*x)^3*(1-7*x)],{x,0,33}],x] (* Vincenzo Librandi, Aug 27 2025 *)

my(N=30, x='x+O('x^N)); Vec(1/sqrt((1-3*x)^3*(1-7*x)))

n*a(n) = (10*n-2)*a(n-1) - 21*n*a(n-2) for n > 1.
a(n) = (1/4)^n * Sum_{k=0..n} 3^k * 7^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(2*(n-k),n-k).
a(n) = Sum_{k=0..n} (-1)^k * 7^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(n+1,n-k).
a(n) = Sum_{k=0..n} 3^(n-k) * binomial(2*k,k) * binomial(n+1,n-k).

A132893 Triangle read by rows: T(n,k) is the number of paths of length n with steps U=(1,1), D=(1,-1) and H=(1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having k peaks (i.e., UDs), 0 <= k <= floor(n/2).

Original entry on oeis.org

1, 2, 4, 1, 9, 4, 21, 13, 1, 50, 40, 6, 121, 118, 27, 1, 296, 340, 106, 8, 730, 965, 381, 46, 1, 1812, 2708, 1296, 220, 10, 4521, 7535, 4241, 935, 70, 1, 11328, 20828, 13482, 3676, 395, 12, 28485, 57266, 41916, 13658, 1940, 99, 1

Offset: 0

Emeric Deutsch, Oct 08 2007

Row n has 1 + floor(n/2) terms.

Row sums yield A005773.

			T(3,1)=4 because we have HUD, UDH, UDU and UUD.
Triangle starts:
    1;
    2;
    4,   1;
    9,   4;
   21,  13,   1;
   50,  40,   6;
  121, 118,  27,   1;

Alois P. Heinz, Rows n = 0..200, flattened
Marilena Barnabei, Flavio Bonetti, and Niccolò Castronuovo, Motzkin and Catalan Tunnel Polynomials, J. Int. Seq., Vol. 21 (2018), Article 18.8.8.

Cf. A005773, A091964, A132894.

G:=((-1+3*z-z^2+t*z^2+sqrt((1+z+z^2-t*z^2)*(1-3*z+z^2-t*z^2)))*1/2)/(z*(1-3*z+z^2-t*z^2)): Gser:=simplify(series(G,z=0,15)): for n from 0 to 12 do P[n]:=sort(coeff(Gser,z,n)) end do: for n from 0 to 12 do seq(coeff(P[n],t,j), j=0..floor((1/2)*n)) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(x, y, t) option remember; expand(`if`(y<0, 0,
     `if`(x=0, 1, b(x-1, y, 1)+b(x-1, y-1, 1)*t+b(x-1, y+1, z))))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0, 1)):
seq(T(n), n=0..15);  # Alois P. Heinz, Feb 01 2019

b[x_, y_, t_] := b[x, y, t] = Expand[If[y < 0, 0, If[x == 0, 1, b[x - 1, y, 1] + b[x - 1, y - 1, 1]*t + b[x - 1, y + 1, z]]]];
T[n_] := Function[p, Table[Coefficient[p, z, i], {i, 0, Exponent[p, z]}]] @ b[n, 0, 1];
T /@ Range[0, 15] // Flatten (* Jean-François Alcover, Oct 06 2019, after Alois P. Heinz *)

T(n,0) = A091964(n).
Sum_{k=0..floor(n/2)} k*T(n,k) = A132894(n-1).
G.f.: G = G(t,z) satisfies z(1 - 3z + z^2 - tz^2)G^2 + (1 - 3z + z^2 - tz^2)G - 1 = 0 (see the Maple program for the explicit expression of G).

A137213 First differences of A137212.

Original entry on oeis.org

0, 1, 4, 15, 52, 173, 560, 1779, 5576, 17305, 53308, 163287, 497980, 1513541, 4587944, 13878075, 41910032, 126395953, 380795380, 1146267039, 3448170436, 10367130845, 31156000928, 93599839107, 281117798360, 844121793481

Offset: 0

Paul Curtz, Mar 06 2008

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-5,-3).

Cf. A000129, A132894, A137212.

[n le 3 select (n-1)^2 else 5*Self(n-1) -5*Self(n-2) -3*Self(n-3): n in [1..31]]; // G. C. Greubel, Jan 05 2022

LinearRecurrence[{5,-5,-3},{0,1,4},30] (* Harvey P. Dale, Apr 18 2019 *)

[3^n - lucas_number1(n+1,2,-1) for n in (0..30)] # G. C. Greubel, Jan 05 2022

From R. J. Mathar, Mar 17 2008: (Start)
O.g.f.: 1/(1-3*x) - 1/(1-2*x-x^2) = x*(1-x)/( (1-3*x)*(1-2*x-x^2) ).
a(n) = 3^n - A000129(n+1). (End)

More terms from R. J. Mathar, Mar 17 2008

A263841 Expansion of (1 - 2x - x^2)/(sqrt(1+x)(1-3x)^(3/2)2x) - 1/(2x).

Original entry on oeis.org

1, 3, 9, 28, 87, 271, 843, 2619, 8123, 25153, 77763, 240054, 740017, 2278329, 7006093, 21520872, 66039651, 202462113, 620164491, 1898109900, 5805127269, 17741909157, 54188530641, 165405964227, 504601360389, 1538559689751, 4688812503053, 14282580916834, 43486805133903

Offset: 0

N. J. A. Sloane, Nov 02 2015

nonn

A. Asinowski and G. Rote, Point sets with many non-crossing matchings, arXiv preprint arXiv:1502.04925 [cs.CG], 2015. See Table 1.

Cf. A005773, A132894, A189911.

A263841 := n -> add((k+1)*binomial(n, k)*binomial(n-k, iquo(n-k,2)), k = 0 .. n):
seq(A263841(n), n = 0 .. 28); # Mélika Tebni, Jan 25 2024

CoefficientList[Series[(1-2x-x^2)/(Sqrt[1+x] (1-3x)^(3/2) 2x)-1/(2x),{x,0,30}],x] (* Harvey P. Dale, Aug 21 2017 *)

my(x='x+O('x^40)); Vec((1-2*x-x^2)/(sqrt(1+x)*(1-3*x)^(3/2)*2*x)-1/(2*x)) \\ Altug Alkan, Nov 10 2015

D-finite with recurrence: -(n+1)*(n^2+n-3)*a(n) + 2*(n^3+3*n^2-4*n-3)*a(n-1) + 3*(n-1)*(n^2+3*n-1)*a(n-2) = 0. - R. J. Mathar, Feb 17 2016
From Mélika Tebni, Jan 24 2024: (Start)
a(n) = A005773(n+1) + A132894(n).
E.g.f.: (1+x)*exp(x)*(BesselI(0,2*x) + BesselI(1,2*x)). (End)
From Mélika Tebni, Jan 25 2024: (Start)
a(n) = Sum_{k=0..n} A189911(k)*binomial(n,k).
a(n) = Sum_{k=0..n} (k+1)*binomial(n,k)*binomial(n-k,floor((n-k)/2)). (End)

A337991 Triangle read by rows: T(n,m) = Sum_{i=1..n} C(n,i-m)C(n+m-i,i-1)C(n+m-i,m)/n, with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 5, 4, 1, 4, 13, 15, 7, 1, 9, 35, 52, 36, 11, 1, 21, 96, 175, 160, 75, 16, 1, 51, 267, 576, 655, 415, 141, 22, 1, 127, 750, 1869, 2541, 2030, 952, 245, 29, 1, 323, 2123, 6000, 9492, 9156, 5488, 1988, 400, 37, 1, 835, 6046, 19107, 34476, 38976, 28476, 13356, 3852, 621, 46, 1

Offset: 0

Vladimir Kruchinin, Oct 06 2020

			Triangle begins as:
   1;
   1,   1;
   1,   2,   1;
   2,   5,   4,   1;
   4,  13,  15,   7,   1;
   9,  35,  52,  36,  11,   1;
  21,  96, 175, 160,  75,  16,  1;
  51, 267, 576, 655, 415, 141, 22,  1;
  ...

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A001006, A005773, A086246, A132894.
Diagonals include: A000124, A006008.
Sums include: A000007 (signed row), A019590 (signed diagonal), A025227 (row), A102407 (diagonal).

B:=Binomial;
A337991:= func< n,k | n eq 0 select 1 else (1/n)*(&+[B(n, j-k)*B(n+k-j, j-1)*B(n+k-j, k): j in [1..n]]) >;
[A337991(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 31 2024

T[0, 0] = 1; T[n_, m_] := Sum[Binomial[n, i - m] * Binomial[n + m - i, i - 1] * Binomial[n + m - i, m]/n, {i, 1, n}]; Table[T[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Amiram Eldar, Oct 06 2020 *)

T(n,m):=if m=n then 1 else if n=0 then 0 else sum(binomial(n,i-m)*binomial(n+m-i,i-1)*binomial(n+m-i,m),i,1,n)/n;

def A337991(n,k):
    b=binomial
    if n==0: return 1
    else: return (1/n)*sum(b(n, j-k)*b(n+k-j, j-1)*b(n+k-j, k) for j in range(1,n+1))
# SageMath
flatten([[A337991(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 31 2024

G.f.: ( 1 - x*(y-1)- sqrt(x^2*(y^2-2*y-3) - 2*x*(y+1) + 1) )/(2*x).
From G. C. Greubel, Oct 31 2024: (Start)
T(n, k) = binomial(n, 1-k)*binomial(n+k-1, k)*Hypergeometric3F2([1-n, (1 -n -k)/2, (2-n-k)/2], [2-k, 1-n-k], 4), with T(0, 0) = 1.
T(n, 0) = A086246(n+1).
T(n, n-1) = A000124(n-1), n >= 1.
T(n, n-2) = A006008(n-1), n >= 2.
T(n, n-3) = (1/72)*(n^4 -6*n^3 +47*n^2 -114*n +144)*binomial(n-1,2), n >= 3.
T(n, n-4) = (1/480)*(n-2)*(n^4 -8*n^3 +99*n^2 -332*n +960)*binomial(n-1,3), n >= 4.
Sum_{k=0..n} T(n, k) = A025227(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A102407(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A019590(n+1). (End)

A359087 a(n) is equal to the last point of a reverse pyramid summation with base 1, 2, 3, ..., n-2, n-1, n, n-1, n-2, ..., 3, 2, 1.

Original entry on oeis.org

1, 4, 19, 78, 301, 1108, 3951, 13758, 47049, 158616, 528619, 1745098, 5715429, 18593032, 60136183, 193525002, 620046513, 1978886448, 6293809971, 19955385762, 63094947981, 198990438408, 626141673375, 1966085927898, 6161660863929, 19276374528468, 60206635741131

Offset: 1

Moosa Nasir, Dec 15 2022

nonn

Each element in the pyramid below the base is equal to the sum of the top left, top, and top right elements.
Each row has 2*n-(1+2*r) elements where r is the row number starting from 0.
The sum of elements in the first row is n^2.
The total number of elements in the pyramid is n^2.

			For n = 3:
  1  2  3  2  1
     6  7  6
       19
so a(3) = 19.
For n = 4:
  1   2   3   4   3   2   1
      6   9  10   9   6
         25  28  25
             78
so a(4) = 78.

Cf. A004737, A027907, A132894.

unsigned long tri(int n, int k)
{
    if (n == 0 && k == 0) return 1;
    if(k < -n || k > n) return 0;
    return tri(n - 1, k - 1) + tri(n - 1, k) + tri(n - 1, k + 1);
}
unsigned long a(int n)
{
    unsigned long sum = 0;
    sum += tri(n - 1,0) * n;
    for (int i = 1; i < n; i++)
    {
        sum += 2 * tri(n - 1,n - i) * i;
    }
    return sum;
}

f:= proc(n) local L,i;
  L:= [seq(i,i=1..n),seq(n-i,i=1..n-1)];
  for i from 1 to n-1 do
    L:= L[1..-3] + L[2..-2] + L[3..-1]
  od;
  op(L)
end proc:
map(f, [$1..30]); # Robert Israel, Dec 17 2022

f[n_] := Module[{L, i}, L = Range[n]~Join~Table[n-i, {i, 1, n-1}]; For[i = 1, i <= n-1, i++, L = L[[1;;-3]] + L[[2;;-2]] + L[[3;;-1]]]; L[[1]]];
f /@ Range[30] (* Jean-François Alcover, Jan 25 2023, after Robert Israel *)

a(n) = Sum_{k=1..2*n-1} A004737(k + (n-1)^2) * A027907(k + (n-1)^2 - 1).
Empirical g.f.: x/(1-3*x)^2 - 2*x^2/((1+x)^(1/2)*(1-3*x)^(3/2)). - Robert Israel, Dec 17 2022
a(n) = n*3^(n-1) - 2*A132894(n-1) (conjectured). - Bernard Schott, Dec 20 2022

A378810 Number of horizontal steps in all peak and valleyless Motzkin meanders of length n.

Original entry on oeis.org

0, 1, 4, 13, 39, 110, 300, 801, 2106, 5473, 14097, 36056, 91697, 232108, 585212, 1470557, 3684682, 9209417, 22967446, 57167993, 142051519, 352427720, 873157093, 2160579740, 5340150100, 13185150903, 32523933395, 80156852042, 197391001215, 485723767342

Offset: 0

John Tyler Rascoe, Dec 08 2024

Motzkin meanders are lattice paths starting at (0,0) with steps Up (0,1), Horizontal (1,0), and Down (0,-1) that stay weakly above the x-axis. Peak and valleyless Motzkin meanders avoid UD and DU.

			For n = 3 we have meanders, UUU, UUH, UHU, UHD, HUU, UHH, HHU, HUH, HHH; giving a total of a(3) = 13 H steps.

Cf. A005773, A132894, A308435, A378809.

A088855(n,k) = {binomial(floor((n-1)/2), floor((k-1)/2))*binomial(ceil((n-1)/2),ceil((k-1)/2))}
A_xy(N) = {my(x='x+O('x^N), h = sum(n=0,N, (1/(1-y*x)^(n+1)) * (if(n<1,1,0) + sum(k=1,n, A088855(n,k)*x^(n+k-1)*(y^(k-1)) )) )); h}
P_xy(N) = Pol(A_xy(N), {x})
A_x(N) = {my(px = deriv(P_xy(N),y), y=1); Vecrev(eval(px))}
A_x(20)

a(n) = Sum_{k=1..n} A378809(n,k)*k.

cp's OEIS Frontend

A005430 Apéry numbers: n*C(2*n,n).

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A328347 Number T(n,k) of n-step walks on cubic lattice starting at (0,0,0), ending at (0,k,n-k) and using steps (0,0,1), (0,1,0), (1,0,0), (-1,1,1), (1,-1,1), and (1,1,-1); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A107230 A number triangle of inverse Chebyshev transforms.

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A383948 Expansion of 1/sqrt((1-3*x)^3 * (1-7*x)).

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A132893 Triangle read by rows: T(n,k) is the number of paths of length n with steps U=(1,1), D=(1,-1) and H=(1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having k peaks (i.e., UDs), 0 <= k <= floor(n/2).

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A137213 First differences of A137212.

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A005430 Apéry numbers: nC(2n,n).

A383948 Expansion of 1/sqrt((1-3x)^3 (1-7*x)).

A263841 Expansion of (1 - 2x - x^2)/(sqrt(1+x)(1-3x)^(3/2)2x) - 1/(2x).

A337991 Triangle read by rows: T(n,m) = Sum_{i=1..n} C(n,i-m)C(n+m-i,i-1)C(n+m-i,m)/n, with T(0,0)=1.