A027307 -id:A027307 - cp's OEIS frontend

A108441 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and having k U=(1,2) steps among the steps leading to the first d step.

1, 1, 1, 3, 6, 1, 15, 39, 11, 1, 97, 284, 100, 16, 1, 721, 2249, 888, 186, 21, 1, 5827, 18890, 7977, 1952, 297, 26, 1, 49759, 165519, 72991, 19731, 3601, 433, 31, 1, 441729, 1496696, 680096, 196864, 40586, 5960, 594, 36, 1, 4035937, 13865297, 6439656

Offset: 0

Emeric Deutsch, Jun 08 2005

			T(2,1)=6 because we have uUddd, Uddud, UddUdd, Ududd, UdUddd and Uuddd.
Triangle begins:
1;
1,1;
3,6,1;
15,39,11,1;
97,284,100,16,1;

Alois P. Heinz, Rows n = 0..140, flattened
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Row sums yield A027307. Column 0 yields A108442.

A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: G:=1/(1-z*A-t*z*A^2): Gser:=simplify(series(G,z=0,12)): P[0]:=1: for n from 1 to 9 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 9 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields sequence in triangular form
# second Maple program:
b:= proc(x, y, t) option remember; expand(`if`(y<0 or y>x, 0,
      `if`(x=0, 1, b(x-1, y-1, false)+b(x-1, y+2, t)*
      `if`(t, z, 1)+b(x-2, y+1, t))))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..n))(b(3*n, 0, true)):
seq(T(n), n=0..10);  # Alois P. Heinz, Oct 06 2015

b[x_, y_, t_] := b[x, y, t] = Expand[If[y<0 || y>x, 0, If[x==0, 1, b[x-1, y - 1, False] + b[x-1, y+2, t]*If[t, z, 1] + b[x-2, y+1, t]]]]; T[n_] := Function[p, Table[Coefficient[p, z, i], {i, 0, n}]][ b[3*n, 0, True]]; Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jan 29 2016, after Alois P. Heinz *)

G.f.: G(t, z)=1/(1-zA-tzA^2)-1, where A=1+zA^2+zA^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).

A108442 Number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and having only u steps among the steps leading to the first d step.

Original entry on oeis.org

1, 1, 3, 15, 97, 721, 5827, 49759, 441729, 4035937, 37702723, 358474735, 3457592161, 33748593841, 332730216579, 3308635650495, 33145196426753, 334193815799233, 3388807714823043, 34537227997917391, 353578650475659617, 3634495706671023505, 37496621681376849219, 388135791657414454815

Offset: 0

Emeric Deutsch, Jun 08 2005

nonn

			a(2)=3 because we have udud, udUdd and uudd.

G. C. Greubel, Table of n, a(n) for n = 0..955
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Column 0 of A108441.

Cf. A027307, A108441.

A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: gser:=series(1/(1-z*A),z=0,30): 1,seq(coeff(gser,z^n),n=1..25);

Flatten[{1,Table[Sum[k*Sum[Binomial[2*n-k, i]*Binomial[3*n-2*k-i-1, 2*n-k-1], {i, 0, n-k}]/(2*n-k), {k, 1, n}],{n,1,20}]}] (* Vaclav Kotesovec, Mar 17 2014, after Vladimir Kruchinin *)

a(n):=if n=0 then 1 else sum((k*sum(binomial(2*n-k,i)*binomial(3*n-2*k-i-1,2*n-k-1),i,0,n-k))/(2*n-k),k,1,n); /* Vladimir Kruchinin, Oct 23 2011 */

G.f.: 1/(1-z*A), where A = 1 + z*A^2 + z*A^3 = (2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
a(n) = Sum_{k=1..n} (k*(Sum_{i=0..n-k} binomial(2*n-k, i)*binomial(3*n-2*k-i-1, 2*n-k-1))/(2*n-k)), n > 0, a(0)=1. - Vladimir Kruchinin, Oct 23 2011
G.f. y(x) satisfies: (3+x)*y*(1-y) + (1+x^2)*y^3 = 1. - Vaclav Kotesovec, Mar 17 2014
a(n) ~ (11+5*sqrt(5))^n / (5^(5/4) * sqrt(Pi) * n^(3/2) * 2^(n+1)). - Vaclav Kotesovec, Mar 17 2014
D-finite with recurrence (2*n-1)*(n-1)*a(n) +6*(n^2-10*n+13)*a(n-1) +(-310*n^2+1869*n-2759)*a(n-2) +48*(-n+3)*a(n-3) +(-310*n^2+1851*n-2705)*a(n-4) +6*(-n^2+2*n+11)*a(n-5) +(n-5)*(2*n-11)*a(n-6)=0. - R. J. Mathar, Jul 26 2022

A108448 Number of peaks of the form ud in all paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1).

Original entry on oeis.org

1, 7, 61, 575, 5641, 56695, 579125, 5984767, 62390545, 654862247, 6911195501, 73265596607, 779594526361, 8321683861015, 89070157349221, 955598531432447, 10273391096237089, 110647714508386375, 1193641560393864605

Offset: 1

Emeric Deutsch, Jun 10 2005

a(n) = Sum_{k=1..n} k*A108446(n,k). Example: a(3) = 1*32 + 2*13 + 3*1 = 61.

			a(2) = 7 because in the ten paths (ud)(ud), (ud)Udd, u(ud)d, uUddd, Udd(ud), UddUdd, Ud(ud)d, UdUddd, U(ud)dd and UUdddd (see A027307) we have 7 ud's (shown between parentheses).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Emeric Deutsch, Problem 10658, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A108446, A108426, A108427.

A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: G:=z*A/(1-2*z*A-3*z*A^2): Gser:=series(G,z=0,25): seq(coeff(Gser,z^n),n=1..23);

RecurrenceTable[{(n-1)*(2*n-1)*a[n]==(18*n^2-26*n+1)*a[n-1] +(46*n^2-225*n+276)*a[n-2]+2*(n-3)*(2*n-5)*a[n-3], a[1]==1, a[2]==7, a[3]==61},a,{n,20}] (* Vaclav Kotesovec, Oct 18 2012 *)

G.f.: z*A/(1-2*z*A-3*z*A^2), where A=1+z*A^2+z*A^3 or, equivalently, A=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
Recurrence: (n-1)*(2*n-1)*a(n) = (18*n^2-26*n+1)*a(n-1) + (46*n^2-225*n+276)*a(n-2) + 2*(n-3)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ sqrt(70*sqrt(5)-150)*((11+5*sqrt(5))/2)^n/(20*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 18 2012. Equivalently, a(n) ~ phi^(5*n - 2) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
a(n) = Sum_{k=1..n} k*C(n-1,k-1)*C(2*n+k-1,n)/(n+k). - Vladimir Kruchinin, Mar 03 2014
a(n) = P(n-1,n,0,3), where P is the Jacobi Polynomial. - Richard Turk, Jun 27 2018
From Peter Bala, Feb 08 2024: (Start)
a(n) = Sum_{k = 0..n-1} binomial(2*n-1, k)*binomial(n-1, k)*2^k.
(n - 1)*(2*n - 1)*(10*n - 17)*a(n) = (220*n^3 - 814*n^2 + 950*n - 341)*a(n-1) + (n - 2)*(2*n - 3)*(10*n - 7)*a(n-2) with a(1) = 1 and a(2) = 7.. (End)

A108450 Number of pyramids in all paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) (a pyramid is a sequence u^pd^p or U^pd^(2p) for some positive integer p, starting at the x-axis).

Original entry on oeis.org

2, 10, 58, 402, 3122, 26010, 227050, 2049186, 18964194, 178976426, 1715905050, 16665027378, 163611970066, 1621103006010, 16189480081354, 162791835045698, 1646810150270914, 16748008972020554, 171135004105459194

Offset: 1

Emeric Deutsch, Jun 11 2005

nonn

A108450(n)=sum(k*A108445(k),k=1..n) (for example, A108450(3)=1*18+2*8+3*8=58). A108450(n)=2*A108453(n). A108450 =2*partial sums of A032349.

			a(2)=10 because in the A027307(2)=10 paths we have altogether 10 pyramids (shown between parentheses): (ud)(ud), (ud)(Udd), (uudd), uUddd, (Udd)(ud), (Udd)(Udd), Ududd, UdUddd, Uuddd, (UUdddd).

Vaclav Kotesovec, Table of n, a(n) for n = 1..140
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A108445, A108453, A032349.

A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: g:=2*z*A^2/(1-z): gser:=series(g,z=0,25): seq(coeff(gser,z^n),n=1..22);

Table[2 Sum[Sum[Binomial[2 k + 2, k - i] Binomial[2 k + i + 1, 2 k + 1], {i, 0, k}]/(k + 1), {k, 0, n - 1}], {n, 19}] (* Michael De Vlieger, Feb 29 2016 *)

a(n):=2*sum(sum(binomial(2*k+2,k-i)*binomial(2*k+i+1,2*k+1),i,0,k)/(k+1),k,0,n-1);
/* Vladimir Kruchinin, Feb 29 2016 */

{a(n)=local(y=2*x); for(i=1, n, y=(2*x*(2+y-x*y)^2)/((1-x)*(2-y+x*y)^2) + (O(x^n))^3); polcoeff(y, n)}
for(n=1, 20, print1(a(n), ", ")) \\ Vaclav Kotesovec, Mar 17 2014

G.f.: 2*z*A^2/(1-z), where A=1+z*A^2+z*A^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
G.f. y(x) satisfies: y = (2*x*(2+y-x*y)^2)/((1-x)*(2-y+x*y)^2). - Vaclav Kotesovec, Mar 17 2014
a(n) ~ (3*sqrt(5)-1) * ((11+5*sqrt(5))/2)^n /(11*5^(1/4)*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 17 2014
a(n) = 2*Sum_{k=0..n-1}(Sum_{i=0..k}(binomial(2*k+2,k-i)* binomial(2*k+i+1,2*k+1))/(k+1)). - Vladimir Kruchinin, Feb 29 2016
D-finite with recurrence n*(2*n-1)*a(n) +6*-(n-1)*(5*n-6)*a(n-1) +4*(23*n^2-97*n+111)*a(n-2) +2*(-29*n^2+142*n-174)*a(n-3) -3*(2*n-5)*(n-4)*a(n-4)=0. - R. J. Mathar, Jul 26 2022

A108453 Number of pyramids of the first kind in all paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) (a pyramid of the first kind is a sequence u^pd^p for some positive integer p, starting at the x-axis).

Original entry on oeis.org

1, 5, 29, 201, 1561, 13005, 113525, 1024593, 9482097, 89488213, 857952525, 8332513689, 81805985033, 810551503005, 8094740040677, 81395917522849, 823405075135457, 8374004486010277, 85567502052729597, 878066090712156521

Offset: 1

Emeric Deutsch, Jun 11 2005

nonn

A108453(n)=sum(k*A108451(k),k=1..n) (for example, A108453(3)=1*16+2*5+3*1=29). A108453(n)=(1/2)*A108450(n). A108453 = partial sums of A032349.

			a(2)=5 because in the A027307(2)=10 paths we have altogether 5 pyramids of the first kind (shown between parentheses): (ud)(ud), (ud)Udd, (uudd), uUddd, Udd(ud), UddUdd, Ududd, UdUddd, Uuddd, UUdddd.

Vaclav Kotesovec, Table of n, a(n) for n = 1..160
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A108450, A108451, A032349.

A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: g:=z*A^2/(1-z): gser:=series(g,z=0,25): seq(coeff(gser,z^n),n=1..22);

{a(n)=local(y=x); for(i=1, n, y=x*(1+y-x*y)^2/((1-x)*(1-y+x*y)^2) + (O(x^n))^3); polcoeff(y, n)}
for(n=1, 20, print1(a(n), ", ")) \\ Vaclav Kotesovec, Mar 17 2014

G.f.=zA^2/(1-z), where A=1+zA^2+zA^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
G.f. y(x) satisfies: x*(1+y-x*y)^2 = (1-x)*y*(1-y+x*y)^2. - Vaclav Kotesovec, Mar 17 2014
a(n) ~ sqrt(23*sqrt(5)-15) * (11+5*sqrt(5))^n / (11* sqrt(5*Pi) * n^(3/2) * 2^(n+1/2)). - Vaclav Kotesovec, Mar 17 2014
D-finite with recurrence n*(2*n-1)*a(n) -6*(n-1)*(5*n-6)*a(n-1) +4*(23*n^2-97*n+111)*a(n-2) +2*(-29*n^2+142*n-174)*a(n-3) -3*(2*n-5)*(n-4)*a(n-4)=0. - R. J. Mathar, Jul 26 2022

A361638 Expansion of g.f. A(x) satisfying A(x) = 1 + x * A(x)^2 * (1 + A(x)^3).

Original entry on oeis.org

1, 2, 14, 142, 1690, 21994, 303126, 4348102, 64235570, 970695442, 14934154334, 233133082494, 3683546302538, 58794776161274, 946619511627622, 15355445768326710, 250717346336174690, 4117189670041072930, 67956239699290313646, 1126763233375565370990

Offset: 0

Seiichi Manyama, Jul 13 2023

Cf. A027307, A219534.

a(n) = sum(k=0, n, binomial(n, k)*binomial(2*n+3*k+1, n)/(2*n+3*k+1));

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(2*n+3*k+1,n)/(2*n+3*k+1).

A108434 Number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have no hills of the form ud (a hill is either a ud or a Udd starting at the x-axis).

Original entry on oeis.org

1, 1, 7, 47, 361, 2977, 25775, 231103, 2127409, 19990241, 190957559, 1848911279, 18104425561, 178975914433, 1783843502047, 17906040994559, 180858717257185, 1836792828317761, 18745545101801063, 192145823547338927

Offset: 0

Emeric Deutsch, Jun 03 2005

nonn

Column 0 of A108433.

The radius of convergence of g.f. y(x) is r = (5*sqrt(5)-11)/2, with y(r) = (2+sqrt(5))/3. - Vaclav Kotesovec, Mar 17 2014

			a(2)=7 because we have uudd, uUddd, UddUdd, Ududd, UdUddd, Uuddd and UUdddd.

Vaclav Kotesovec, Table of n, a(n) for n = 0..200
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A108431, A108432, A108433.

g:=1/(1+z-z*A-z*A^2): A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3:gser:=series(g,z=0,27): 1,seq(coeff(gser,z^n),n=1..24);

{a(n)=local(y=1+x); for(i=1, n, y=-(-1 + 3*x*y - 3*x*(1+x)*y^2 + x*(-1+2*x+x^2)*y^3) + (O(x^n))^3); polcoeff(y, n)}
for(n=0, 20, print1(a(n), ", ")) \\ Vaclav Kotesovec, Mar 17 2014

G.f. = 1/(1+z-zA-zA^2), where A=1+zA^2+zA^3 or, equivalently, A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
G.f. y(x) satisfies: -1+y + 3*x*y - 3*x*(1+x)*y^2 + x*(-1+2*x+x^2)*y^3 = 0. - Vaclav Kotesovec, Mar 17 2014
a(n) ~ (11+5*sqrt(5))^n * sqrt(123 + 55*sqrt(5)) / (9 * 5^(1/4) * sqrt(Pi) * n^(3/2) * 2^(n+3/2)). - Vaclav Kotesovec, Mar 17 2014
a(n) ~ phi^(5*n + 5) / (18 * 5^(1/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Sep 23 2017
D-finite with recurrence n*(2*n+1)*(n-2)*a(n) +2*(-13*n^3+36*n^2-29*n+9)*a(n-1) +4*(n-1)*(10*n^2-20*n+9)*a(n-2) +2*(13*n^3-42*n^2+41*n-9)*a(n-3) +n*(n-2)*(2*n-5)*a(n-4)=0. - R. J. Mathar, Jul 26 2022

A108435 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have k returns to the x-axis.

Original entry on oeis.org

2, 6, 4, 34, 24, 8, 238, 172, 72, 16, 1858, 1360, 624, 192, 32, 15510, 11444, 5520, 1952, 480, 64, 135490, 100520, 50040, 19136, 5600, 1152, 128, 1223134, 911068, 463512, 186416, 60320, 15168, 2688, 256, 11320066, 8457504, 4371808, 1821312, 629440, 178176, 39424, 6144, 512

Offset: 1

Emeric Deutsch, Jun 04 2005

Number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have k steps up to the first peak. Example: T(2,2)=4 because we have uudd, uUddd, Uuddd and UUdddd. Row sums yield A027307. T(n,1)=A108424(n). T(n,n)=2^n.

			T(2,2)=4 because u(d)u(d), u(d)Ud(d), Ud(d)u(d) and Ud(d)Ud(d) (the steps d that return to the x-axis are shown between parentheses).
Triangle begins:
  2;
  6,4;
  34,24,8;
  238,172,72,16;
  1858,1360,624,192,32;
  ...

Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A108424, A108436.

Maple
```
T:=proc(n,k) if k
				
```

T[n_, k_] := Which[k < n, (k/(n - k))*(3*2^k*Binomial[n - 1, k] + Sum[2^(n - 1 - j)*(5*n - 2*k + j + 1)*Binomial[n - 1, j]*Binomial[2*n - k - 1, n + j]/(n + j + 1), {j, 0, n - k - 2}]), k == n, 2^n, True, 0];
Table[T[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 11 2024, after Maple code. *)

T(n, k)=[k/(n-k)][3*2^k*binomial(n-1, k)+sum(2^(n-1-j)*(5n-2k+j+1)binomial(n-1, j)binomial(2n-k-1, n+j)/(n+j+1), j=0..n-k-2)] if kA027307).

Keyword tabf changed to tabl by Michel Marcus, Apr 09 2013

A108443 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and have k triple descents (i.e., ddd's).

Original entry on oeis.org

1, 2, 6, 3, 1, 21, 24, 15, 5, 1, 80, 150, 145, 84, 31, 7, 1, 322, 857, 1145, 949, 528, 202, 53, 9, 1, 1347, 4692, 8096, 8801, 6598, 3551, 1394, 398, 81, 11, 1, 5798, 25102, 53457, 72338, 68594, 47805, 25092, 10019, 3040, 692, 115, 13, 1, 25512, 132484, 337132

Offset: 0

Emeric Deutsch and Paul D. Hanna, Jun 10 2005

Row n has 2n-1 terms (n >= 1). Row sums yield A027307. Column 0 yields A106228.

			T(2,1) = 3 because we have uUddd, Uuddd and UdUddd.
Triangle begins:
    1;
    2;
    6,    3,    1;
   21,   24,   15,    5,    1;
   80,  150,  145,   84,   31,    7,    1;
  322,  857, 1145,  949,  528,  202,   53,    9,    1;

Alois P. Heinz, Rows n = 0..100, flattened
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A106228.

b:= proc(x, y, t) option remember; expand(`if`(y<0 or y>x, 0,
      `if`(x=0, 1, b(x-1, y-1, min(2, t+1))*`if`(t=2, z, 1)+
       b(x-1, y+2, 0)+b(x-2, y+1, 0))))
    end:
T:= n->(p->seq(coeff(p, z, i), i=0..degree(p)))(b(3*n, 0$2)):
seq(T(n), n=0..8);  # Alois P. Heinz, Oct 06 2015

b[x_, y_, t_] := b[x, y, t] = Expand[If[y < 0 || y > x, 0, If[x == 0, 1, b[x - 1, y - 1, Min[2, t + 1]]*If[t == 2, z, 1] + b[x - 1, y + 2, 0] + b[x - 2, y + 1, 0]]]]; T[n_] := Function[p, Table[Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][b[3*n, 0, 0]]; Table[T[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Dec 21 2016, after Alois P. Heinz *)

{T(n,k)=local(G=1+z*O(z^n)+t*O(t^k));for(k=1,n, G=1+z*(t+z-t*z)^2*G^3+z*(2-t)*(t+z-t*z)*G^2+2*z*(1-t)*G); polcoeff(polcoeff(G,n,z),k,t)}

G.f. G = G(t,z) satisfies G = 1 + z(t + z - tz)^2*G^3 + z(2-t)(t + z - tz)G^2 + 2z(1-t)G.

A108446 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and have k peaks of the form ud.

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 20, 32, 13, 1, 113, 223, 135, 26, 1, 688, 1620, 1300, 412, 45, 1, 4404, 12064, 12050, 5350, 1030, 71, 1, 29219, 91335, 109134, 62450, 17575, 2247, 105, 1, 199140, 699689, 973077, 682234, 254625, 49210, 4438, 148, 1, 1385904, 5407744

Offset: 0

Emeric Deutsch, Jun 10 2005

Row sums yield A027307. Column 0 yields A108447. T(n,n-1) = A008778(n-1) = n(n^2+6n-1)/6. Number of ud peaks in all paths from (0,0) to (3n,0) is given by A108448.

			T(2,1) = 5 because we have udUdd, uudd, Uddud, Ududd and Uuddd.
Triangle begins:
1;
1,1;
4,5,1;
20,32,13,1;
113,223,135,26,1;

Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A008778, A108447, A108448, A108425, A108426.

T:=proc(n,k) if n=0 and k=0 then 1 elif n=0 then 0 elif k=n then 1 elif k=n then 1 else (1/n)*binomial(n,k)*sum(binomial(n-k,j)*binomial(n+2*j,k+j-1),j=0..n-k) fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[0, 0] = 1; T[n_, k_] := (1/n) Binomial[n, k]*Sum[Binomial[n-k, j]* Binomial[n+2j, k+j-1], {j, 0, n-k}];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 19 2018 *)

T(n,k) = (1/n) binomial(n, k)*sum(binomial(n-k,j)*binomial(n+2j,k+j-1), j=0..n-k).

G.f.: G = G(t,z) satisfies G = 1+z(G-1+t)G+zG^3.

cp's OEIS Frontend

A108441 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and having k U=(1,2) steps among the steps leading to the first d step.

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A108442 Number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and having only u steps among the steps leading to the first d step.

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A108448 Number of peaks of the form ud in all paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1).

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A108450 Number of pyramids in all paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) (a pyramid is a sequence u^pd^p or U^pd^(2p) for some positive integer p, starting at the x-axis).

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A361638 Expansion of g.f. A(x) satisfying A(x) = 1 + x * A(x)^2 * (1 + A(x)^3).

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A108434 Number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have no hills of the form ud (a hill is either a ud or a Udd starting at the x-axis).

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