A110236 -id:A110236 - cp's OEIS frontend

A202411 a(n) = Sum_{k=floor(n/4)..R} C(k, mk - (-1)^n(R - k)) * C(k + 1, m(k + 2) - (-1)^n(R - k + 1)) where m = (n + 1) mod 2 and R = (n + m - 3)/2 for n > 0 and a(0) = 1.

1, 0, 1, 1, 1, 2, 3, 4, 7, 10, 16, 24, 39, 58, 95, 143, 233, 354, 577, 881, 1436, 2204, 3590, 5534, 9011, 13940, 22691, 35213, 57299, 89162, 145043, 226238, 367931, 575114, 935078, 1464382, 2380405, 3734150, 6068745, 9534594, 15492702, 24374230, 39598631

Offset: 0

Peter Luschny, Jan 14 2012

nonn

			Fibonacci meanders classified by maximal run length of 1s (see the link) lead to the triangle
   0,  1;
   1,  1,  0, 1;
   2,  1,  1, 1, 0, 1;
   4,  3,  2, 1, 1, 1, 0, 1;
  10,  7,  4, 3, 2, 1, 1, 1, 0, 1;
  24, 16, 10, 7, 4, 3, 2, 1, 1, 1, 0, 1.

Peter Luschny, Fibonacci meanders.

Cf. A110236, A203611.

A202411 := proc(n) local A, R, B, C, D, Z, H, J; if n = 0 then RETURN(1) fi;
H:=iquo(n,2); A:=iquo(H,2); R:=H-1; B:=A-R/2+1; C:=A+1; D:=A-R; J:=n mod 2; if J = 0 then Z:=`if`(H mod 2 = 1,(H+1)/2,H^2*(H+2)/16) else Z:=`if`(H mod 2 = 1,1, H*(H+2)/4) fi; Z*hypergeom([1,C,C+1,D,D-J],[B,B,B-1/2,B+1/2-J],1/16) end:
seq(simplify(A202411(i)),i=0..42);

A202411[0] = 1; A202411[n_] := Module[{A, R, B, C, D, Z, H, J}, H = Quotient[n, 2]; A = Quotient[H, 2]; R = H-1; B = A - R/2 + 1; C = A+1; D = A - R; J = Mod[n, 2]; If[J == 0, Z = If[Mod[H, 2] == 1, (H+1)/2, H^2*(H + 2)/16], Z = If[Mod[H, 2] == 1, 1, H*(H+2)/4]]; Z*HypergeometricPFQ[{1, C, C + 1, D, D - J}, {B, B, B - 1/2, B + 1/2 - J}, 1/16]]; Table[A202411[n], {n, 0, 42}]
(* Jean-François Alcover, Jan 27 2014, translated from Maple *)

For n > 0 let H = floor(n/2), A = floor(H/2), R = H - 1, B = A - R/2 + 1, C = A + 1, D = A - R, J = n mod 2 and Z = if(H mod 2 = 1, (H + 1)/2, H^2*(H + 2)/16) if J = 0 else Z = if(H mod 2 = 1, 1, H*(H + 2)/4); then:

a(n) = Z*Hypergeometric([1, C, C+1, D, D-J], [B, B, B-1/2, B+1/2-J], 1/16).

A203611 a(n) = Sum_{k=0..n} binomial(k-1,2k-1-n)binomial(k,2*k-n), with a(0) = 1.

Original entry on oeis.org

1, 1, 1, 3, 7, 16, 39, 95, 233, 577, 1436, 3590, 9011, 22691, 57299, 145043, 367931, 935078, 2380405, 6068745, 15492702, 39598631, 101323446, 259522398, 665332007, 1707137941, 4383662419, 11264675925, 28966161253, 74530441162, 191879611399, 494265165151

Offset: 0

Peter Luschny, Jan 14 2012

nonn

For the connection with Fibonacci meanders classified by maximal run length of 1s (see the link).

Apparently the number of grand Motzkin paths of length n+1 that avoid UU. - David Scambler, Jul 04 2013

Michael De Vlieger, Table of n, a(n) for n = 0..2397
Jean-Luc Baril, Sergey Kirgizov, Rémi Maréchal, and Vincent Vajnovszki, Grand Dyck paths with air pockets, arXiv:2211.04914 [math.CO], 2022.
Jean-Luc Baril and José L. Ramírez, Fibonacci and Catalan paths in a wall, 2023.
Peter Luschny, Fibonacci meanders.

Cf. A110236, bisection of A202411.

A203611:= func< n | n eq 0 select 1 else (&+[Binomial(k-1,2*k-n-1)*Binomial(k,2*k-n): k in [0..n]]) >;
[A203611(n): n in [0..40]]; // G. C. Greubel, Mar 12 2025

a := n -> hypergeom([-n/2, 1 - n/2, (1-n)/2, (1-n)/2], [1, -n, 1 - n], 16):
seq(simplify(a(n)), n = 0..31); # Peter Luschny, Mar 24 2023

a[n_] := Module[{a, r, b, c, d, z}, If[n == 0, Return[1]]; a = Quotient[n, 2]; r = n-1; b = a-r/2+1; c = a+1; d = a-r; z = If[Mod[n, 2] == 1, (n+1)/2, n^2*(n+2)/16]; z*HypergeometricPFQ[{1, c, c+1, d, d}, {b, b, b-1/2, b+1/2}, 1/16] ]; Table[a[n], {n, 0, 31}] (* Jean-François Alcover, Jun 27 2013, translated from Maple *)
Table[Sum[Binomial[k-1,2k-1-n]Binomial[k,2k-n],{k,0,n}],{n,0,40}] (* Harvey P. Dale, May 25 2014 *)

my(x='x+O('x^66)); Vec( 2*x/((1+x-x^2) * sqrt((x^2+x+1) * (x^2-3*x+1)) -x^4 +2*x^3 +x^2 +2*x -1) ) \\ Joerg Arndt, May 06 2013

def A203611(n): return 1 if n==0 else sum(binomial(k-1,2*k-n-1)*binomial(k,2*k-n) for k in range(n+1))
print([A203611(n) for n in range(41)]) # G. C. Greubel, Mar 12 2025

For n>0 let A=floor(n/2), R = n-1, B = A - R/2 + 1, C = A+1, D = A-R and Z = (n+1)/2 if n mod 2 = 1, otherwise Z = n^2*(n+2)/16. Then a(n) = Z*Hypergeometric5F4([1,C,C+1,D,D],[B,B,B-1/2,B+1/2],1/16).
G.f.: 2*x/((1+x-x^2)*sqrt((1+x+x^2)*(1-3*x+x^2)) - (1-2*x-x^2-2*x^3+x^4)). - Mark van Hoeij, May 06 2013
a(n) ~ phi^(2*n + 1) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 08 2019
a(n) = hypergeom([-n/2, 1 - n/2, (1-n)/2, (1-n)/2], [1, -n, 1 - n], 16). - Peter Luschny, Mar 24 2023
D-finite with recurrence n*a(n) -(n+1)*a(n-1) -2*(2*n-5)*a(n-2) -(n+3)*a(n-3) +3*(n-5)*a(n-5) -(n-6)*a(n-6) = 0. - R. J. Mathar, Nov 22 2024

A201631 a(n) is the number of Fibonacci meanders of length m*n and central angle 360/m degrees where m = 2.

Original entry on oeis.org

1, 3, 6, 13, 30, 70, 167, 405, 992, 2450, 6090, 15214, 38165, 96069, 242530, 613811, 1556856, 3956316, 10070871, 25674210, 65541142, 167517654, 428635032, 1097874434, 2814611701, 7221917871, 18544968768, 47655572191, 122544150258, 315313433594, 811792614547

Offset: 1

Peter Luschny, Jan 15 2012

nonn

Empirically the partial sums of A051291. - Sean A. Irvine, Jul 13 2022

The above conjecture was proved by Baril et al., which also give a formal definition of the Fibonacci meanders and describe a bijection with a certain class of peakless grand Motzkin paths of length n. - Peter Luschny, Mar 16 2023

			a(3) = 6 = card({100001, 100100, 110000, 111001, 111100, 111111}).

Jean-Luc Baril, Sergey Kirgizov, Rémi Maréchal, and Vincent Vajnovszki, Enumeration of Dyck paths with air pockets, arXiv:2202.06893 [cs.DM], 2022-2023.
Peter Luschny, Fibonacci meanders.

Cf. A110236, A110198, A202411, A203611, A051291, A358734.

Cf. A361574 (case m=3).

A201631 := n -> add(A202411(k),k=0..2*n-1): seq(A201631(i),i=1..9);
# Alternative, using the g.f. of Baril et al.:
S := (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R)*R):
R := (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2): ser := series(S, x, 33):
seq(coeff(ser, x, n), n = 1..31); # Peter Luschny, Mar 16 2023
# Using a recurrence:
a := proc(n) option remember; if n < 5 then return [0, 1, 3, 6, 13][n + 1] fi;
(n*(2*n - 1)*(2*n - 3)*(n - 5)*a(n - 5) - (n - 4)*(2*n - 1)^2*(3*n - 5)*a(n - 4) + (2*n - 5)*(n - 3)*(2*n^2 - 3*n + 2)*a(n - 3) - (2*n - 3)*(n - 2)*(2*n^2 - 3*n + 5)*a(n - 2) + (3*n - 4)*(2*n - 1)*(2*n - 5)*(n - 1)*a(n - 1))/(n*(2*n - 3)*(2*n - 5)*(n - 1)) end: seq(a(n), n = 1..31); # Peter Luschny, Mar 16 2023

a[n_] := Sum[A202411[k], {k, 0, 2 n - 1}];
Array[a, 31] (* Jean-François Alcover, Jun 29 2019 *)

a(n) = Sum_{k=0..2n-1} A202411(k).

a(n) = [x^n] (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R) * R), where R = (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2). (This is Theorem 21 in Baril et al.) - Peter Luschny, Mar 16 2023

A190164 Triangle read by rows: T(n,k) is the number of peakless Motzkin paths of length n having a total of k (1,0)-steps at levels 0,2,4,... .

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 2, 0, 0, 1, 1, 3, 3, 0, 0, 1, 2, 4, 6, 4, 0, 0, 1, 4, 8, 9, 10, 5, 0, 0, 1, 7, 18, 19, 16, 15, 6, 0, 0, 1, 12, 35, 48, 36, 25, 21, 7, 0, 0, 1, 22, 66, 102, 100, 60, 36, 28, 8, 0, 0, 1, 41, 132, 209, 229, 180, 92, 49, 36, 9, 0, 0, 1, 76, 266, 450, 504, 440, 294, 133, 64, 45, 10, 0, 0, 1

Offset: 0

Emeric Deutsch, May 06 2011

Sum of entries in row n is A004148(n) (the RNA secondary structure numbers).
T(n,0)=A190165(n).
Sum_{k>=0} k*T(n,k) = A190166(n).
The trivariate g.f. H(t,s,z), where t (s) marks (1,0)-steps at even (odd) levels and z marks length, satisfies the equation
z^2*(1-tz+z^2)*H^2 - (1-tz+z^2)*(1-sz+z^2)*H + 1-sz+z^2 = 0.

			T(5,2)=3 because we have h'h'uhd, h'uhdh', and uhdh'h', where u=(1,1), h=(1,0), d=(1,-1) (the even-level h-steps are marked).
Triangle starts:
  1;
  0, 1;
  0, 0, 1;
  1, 0, 0, 1;
  1, 2, 0, 0, 1;
  1, 3, 3, 0, 0, 1;

Cf. A004148, A190165, A190166, A110236, A190167.

eq := z^2*(1-t*z+z^2)*G^2-(1-z+z^2)*(1-t*z+z^2)*G+1-z+z^2 = 0: g := RootOf(eq, G): Gser := simplify(series(g, z = 0, 15)): for n from 0 to 13 do P[n] := sort(expand(coeff(Gser, z, n))) end do: for n from 0 to 12 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form

m = 13; G[_] = 0;
Do[G[z_] = -((z^2 G[z]^2 (-t z + z^2 + 1) + z^2 - z + 1)/((z^2 - z + 1)(t z - z^2 - 1))) + O[z]^m, {m}];
CoefficientList[#, t]& /@ CoefficientList[G[z], z] // Flatten (* Jean-François Alcover, Nov 15 2019 *)

G.f.: G = G(t,z) satisfies the equation z^2*(1-tz+z^2)*G^2 - (1-z+z^2)*(1-tz+z^2)*G + 1 - z + z^2 = 0.

A190166 Number of (1,0)-steps at levels 0,2,4,... in all peakless Motzkin paths of length n.

Original entry on oeis.org

0, 1, 2, 3, 6, 14, 34, 83, 202, 495, 1224, 3046, 7616, 19115, 48130, 121527, 307602, 780244, 1982834, 5047377, 12867438, 32847357, 83952780, 214806750, 550170300, 1410412561, 3618785462, 9292203549, 23877482490, 61397367692, 157972743178, 406693829059, 1047585820586, 2699811117189

Offset: 0

Emeric Deutsch, May 06 2011

nonn

a(n)=Sum(k*A190164(n,k),k>=0).

a(n)=A110236(n) - A190169(n).

			a(4)=6 because in h'h'h'h', h'uhd, uhdh', and uhhd, where u=(1,1), h=(1,0), d=(1,-1), we have 4+1+1+0 h-steps at even levels (marked).

Cf. A190164, A110236, A190169, A004148.

G := z/((1-z+z^2)*sqrt((1+z+z^2)*(1-3*z+z^2))): Gser := series(G,z=0,36): seq(coeff(Gser,z,n),n=0..33);

G.f. = z/[(1-z+z^2)sqrt((1+z+z^2)(1-3z+z^2))].
Conjecture: (-n+1)*a(n) +(3*n-4)*a(n-1) +2*(-n+1)*a(n-2) +3*(n-2)*a(n-3) +2*(-n+3)*a(n-4) +(3*n-8)*a(n-5) +(-n+3)*a(n-6)=0. - R. J. Mathar, Apr 09 2019
a(n) ~ phi^(2*n+2) / (4 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, May 29 2022

A190169 Number of (1,0)-steps at levels 1,3,5,... in all peakless Motzkin paths of length n.

Original entry on oeis.org

0, 0, 0, 1, 4, 10, 24, 60, 152, 386, 980, 2488, 6324, 16098, 41032, 104711, 267512, 684138, 1751316, 4487217, 11506792, 29530524, 75841152, 194910254, 501234960, 1289755668, 3320603016, 8553723949, 22044934324, 56841474482, 146626826376, 378392593206, 976884539336, 2522936490418

Offset: 0

Emeric Deutsch, May 06 2011

nonn

a(n)=Sum(k*A190167(n,k),k>=0).

a(n)=A110236(n) - A190166(n).

			a(4)=4 because in hhhh, huh'd, uh'dh, and uh'h'd, where u=(1,1), h=(1,0), d=(1,-1), we have 0+1+1+2 h-steps at odd levels (marked).

Cf. A190167, A110236, A190166

G := ((1-2*z+z^2-2*z^3+z^4)*1/2)/(z*(1-z+z^2)*sqrt((1+z+z^2)*(1-3*z+z^2)))-(1/2)/z: Gser:=series(G,z=0,36): seq(coeff(Gser,z,n),n=0..33);

G.f. = (1-2z+z^2-2z^3+z^4)/[2z(1-z+z^2)sqrt((1+z+z^2)(1-3z+z^2))]-1/(2z).

Conjecture: -(n-1)*(n+1)*a(n) -n*(n-19)*a(n-1) +2*(n-1)*(7*n-40)*a(n-2) -(n-2)*(17*n-97)*a(n-3) +2*(9*n^2-64*n+119)*a(n-4) -17*(n-4)*(n-5)*a(n-5) +(19*n-59)*(n-5)*a(n-6) -2*(8*n-21)*(n-6)*a(n-7) +2*(2*n-5)*(n-7)*a(n-8)=0. - R. J. Mathar, Apr 09 2019

A190172 Triangle read by rows: T(n,k) is the number of peakless Motzkin paths of length n having k UHD's; here U=(1,1), H=(1,0), and D=(1,-1).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 4, 4, 8, 8, 1, 16, 18, 3, 33, 40, 9, 69, 90, 25, 1, 146, 204, 69, 4, 312, 467, 183, 16, 673, 1074, 479, 56, 1, 1463, 2481, 1239, 185, 5, 3202, 5752, 3180, 576, 25, 7050, 13378, 8104, 1734, 105, 1, 15605, 31196, 20544, 5076, 405, 6, 34705, 72912, 51852, 14546, 1451, 36

Offset: 0

Emeric Deutsch, May 06 2011

Number of entries in row n is 1+floor(n/3).
Sum of entries in row n = A004148 (the RNA secondary structure numbers).
T(n,0)=A004149(n).
Sum(k*T(n,k),k>=0)=A110236(n-2) (n>=3).

			T(5,1)=4 because we have HHUHD, HUHDH, UHDH, and UUHDD.
Triangle starts:
1;
1;
1;
1,1;
2,2;
4,4;
8,8,1;
16,18,3;

Cf. A004148, A004149, A110236

eq := G = 1+z*G+z^2*G*(G-1-z+t*z): G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 25)): for n from 0 to 17 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 17 do seq(coeff(P[n], t, k), k = 0 .. floor((1/3)*n)) end do; # yields sequence in triangular form

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

A089742 Number of subwords UHH...HD in all peakless Motzkin paths of length n+3, where U=(1,1), D=(1,-1) and H=(1,0).

Original entry on oeis.org

1, 3, 7, 17, 41, 99, 242, 596, 1477, 3681, 9215, 23155, 58368, 147530, 373768, 948882, 2413264, 6147414, 15682008, 40056238, 102434119, 262228051, 671945055, 1723350315, 4423518544, 11362907022, 29208834520, 75131251334, 193370093508

Offset: 0

Emeric Deutsch, Jan 08 2004

nonn

This sequence can also be easily expressed using RNA secondary structure terminology.

			a(1)=3 because in the four peakless Motzkin paths of length 4, namely HHHH, H(UHD), (UHD)H and (UHHD), we have altogether three subwords of the required form (shown between parentheses).

I. L. Hofacker, P. Schuster and P. F. Stadler, Combinatorics of RNA secondary structures, Discrete Appl. Math., 88, 1998, 207-237.
P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26 (1979), 261-272.
M. Vauchassade de Chaumont and G. Viennot, Polynômes orthogonaux et problèmes d'énumération en biologie moléculaire, Sem. Loth. Comb. B08l (1984) 79-86. [Formerly: Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, p. 79-86.]
M. S. Waterman, Home Page (contains copies of his papers)

Cf. A004148.

Partial sums of A110236.

a(n):=sum(sum(j*sum((binomial(2*j+2*i,i)*sum(binomial(k,m-k-2*j-2*i)*binomial(k+2*j+2*i-1,k)*(-1)^(k-m),k,0,m-2*j-2*i))/(j+i),i,0,m/2-j),j,1,m/2),m,0,n+2); /* Vladimir Kruchinin, Mar 07 2016 */

G.f.= g^2/[(1-z)(1-z^2*g^2)], where g=(1-z+z^2-sqrt(1-2z-z^2-2*z^3+z^4))/(2z^2) is the g.f. of sequence A004148 (RNA secondary structures).
a(n) = Sum_{m=0..n+2 }(Sum_{j=1..m/2}(j*Sum_{i=0..m/2-j} ((binomial(2*j+2*i,i)*Sum_{k=0..m-2*j-2*i}(binomial(k,m-k-2*j-2*i)*binomial(k+2*j+2*i-1,k)*(-1)^(k-m)))/(j+i)))). - Vladimir Kruchinin, Mar 07 2016
D-finite with recurrence (n+2)*a(n) +(-4*n-5)*a(n-1) +(5*n-1)*a(n-2) +(-5*n+7)*a(n-3) +(5*n-3)*a(n-4) +(-5*n+9)*a(n-5) +(4*n-13)*a(n-6) +(-n+4)*a(n-7)=0. - R. J. Mathar, Jul 24 2022

A110235 Triangle read by rows: T(n,k)(1<=k<=n) is the number of peakless Motzkin paths of length n having k (1,0) steps (can be easily translated into RNA secondary structure terminology).

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 0, 6, 0, 1, 0, 6, 0, 10, 0, 1, 1, 0, 20, 0, 15, 0, 1, 0, 10, 0, 50, 0, 21, 0, 1, 1, 0, 50, 0, 105, 0, 28, 0, 1, 0, 15, 0, 175, 0, 196, 0, 36, 0, 1, 1, 0, 105, 0, 490, 0, 336, 0, 45, 0, 1, 0, 21, 0, 490, 0, 1176, 0, 540, 0, 55, 0, 1, 1, 0, 196, 0, 1764, 0, 2520, 0

Offset: 1

Emeric Deutsch, Jul 17 2005

Row sums yield A004148. sum(k*T(n,k),k=1..n)=A110236(n).

			T(5,3)=6 because we have UHDHH, UHHDH, UHHHD, HUHDH, HUHHD and HHUHD, where U=(1,1), D=(1,-1), H=(1,0).
Triangle starts:
1;
0, 1;
1, 0, 1;
0, 3, 0, 1;
1, 0, 6, 0, 1;
0, 6, 0, 10, 0, 1;
1, 0, 20, 0, 15, 0, 1;
0, 10, 0, 50, 0, 21, 0, 1;
1, 0, 50, 0, 105, 0, 28, 0, 1;
0, 15, 0, 175, 0, 196, 0, 36, 0, 1;
1, 0, 105, 0, 490, 0, 336, 0, 45, 0, 1; ...

W. R. Schmitt and M. S. Waterman, Linear trees and RNA secondary structure, Discrete Appl. Math., 51, 317-323, 1994.
P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26 (1978), 261-272.
M. Vauchassade de Chaumont and G. Viennot, Polynômes orthogonaux et problèmes d'énumération en biologie moléculaire, Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, Actes 8e Sem. Lotharingien, pp. 79-86.

Cf. A004148, A110236.

T:=proc(n,k) if n+k mod 2 = 0 then 2*binomial((n+k)/2,k)*binomial((n+k)/2,k-1)/(n+k) else 0 fi end: for n from 1 to 14 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form

T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^2*y^j*x^(m-j))*x^m/m)+O(x^(n+1))),n,x),k,y)
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Oct 21 2012

T(n, k) = [2/(n+k)]binomial((n+k)/2, k)*binomial((n+k)/2, k-1).
G.f.: g=g(t, z) satisfies g=1+tzg+z^2*g(g-1).
G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2 * y^k * x^(n-k)] * x^n/n ). - Paul D. Hanna, Oct 21 2012

A110237 Triangle read by rows: T(n,k) (0 <= k <= ceiling(n/2)-1) is the number of (1,0) steps at level k in all peakless Motzkin paths of length n (can be easily translated into RNA secondary structure terminology).

Original entry on oeis.org

1, 2, 3, 1, 6, 4, 13, 10, 1, 28, 24, 6, 62, 59, 21, 1, 140, 144, 62, 8, 320, 350, 174, 36, 1, 740, 852, 474, 128, 10, 1728, 2077, 1263, 410, 55, 1, 4068, 5072, 3318, 1240, 230, 12, 9645, 12412, 8634, 3608, 835, 78, 1, 23010, 30440, 22314, 10216, 2792, 376, 14

Offset: 1

Emeric Deutsch, Jul 17 2005

Row n has ceiling(n/2) terms. Row sums yield A110236.

			T(5,1)=10 because in the 8 (=A004148(5)) peakless Motzkin paths of length 5, namely HHHHH, U(H)DHH, U(HH)DH, U(HHH)D, HU(H)DH, HU(HH)D, HHU(H)D and UUHDD (where U=(1,1), H=(1,0) and D=(1,-1)), we have altogether 10 H steps at level 1 (shown between parentheses).
Triangle starts:
   1;
   2;
   3,  1;
   6,  4;
  13, 10,  1;

W. R. Schmitt and M. S. Waterman, Linear trees and RNA secondary structure, Discrete Appl. Math., 51, 317-323, 1994.
P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26 (1978), 261-272.
M. Vauchassade de Chaumont and G. Viennot, Polynômes orthogonaux et problèmes d'énumération en biologie moléculaire, Publ. I.R.M.A. Strasbourg, 1984, 229/S-08, Actes 8e Sem. Lotharingien, pp. 79-86.

Cf. A004148, A110236.

g:=(1-z+z^2-sqrt(1-2*z-z^2-2*z^3+z^4))/2/z^2: G:=z*g^2/(1-t*z^2*g^2): Gser:=simplify(series(G,z=0,20)): for n from 1 to 15 do P[n]:=coeff(Gser,z^n) od: for n from 1 to 15 do seq(coeff(t*P[n],t^k),k=1..ceil(n/2)) od;

T(n,m):=(m+1)*sum((binomial(2*m+2*i+2,i)*sum(binomial(k,n-2*m-k-2*i-1)*binomial(2*m+k+2*i+1,k)*(-1)^(n-k-1),k,0,n-2*m-2*i-1))/(m+i+1),i,0,(n-1)/2-m); /* Vladimir Kruchinin, Mar 07 2016 */

G.f.: z*g^2/(1-tz^2*g^2), where g = 1 + zg + z^2*g(g-1) = (1 - z + z^2 - sqrt(1 - 2z - z^2 - 2z^3 + z^4))/(2z^2) is the g.f. of the RNA secondary structure numbers (A004148).

T(n,m) = (m+1)*Sum_{i=0..(n-1)/2-m}((binomial(2*m+2*i+2,i)*Sum_{k=0..n-2*m-2*i-1}(binomial(k,n-2*m-k-2*i-1)*binomial(2*m+k+2*i+1,k)*(-1)^(n-k-1)))/(m+i+1)). - Vladimir Kruchinin, Mar 07 2016

cp's OEIS Frontend

A202411 a(n) = Sum_{k=floor(n/4)..R} C(k, m*k - (-1)^n*(R - k)) * C(k + 1, m*(k + 2) - (-1)^n*(R - k + 1)) where m = (n + 1) mod 2 and R = (n + m - 3)/2 for n > 0 and a(0) = 1.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A203611 a(n) = Sum_{k=0..n} binomial(k-1,2*k-1-n)*binomial(k,2*k-n), with a(0) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

A201631 a(n) is the number of Fibonacci meanders of length m*n and central angle 360/m degrees where m = 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A190164 Triangle read by rows: T(n,k) is the number of peakless Motzkin paths of length n having a total of k (1,0)-steps at levels 0,2,4,... .

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A190166 Number of (1,0)-steps at levels 0,2,4,... in all peakless Motzkin paths of length n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A190169 Number of (1,0)-steps at levels 1,3,5,... in all peakless Motzkin paths of length n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A190172 Triangle read by rows: T(n,k) is the number of peakless Motzkin paths of length n having k UHD's; here U=(1,1), H=(1,0), and D=(1,-1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A089742 Number of subwords UHH...HD in all peakless Motzkin paths of length n+3, where U=(1,1), D=(1,-1) and H=(1,0).

A202411 a(n) = Sum_{k=floor(n/4)..R} C(k, mk - (-1)^n(R - k)) * C(k + 1, m(k + 2) - (-1)^n(R - k + 1)) where m = (n + 1) mod 2 and R = (n + m - 3)/2 for n > 0 and a(0) = 1.

A203611 a(n) = Sum_{k=0..n} binomial(k-1,2k-1-n)binomial(k,2*k-n), with a(0) = 1.