A002002 -id:A002002 - cp's OEIS frontend

A035029 a(n) = Sum_{k=0..n} (k+1) * Sum_{j=0..n} 2^jbinomial(n,j)binomial(n-k,j).

1, 5, 26, 138, 743, 4043, 22180, 122468, 679757, 3789297, 21199998, 118973550, 669447123, 3775577367, 21336790152, 120795829128, 684962855705, 3889578815453, 22115533878178, 125892252068498, 717400693313471, 4092099111728355, 23362391663233196, 133488737662062188, 763310051648602213

Offset: 0

N. J. A. Sloane

nonn

Number of regions in all the dissections of a convex (n+3)-gon by non-intersecting diagonals. a(1)=5 because in the three dissections of a square we have altogether five regions: one in the "no-diagonals" dissection and two in each of the dissections by one of the two diagonals of the square. - Emeric Deutsch, Dec 28 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions
D. Merlini, R. Sprugnoli and M. C. Verri, Waiting patterns for a printer, FUN with algorithm'01, Isola d'Elba, 2001.

Cf. A001003, A001850, A002002, A049600.

I:=[1,5,26]; [n le 3 select I[n] else ( (7*n+1)*Self(n-1) - (7*n-15)*Self(n-2) + (n-3)*Self(n-3) )/(n+1): n in [1..30]]; // G. C. Greubel, Oct 20 2022

CoefficientList[Series[(1-x)^2/(8*x^2*Sqrt[1-6*x+x^2])-(1+x)/(8*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)
With[{P = LegendreP}, Table[((n+1)*(n+3)*P[n+3,3] -(6*n^2+22*n+17)*P[n+2,3] +(n+ 2)*(5*n+8)*P[n+1,3])/(8*(n+1)*(n+2)), {n,0,40}]] (* G. C. Greubel, Oct 20 2022 *)

def A001850(n): return gen_legendre_P(n,0,3)
def A035029(n): return ((n+1)*(n+3)*A001850(n+3) - (6*n^2 +22*n +17)*A001850(n+2) + (n+2)*(5*n+8)*A001850(n+1))/(8*(n+1)*(n+2))
[A035029(n) for n in range(40)] # G. C. Greubel, Oct 20 2022

a(n) = (1/4)*(A002002(n+2) - A002002(n+1)).
G.f.: (1-z)^2/(8*z^2*sqrt(1-6*z+z^2))-(1+z)/(8*z^2). - Emeric Deutsch, Dec 28 2003
a(n) = T(n+1, n+2), array T as in A049600.
Form an array with the m(n,1)=1 and m(1,n) = n*(n+1)/2 for n=1,2,3... The interior terms m(i,j) = m(i,j-1) + m(i-1,j-1) + m(i-1,j). The terms in the main diagonal = {a(n)}. - J. M. Bergot, Dec 01 2012
D-finite with recurrence: (n+2)*a(n) + (7*n+8)*a(n-1) - (7*n-8)*a(n-2) + (n-2)*a(n-3). - R. J. Mathar, Dec 03 2012
a(n) ~ (3+2*sqrt(2))^(n+3/2) / (2^(9/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 12 2014
a(n) = Jypergeometric2F1([-n, n+3]; [1]; -1), which satisfies the recurrence. - Benedict W. J. Irwin, Oct 14 2016

A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)binomial(n-1,k)binomial(n+k,k)*binomial(n-1+k,k).

Original entry on oeis.org

0, 1, 13, 253, 5741, 142001, 3713305, 100961365, 2825369965, 80843126905, 2354354542013, 69555586698533, 2079506849058809, 62797822901091409, 1912714261589863633, 58691012753161717253, 1812592783819686728045, 56298976785759622597385, 1757493098495181029912485

Offset: 0

Peter Bala, Apr 06 2022

Compare with A005259(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2.
Main superdiagonal (or main subdiagonal) of the symmetric square array A143007.
This is the sequence (A(n,n-1,n,n-1)) in the notation of Straub 2014, where it is proved that the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.
More generally, for nonnegative integers A, B and C with A >= 2 let S(n;A,B,C) = Sum_{k = 1..n} binomial(n,k)^A * binomial(n+k-1,k-1)^B * binomial(2*k,n)^C. Then we conjecture the sequence (S(n;A,B,C))n>=1 satisfies the same supercongruences. This is the case A = B = 2, C = 0. Compare with Osburn et al., Theorem 1.2.

			Examples of supercongruences:
a(11) - a(1) = 69555586698533 - 1 = (2^2)*(11^3)*1321*2521*3923 == 0 (mod 11^3).
a(2*7) - a(2) = 1912714261589863633 - 13 = (2^2)*(3^2)*5*(7^3)*776221* 39911503 == 0 (mod 7^3).

Robert Osburn, Brundaban Sahu and Armin Straub, Supercongruences for sporadic sequences, arXiv:1312.2195 [math.NT], 2013-2014.
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.
Eric Weisstein's World of Mathematics, Strehl identities

Cf. A002002, A005259, A108628, A143007, A177316.

seq(add(binomial(n,k)^2*binomial(n+k-1,k-1)^2, k = 1..n), n = 0..20);

a[n_] := n^2 HypergeometricPFQ[{1 - n, 1 - n, 1 + n, 1 + n}, {1, 2, 2}, 1];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Apr 17 2022 *)

a(n) = sum(k=1, n, binomial(n,k)^2*binomial(n+k-1,k-1)^2); \\ Michel Marcus, Apr 19 2022

def A352653(n):
    if n == 0: return 0
    m, g = 1, 0
    for k in range(n+1):
        g += m*n**2//(n+k)**2
        m *= ((n+k+1)*(n-k))**2
        m //= (k+1)**4
    return g>>1 # Chai Wah Wu, Oct 03 2022

The sequence can be extended to negative values of n:
a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+k,k)*binomial(-n-1,k)*binomial(-n-1+k,k) = a(n), since binomial(-m,k) = (-1)^k*binomial(m+k-1,k) for nonnegative k.
a(n) = A177316(n)/2 for n >= 1.
a(n) = Sum_{k = 1..n} binomial(n,k)^2*binomial(n+k-1,k-1)^2.
a(n) = (1/2)*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k-1,k)^2 for n >= 1.
a(n) = hypergeom([n, n+1, -n, -n+1], [1, 1, 1], 1) for n >= 1.
a(n) = (1/2)*hypergeom([n, n, -n, -n], [1, 1, 1], 1) for n >= 1.
a(n) = (1/4)*binomial(2*n,n)^2*hypergeom([-n+1, -n+1, -n, -n], [1, -2*n+1, -2*n+1], 1) for n >= 1
a(n) = [x^n*y^(n-1)*z^n*t^(n-1)] 1/((1 - x - y)(1 - z - t) - x*y*z*t).
a(n) = [x^n] 1/(1 - x) * P(n-1,(1 + x)/(1 - x))^2 = [x^(n-1)] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2 for n >= 1, where P(n,x) denotes the n-th Legendre polynomial. Cf. A005259(n) = [x^n] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2.
(n + 1)^3*(2*n - 1)(3*n^2 - 3*n + 1)*a(n+1) = 2*(102*n^6 - 68*n^4 + 21*n^2 - 3)*a(n) - (n - 1)^3*(2*n + 1)*(3*n^2 + 3*n + 1)*a(n-1) with a(0) = 0 and a(1) = 1.
The Gauss congruences a(n*p^r) == a(n*p^(r-1)) (mod p^r) hold for all primes p and all positive integers n and r. It follows that the expansion of exp( Sum_{n >= 1} a(n)/n*x^n ) = 1 + x + 7*x^2 + 91*x^3 + 1544*x^4 + 30448*x^5 + 661506*x^6 + 15377010*x^7 + ... has integer coefficients.
a(n) = n^2*hypergeom([1 - n, 1 - n, 1 + n, 1 + n], [1, 2, 2], 1). - Peter Luschny, Apr 17 2022
From Peter Bala, Mar 18 2023: (Start)
a(n) = Sum_{k = 0..n-1} (n-k)/(n+k) * binomial(n,k)^2 * binomial(n+k,k)^2.
a(n) = 1/6*(A005259(n) + A005259(n-1)). (End)
a(n) ~ (1 + sqrt(2))^(4*n) / (2^(9/4) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 29 2023
a(n) = (1/3)*Sum_{k = 0..n} binomial(n,k)*binomial(n+k-1,k)*f(k) for n >= 1, where f(n) = A000172(n) denotes the n-th Franel number. Cf. A005259(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*f(k) (the second Strehl identity). - Peter Bala, Jun 26 2023
a(n) = Sum_{0 <= j <= k <= n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)* binomial(n+j-1, j)^2*binomial(n-1, k-j) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, k). Cf. Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k) = A005259(n). - Peter Bala, Sep 22 2024

A378612 a(n) = Sum_{k=0..n} binomial(3n+k-1,k) binomial(n-1,n-k).

Original entry on oeis.org

1, 3, 27, 264, 2703, 28443, 304740, 3306852, 36225519, 399755001, 4437142467, 49485052224, 554059164036, 6224177431332, 70120015345512, 791898021185484, 8962485528377583, 101626868754849381, 1154295872365035537, 13130360954151723480, 149562006735075309783

Offset: 0

Seiichi Manyama, Dec 01 2024

nonn

Cf. A002002, A378611, A378613.

Cf. A369012.

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

a(n) = [x^n] 1/(1 - x/(1 - x))^(3*n).

a(n) = (1/4)^n * [x^(3*n)] 3/(1 - x/(1 - x))^n for n > 0.

A104550 Number of horizontal segments in all Schroeder paths of length 2n (a horizontal segment is a maximal string of horizontal steps).

Original entry on oeis.org

1, 4, 20, 104, 552, 2972, 16172, 88720, 489872, 2719028, 15157188, 84799992, 475894200, 2677788492, 15102309468, 85347160608, 483183316512, 2739851422820, 15558315261812, 88462135512712, 503569008273992, 2869602773253884, 16368396446913420, 93449566652932784, 533954950648248752

Offset: 1

Emeric Deutsch, Mar 14 2005

nonn

A Schroeder path is a lattice path starting from (0,0), ending at a point on the x-axis, consisting only of steps U=(1,1), D=(1,-1) and H=(2,0) and never going below the x-axis. Schroeder paths are counted by the large Schroeder numbers (A006318).

			a(2)=4 because we have (HH),(H)UD,UD(H),U(H)D,UDUD and UUDD; the 4 horizontal segments are shown between parentheses.

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A006318, A104549, A002002 (partial sums).

Cf. A035028.

G:=(1-z)*(1-z-sqrt(1-6*z+z^2))/2/sqrt(1-6*z+z^2): Gser:=series(G,z=0,28): seq(coeff(Gser,z^n),n=1..24);
a := n -> hypergeom([-n, n-1], [1], -1);
seq(round(evalf(a(n),36)),n=1..23); # Peter Luschny, Aug 02 2014

Rest[CoefficientList[Series[(1-x)*(1-x-Sqrt[1-6*x+x^2])/ (2*Sqrt[1 -6*x+x^2]), {x, 0, 20}], x]] (* Vaclav Kotesovec, Oct 17 2012 *)

a(n):=sum(binomial(n+1,k)*binomial(n+k-1,k),k,0,n+1); /* Vladimir Kruchinin, Jun 15 2020 */

x='x+O('x^66); Vec((1-x)*(1-x-sqrt(1-6*x+x^2))/(2*sqrt(1-6*x+x^2))) \\ Joerg Arndt, May 13 2013

G.f.: (1-x)*(1-x-sqrt(1-6*x+x^2))/(2*sqrt(1-6*x+x^2)).
a(n) = Jacobi_P(n+1,-1,-2,3). [Paul Barry, Sep 27 2009]
Recurrence: n*a(n) = (7*n-6)*a(n-1) - (7*n-22)*a(n-2) + (n-4)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(6*sqrt(2)-8)*(3+2*sqrt(2))^n/(2*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 17 2012
a(n) = Hyper2F1([-n, n-1], [1], -1). - Peter Luschny, Aug 02 2014
a(n) = Sum_{k=0..n+1} C(n+1,k)*C(n+k-1,k). - Vladimir Kruchinin, Jun 15 2020

A110127 Number of EE's crossing the line y = x (i.e., two consecutive E steps from the line y = x+1 to the line y = x-1) in all Delannoy paths of length n.

Original entry on oeis.org

0, 0, 1, 10, 75, 508, 3277, 20566, 126871, 773688, 4679769, 28136546, 168395235, 1004239156, 5971820709, 35429993390, 209800355631, 1240361694064, 7323260678065, 43187703202234, 254439363998587, 1497730375793004

Offset: 0

Emeric Deutsch, Jul 13 2005

nonn

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).

{A110127}[n+2] = conv({0, {A002002})[n]. - Tilman Neumann, Feb 05 2009

			a(2) = 1 because, among the 13 (=A001850(2)) Delannoy paths of length 2, only NEEN has an EE crossing the line y = x.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Robert A. Sulanke, Objects Counted by the Central Delannoy Numbers, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.

Cf. A001850, A110121.

R:=(1-z-sqrt(1-6*z+z^2))/2/z: G:=z^2*R^2/(1-6*z+z^2): Gser:=series(G,z=0,27): 0,seq(coeff(Gser,z^n),n=1..24);

CoefficientList[Series[x^2*((1-x-Sqrt[1-6*x+x^2])/2/x)^2/(1-6*x+x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 24 2012 *)

a(n) = Sum_{k=0..floor(n/2)} k*A110121(n,k).
G.f.: z^2*R^2/(1-6z+z^2), where R = 1+zR+zR^2 = [1-z-sqrt(1-6z+z^2)]/(2z) is the g.f. of the large Schroeder numbers (A006318).
Recurrence: n*(2*n-5)*a(n) = 6*(4*n^2 - 13*n + 8)*a(n-1) - 4*(19*n^2 - 76*n + 75)*a(n-2) + 6*(4*n^2 - 19*n + 20)*a(n-3) - (n-4)*(2*n-3)*a(n-4). - Vaclav Kotesovec, Oct 24 2012
a(n) ~ 1/8*sqrt(2)*(3+2*sqrt(2))^n. - Vaclav Kotesovec, Oct 24 2012

A259554 a(n) = Sum_{i=0..n} (2^(i)(-1)^(i+n)C(n,i)C(2n+i-1,n-1)).

Original entry on oeis.org

1, 7, 52, 403, 3206, 25954, 212738, 1760035, 14666470, 122920642, 1035046816, 8749594462, 74207078908, 631140253072, 5381022869822, 45975731083555, 393556869530630, 3374504760608026, 28977403637496104, 249167023897718138

Offset: 1

Vladimir Kruchinin, Jun 30 2015

G. C. Greubel, Table of n, a(n) for n = 1..1000
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A002002, A003168.

a := n -> n*hypergeom([2*n+1, -n+1], [2], -1):
seq(simplify(a(n)), n=1..9); # Peter Luschny, Oct 07 2016

Table[Sum[2^i * (-1)^(i+n) * Binomial[n, i] * Binomial[2*n+i-1, n-1], {i, 0, n}], {n,1,20}] (* Vaclav Kotesovec, Jul 01 2015 *)

a(n):=sum(2^(i)*(-1)^(i+n)*binomial(n,i)*binomial(2*n+i-1,n-1),i,0,n);

a(n) = sum(i=0, n, 2^i*(-1)^(i+n)*binomial(n, i)*binomial(2*n+i-1, n-1)); \\ Michel Marcus, Jul 02 2015

G.f.: A(x) = x*B(x)'/B(x), where B(x) is g.f. of A003168.
Recurrence: 4*n*(2*n-1)*(17*n^2 - 51*n + 38)*a(n) = (1207*n^4 - 4828*n^3 + 6659*n^2 - 3662*n + 672)*a(n-1) - 2*(n-2)*(2*n-3)*(17*n^2 - 17*n + 4)*a(n-2). - Vaclav Kotesovec, Jul 01 2015
a(n) ~ (71+17*sqrt(17))^n / (17^(1/4) * sqrt(Pi*n) * 2^(4*n+1)). - Vaclav Kotesovec, Jul 01 2015
a(n) = (1/2)*Sum_{k = 0..n} binomial(n-1,n-k)*binomial(2*n+k-1,k). - Vladimir Kruchinin, Oct 07 2016
a(n) = n*hypergeom([2*n+1, -n+1], [2], -1). - Peter Luschny, Oct 07 2016
From Peter Bala, Nov 08 2022: (Start)
a(n) = (1/2)*[x^n] ( (1 - x)/(1 - 2*x) )^(2*n). Cf. A002002(n) = [x^n] ( (1 - x)/(1 - 2*x) )^n.
a(n) = (1/2)*Sum_{k = 0..n} (-1)^(n-k)*2^k*binomial(2*n,n-k)*binomial(2*n+k-1,k).
a(n) = (1/2)*(-1)^n*binomial(2*n,n)*hypergeom( [-n, 2*n], [n+1], 2).
The Gauss congruences hold: a(n*p^r) == a(n^p^(r-1)) (mod p^r) for all primes p >= 3 and all positive integers n and r. (End)

A346540 Number A(n,k) of walks on square lattice from (n,k) to (0,0) using steps that decrease the Euclidean distance to the origin and increase the Euclidean distance to (n,k) and that change each coordinate by at most 1; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 3, 3, 3, 7, 5, 5, 7, 19, 24, 13, 24, 19, 51, 81, 25, 25, 81, 51, 141, 298, 173, 63, 173, 298, 141, 393, 1070, 739, 129, 129, 739, 1070, 393, 1107, 3868, 3423, 1210, 321, 1210, 3423, 3868, 1107, 3139, 13960, 15363, 6273, 681, 681, 6273, 15363, 13960, 3139

Offset: 0

Alois P. Heinz, Sep 16 2021

Lattice points may have negative coordinates, and different walks may differ in length. All walks are self-avoiding.

			Square array A(n,k) begins:
    1,    1,     3,     7,    19,    51,   141,   393, ...
    1,    3,     5,    24,    81,   298,  1070,  3868, ...
    3,    5,    13,    25,   173,   739,  3423, 15363, ...
    7,   24,    25,    63,   129,  1210,  6273, 34318, ...
   19,   81,   173,   129,   321,   681,  8371, 51727, ...
   51,  298,   739,  1210,   681,  1683,  3653, 57644, ...
  141, 1070,  3423,  6273,  8371,  3653,  8989, 19825, ...
  393, 3868, 15363, 34318, 51727, 57644, 19825, 48639, ...
  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
Alois P. Heinz, Animation of A(2,4) = 173 walks
Wikipedia, Counting lattice paths
Wikipedia, Self-avoiding walk

Columns (or rows) k=0-1 give: A002426, A347947.
Main diagonal gives A001850.
A(2n,n) gives A346541.
A(n-1,n) gives A002002 for n >= 1.

s:= proc(n) option remember;
     `if`(n=0, [[]], map(x-> seq([x[], i], i=-1..1), s(n-1)))
    end:
b:= proc(l, v) option remember; (n-> `if`(l=[0$n], 1, add((h-> `if`(
      add(i^2, i=h)add(i^2, i=v-l)
      , b(h, v), 0))(l+x), x=s(n))))(nops(l))
    end:
A:= (n, k)-> b(sort([n, k])$2):
seq(seq(A(n, d-n), n=0..d), d=0..12);

s[n_] := s[n] = If[n == 0, {{}}, Sequence @@
     Table[Append[#, i], {i, -1, 1}]& /@ s[n-1]];
b[l_, v_] := b[l, v] = With[{n = Length[l]},
     If[l == Table[0, {n}], 1, Sum[With[{h = l+x},
     If[h.h(v-l).(v-l), b[h, v], 0]], {x, s[n]}]]];
A[n_, k_] := b[Sort[{n, k}], Sort[{n, k}]];
Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Nov 04 2021, after Alois P. Heinz *)

A(n,k) = A(k,n).

A378567 a(n) = Sum_{k=0..n} binomial(n+k-1,k) * binomial(n+3*k-1,n-k).

Original entry on oeis.org

1, 1, 11, 88, 715, 5951, 50288, 429696, 3702987, 32125390, 280211701, 2454992618, 21588647392, 190444368401, 1684556756320, 14935618142768, 132695019071499, 1181070210132582, 10529299131757754, 94005323670592130, 840373149466892965, 7521508912742542806

Offset: 0

Seiichi Manyama, Dec 01 2024

nonn

Cf. A002002, A378565, A378566.

Cf. A055991, A367234.

a[n_]:=SeriesCoefficient[ 1/(1 - x/(1 - x)^4)^n,{x,0,n}]; Array[a,22,0] (* Stefano Spezia, Dec 01 2024 *)

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

a(n) = [x^n] 1/(1 - x/(1 - x)^4)^n.

A104553 Sum of trapezoid weights of all Schroeder paths of length 2n.

Original entry on oeis.org

1, 7, 38, 198, 1039, 5533, 29852, 162716, 893997, 4942723, 27466082, 153264066, 858230875, 4820155001, 27141345912, 153168964216, 866086326425, 4905744855359, 27830459812830, 158102366711550, 899290473825511, 5120997554408597, 29191620055374228, 166560724629655188

Offset: 1

Emeric Deutsch, Mar 14 2005

nonn

A Schroeder path is a lattice path starting from (0,0), ending at a point on the x-axis, consisting only of steps U=(1,1), D=(1,-1) and H=(2,0) and never going below the x-axis. Schroeder paths are counted by the large Schroeder numbers (A006318). A trapezoid in a Schroeder path is a factor of the form U^i H^j D^i (i>=1, j>=0), i being the height of the trapezoid. A trapezoid in a Schroeder path w is maximal if, as a factor in w, it is not immediately preceded by a U and immediately followed by a D. The trapezoid weight of a Schroeder path is the sum of the heights of its maximal trapezoids. For example, in the Schroeder path w=UH(UHD)D(UUDD) we have two trapezoids (shown between parentheses) of heights 1 and 2, respectively. The trapezoid weight of w is 1+2=3. This concept is an analogous to the concept of pyramid weight in a Dyck path (see the Denise-Simion paper). Partial sums of A047665 which, in turn, are the partial sums of A002002.

			a(2) = 7 because the six Schroeder paths of length 4, namely HH,(UD)H,H(UD),(UHD), (UD)(UD) and (UUDD), have trapezoid weights 0,1,1,1,2 and 2, respectively; the maximal trapezoids are shown between parentheses.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
A. Denise and R. Simion, Two combinatorial statistics on Dyck paths, Discrete Math., 137, 1995, 155-176.

Cf. A006318, A047665, A002002, A104552.

G:=(1-z-sqrt(1-6*z+z^2))/2/(1-z)^2/sqrt(1-6*z+z^2):Gser:=series(G,z=0,28): seq(coeff(Gser,z^n),n=1..25);

CoefficientList[Series[(1 - x - Sqrt[1 - 6 x + x^2]) / x /(2 (1 - x)^2 Sqrt[1 - 6 x + x^2]), {x, 0, 30}], x] (* Harvey P. Dale, May 26 2011 *)

x='x+O('x^66); Vec((1-x-sqrt(1-6*x+x^2))/(2*(1-x)^2*sqrt(1-6*x+x^2))) \\ Joerg Arndt, May 13 2013

G.f.: (1-x-sqrt(1-6*x+x^2))/(2*(1-x)^2*sqrt(1-6*x+x^2)).
Recurrence: n*(2*n-3)*a(n) = 2*(8*n^2 - 15*n + 5)*a(n-1) - 2*(14*n^2 - 28*n + 11)*a(n-2) + 2*(8*n^2 - 17*n + 7)*a(n-3) - (n-2)*(2*n-1)*a(n-4). - Vaclav Kotesovec, Oct 24 2012
a(n) ~ sqrt(48+34*sqrt(2))*(3+2*sqrt(2))^n/(16*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 24 2012

Typo in Mma program fixed by Vincenzo Librandi, May 13 2013

A180364 a(n) = sum_{k=0..n} C(n,k)C(n+k,k)(2k+1)^2, where C(m,k) denotes the binomial coefficient m!/(k!(m-k)!).

Original entry on oeis.org

1, 19, 205, 1839, 14961, 114483, 839917, 5975455, 41524897, 283272723, 1903686093, 12636115407, 83007985425, 540484102707, 3492471392493, 22418010385983, 143062290575937, 908253002030355, 5739641232682957, 36121371405797743, 226475167518421681

Offset: 0

Zhi-Wei Sun, Aug 29 2014

nonn

For any n > 0, we have a(0)+ ... + a(n-1) = n*sum_{k=0}^{n-1} (2*k+1)*C(n-1,k)*C(n+k,k) = n^2*A002002(n). The first equality can be easily deduced, and the second equality holds since both sides satisfy the same recurrence by the Zeilberger algorithm.

Conjecture: The sequence a(n+1)/a(n) (n = 0,1,...) is strictly decreasing to the limit 3+2*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,..) is strictly increasing to the limit 1.

			a(1) = 19 since sum_{k=0,1} C(1,k)*C(1+k,k)*(2k+1)^2 = 1 + 2*3^2 = 19.

Zhi-Wei Sun, Table of n, a(n) for n = 0..200
Zhi-Wei Sun, Two kinds of numbers and their arithmetic properties, arXiv:1408.5381, 2014.

Cf. A002002, A246460, A246542, A246543.

a[n_] := Sum[Binomial[n, k] Binomial[n + k, k] (2 k + 1)^2, {k, 0, n}]
Table[a[n], {n, 0, 20}]

Recurrence (obtained via the Zeilberger algorithm):
-(n+1)*(2*n^2+10*n+11)*a(n) + (2*n+3)*(6*n^2+18*n-7)*a(n+1) - (n+2)*(2*n^2+2*n-1)*a(n+2) = 0.
a(n) ~ sqrt(8+6*sqrt(2)) * (3+2*sqrt(2))^n * n^(3/2) / (2*sqrt(Pi)). - Vaclav Kotesovec, Sep 02 2014

cp's OEIS Frontend

A035029 a(n) = Sum_{k=0..n} (k+1) * Sum_{j=0..n} 2^j*binomial(n,j)*binomial(n-k,j).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)*binomial(n-1,k)*binomial(n+k,k)*binomial(n-1+k,k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A378612 a(n) = Sum_{k=0..n} binomial(3*n+k-1,k) * binomial(n-1,n-k).

Views

Author

Keywords

Crossrefs

Programs

PARI

Formula

A104550 Number of horizontal segments in all Schroeder paths of length 2n (a horizontal segment is a maximal string of horizontal steps).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

Formula

A110127 Number of EE's crossing the line y = x (i.e., two consecutive E steps from the line y = x+1 to the line y = x-1) in all Delannoy paths of length n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A259554 a(n) = Sum_{i=0..n} (2^(i)*(-1)^(i+n)*C(n,i)*C(2*n+i-1,n-1)).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

Formula

A346540 Number A(n,k) of walks on square lattice from (n,k) to (0,0) using steps that decrease the Euclidean distance to the origin and increase the Euclidean distance to (n,k) and that change each coordinate by at most 1; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

A035029 a(n) = Sum_{k=0..n} (k+1) * Sum_{j=0..n} 2^jbinomial(n,j)binomial(n-k,j).

A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)binomial(n-1,k)binomial(n+k,k)*binomial(n-1+k,k).

A378612 a(n) = Sum_{k=0..n} binomial(3n+k-1,k) binomial(n-1,n-k).

A259554 a(n) = Sum_{i=0..n} (2^(i)(-1)^(i+n)C(n,i)C(2n+i-1,n-1)).

A180364 a(n) = sum_{k=0..n} C(n,k)C(n+k,k)(2k+1)^2, where C(m,k) denotes the binomial coefficient m!/(k!(m-k)!).