A071356 -id:A071356 - cp's OEIS frontend

A110681 A convolution triangle of numbers based on A071356.

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 72, 64, 30, 8, 1, 272, 260, 140, 48, 10, 1, 1064, 1072, 636, 256, 70, 12, 1, 4272, 4480, 2856, 1288, 420, 96, 14, 1, 17504, 18944, 12768, 6272, 2320, 640, 126, 16, 1, 72896, 80928, 57024, 29952, 12192, 3852, 924, 160, 18, 1

Offset: 0

Philippe Deléham, Sep 14 2005

			Triangle starts:
   1;
   2,  1;
   6,  4,  1;
  20, 16,  6,  1;
  72, 64, 30,  8,  1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..11324 (rows 0 <= n <= 150).
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973), p. 32-33.
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)

Cf. A071356 (1st column), A071357 (2nd column).

Cf. A052705 (diagonal sums), A001003 (row sums).

T[n_, k_] := T[n, k] = Which[n == k == 0, 1, n == 0, 0, k == 0, 0, k > n, 0, True, T[n - 1, k - 1] + 2 T[n - 1, k] + 2 T[n - 1, k + 1]]; Table[T[n, k], {n, 0, 10}, {k, n}] // Flatten (* Michael De Vlieger, Nov 05 2017 *)

T(0, 0) = 1; T(n, k) = 0 if k<0 or if k>n; T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + 2*T(n-1, k+1).
Sum_{k, k>=0} T(m, k)*T(n, k)*2^k = T(m+n, 0) = A071356(m+n).
Sum_{k, k>=0} T(n, k)*(2^(k+1) - 1) = 5^n.
Sum_{k, k>=0} (-1)^(n-k)*T(n, k)*(2^(k+1) - 1) = 1.

A068764 Generalized Catalan numbers 2xA(x)^2 -A(x) +1 -x =0.

Original entry on oeis.org

1, 1, 4, 18, 88, 456, 2464, 13736, 78432, 456416, 2697088, 16141120, 97632000, 595912960, 3665728512, 22703097472, 141448381952, 885934151168, 5575020435456, 35230798994432, 223485795258368, 1422572226146304, 9083682419818496, 58169612565614592, 373486362257899520, 2403850703479816192

Offset: 0

Wolfdieter Lang, Mar 04 2002

a(n) = K(2,2; n)/2 with K(a,b; n) defined in a comment to A068763.

Hankel transform is A166232(n+1). - Paul Barry, Oct 09 2009

			G.f. = 1 + x + 4*x^2 + 18*x^3 + 88*x^4 + 456*x^5 + 2464*x^6 + 13736*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A025227, A025228, A025229, A025230.

Cf. A071356, A001003, A025235.

Table[SeriesCoefficient[(1-Sqrt[1-8*x*(1-x)])/(4*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 13 2012 *)
Round@Table[4^(n-1) Hypergeometric2F1[(1-n)/2, 1-n/2, 2, 1/2] + KroneckerDelta[n]/Sqrt[2], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 07 2015 *)
a[ n_] := If[ n < 1, Boole[n == 0], 4^(n - 1) Hypergeometric2F1[ (1 - n)/2, (2 - n)/2, 2, 1/2]]; (* Michael Somos, Nov 08 2015 *)

a(n):=sum(binomial(n-1,k-1)*1/k*sum(binomial(k,j)*binomial(k+j,j-1),j,1,k),k,1,n); /* Vladimir Kruchinin, Aug 11 2010 */

{a(n) = my(A); if( n<1, n==0, n--;  A = x * O(x^n); n! * simplify( polcoeff( exp(4*x + A) * besseli(1, 2*x * quadgen(8) + A), n)))}; /* Michael Somos, Mar 31 2007 */

x='x+O('x^66); Vec((1-sqrt(1-8*x*(1-x)))/(4*x)) \\ Joerg Arndt, May 06 2013

G.f.: (1-sqrt(1-8*x*(1-x)))/(4*x).
a(n+1) = 2*sum(a(k)*a(n-k), k=0..n), n>=1, a(0) = 1 = a(1).
a(n) = (2^n)*p(n, -1/2) with the row polynomials p(n, x) defined from array A068763.
E.g.f. (offset -1) is exp(4*x)*BesselI(1, 2*sqrt(2)*x)/(sqrt(2)*x). - Vladeta Jovovic, Mar 31 2004
The o.g.f. satisfies A(x) = 1 + x*(2*A(x)^2 - 1), A(0) = 1. - Wolfdieter Lang, Nov 13 2007
a(n) = subs(t=1,(d^(n-1)/dt^(n-1))(-1+2*t^2)^n)/n!, n >= 2, due to the Lagrange series for the given implicit o.g.f. equation. This formula holds also for n=1 if no differentiation is used. - Wolfdieter Lang, Nov 13 2007, Feb 22 2008
1/(1-x/(1-x-2x/(1-x/(1-x-2x/(1-x/(1-x-2x/(1-..... (continued fraction). - Paul Barry, Jan 29 2009
a(n) = A166229(n)/(2-0^n). - Paul Barry, Oct 09 2009
a(n) = sum(binomial(n-1,k-1)*1/k*sum(binomial(k,j)*binomial(k+j,j-1),j,1,k),k,1,n), n>0. - Vladimir Kruchinin, Aug 11 2010
D-finite with recurrence: (n+1)*a(n) = 4*(2*n-1)*a(n-1) - 8*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 13 2012
a(n) ~ sqrt(1+sqrt(2))*(4+2*sqrt(2))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 13 2012
a(n) = 4^(n-1)*hypergeom([(1-n)/2,1-n/2], [2], 1/2) + 0^n/sqrt(2). - Vladimir Reshetnikov, Nov 07 2015
0 = a(n)*(+64*a(n+1) - 160*a(n+2) + 32*a(n+3)) + a(n+1)*(+32*a(n+1) + 48*a(n+2) - 20*a(n+3)) + a(n+2)*(+4*a(n+2) + a(n+3)) for all n>=0. - Michael Somos, Nov 08 2015
a(n) = (-1)^n * Sum_{k=0..n} (-2)^k * binomial(n,k) * binomial(2*k+1,n) / (2*k+1). - Seiichi Manyama, Jul 24 2023

A036774 Number of labeled rooted unordered binary trees (each node has out-degree <= 2).

Original entry on oeis.org

0, 1, 2, 9, 60, 540, 6120, 83790, 1345680, 24811920, 516650400, 11992503600, 307069963200, 8598348158400, 261387760233600, 8573572885878000, 301809119163552000, 11349727401396384000, 454104511068656448000, 19261139319649202976000

Offset: 0

N. J. A. Sloane

nonn

Fung Lam, Table of n, a(n) for n = 0..394
B. Otto, Coalescence under Preimage Constraints, arXiv:1903.00542 [math.CO], 2019.
L. Takacs, Enumeration of rooted trees and forests, Math. Scientist 18 (1993), 1-10, esp. Eq. (14) with r = 2.
Index entries for sequences related to rooted trees

A071356(n) = a(n + 1) * 2^n/(n + 1)!.

Cf. A036775 (outdegree <= r = 3), A036776 (out-degree <= r = 4), A036777 (out-degree <= r = 5).

# This is a crude Maple program based on Eq. (14), p. 4, in Takacs (1993). It calculates a(n) for n >= 2. Here, r = 2 is the maximum out-degree of each node.
ff := proc(r, n) simplify(subs(x = 0, diff(sum(x^k/k!, k = 0 .. r)^n, x$(n - 1)))); end;
seq(ff(2, i), i = 2 .. 40); # Petros Hadjicostas, Jun 09 2019

Range[0, 20]! CoefficientList[Series[(1 - x - ((x - 1)^2 - 2 x^2)^(1/2))/x, {x, 0, 20}], x] (* Geoffrey Critzer, Nov 22 2011 *)
f[r_, n_][x_] := Sum[x^k/k!, {k, 0, r}]^n;
a[n_] := If[n == 1, 1, Derivative[n - 1][f[2, n]][0]];
a /@ Range[0, 19] (* Jean-François Alcover, Apr 20 2020, after Petros Hadjicostas *)
a[n_] := n! Hypergeometric2F1[1/2 - n/2, 1 - n/2, 2, 2]; a[0] = 0;
Array[a, 20, 0] (* Peter Luschny, Apr 20 2020 *)

makelist(n!*sum(binomial(n-1,2*k)*binomial(2*k,k)/(2^k*(k+1)),k,0,floor((n-1)/2)),n,0,20); /* Emanuele Munarini, Feb 06 2013 */

a(n)=if(n<1,0,n!*polcoeff(2*x/(1-x+sqrt(1-2*x-x^2+O(x^n))),n))

a(n)=if(n<1,0,n!*polcoeff(serreverse(2*x/(2+2*x+x^2)+x*O(x^n)),n))

E.g.f.: (1 - x - sqrt(1-2*x-x^2))/x.
E.g.f. A(x) satisfies x*A(x)^2 + 2*(x-1)*A(x) + 2*x = 0, A(0)=0 and A(x) = x/(1-x-(x/2)*A(x)). - Michael Somos, Sep 06 2003
a(n) = n!*Sum_{k=0..floor((n-1)/2)} binomial(n-1, 2k)*binomial(2k, k)/(2^k*(k+1)). - Emanuele Munarini, Feb 06 2013
a(n) ~ sqrt(2-sqrt(2))*n^(n-1)/(exp(n)*(sqrt(2)-1)^(n+1)). - Vaclav Kotesovec, Sep 24 2013
Recurrence: (n+1)*a(n) = n*(n-1)*(n-2)*a(n-2) + n*(2*n-1)*a(n-1), n >= 3, a(1)=1, a(2)=2. - Fung Lam, Feb 24 2014
a(n) = n!*hypergeom([(1 - n)/2, 1 - n/2], [2], 2) for n >= 1. Peter Luschny, Apr 20 2020

Better description and formula from Christian G. Bower, Nov 29 2001

"unordered" added to the name by David Callan, Apr 22 2012

A062991 Coefficient triangle for certain polynomials N(2; n,x) (rising powers of x).

Original entry on oeis.org

1, 2, -1, 5, -6, 2, 14, -28, 20, -5, 42, -120, 135, -70, 14, 132, -495, 770, -616, 252, -42, 429, -2002, 4004, -4368, 2730, -924, 132, 1430, -8008, 19656, -27300, 23100, -11880, 3432, -429, 4862, -31824, 92820, -157080, 168300, -116688, 51051, -12870, 1430

Offset: 0

Wolfdieter Lang, Jul 12 2001

The g.f. for the sequence of column m of triangle A009766(n,m) (or Catalan A033184(n,n-m) diagonals) is N(2; m-1,x)*(x^m)/(1-x)^(m+1), m >= 1, with N(2; n,x) = Sum_{k=0..n} T(n,k)*x^k.
For k=0..1 the column sequences give A000108(n+1) (Catalan), -A002694. The row sums give A000012 (powers of 1) and (unsigned) A062992.
Another version of [1, 1, 1, 1, 1, 1, 1, 1, ...] DELTA [0, -1, -1, -1, -1, -1, -1, -1, ...] = 1; 1, 0; 2, -1, 0; 5, -6, 2, 0; 14, -28, 20, -5, 0; 42, -120, 135, -70, 14, 0; ... where DELTA is Deléham's operator defined in A084938.
The positive triangle is |T(n,k)| = binomial(2*n+2, n-k)*binomial(n+k, k)/(n+1). - Paul Barry, May 11 2005

			The triangle N2 = {a(n,k)} begins:
n\k      0       1      2       3       4       5      6       7     8     9
----------------------------------------------------------------------------
0:       1
1:       2      -1
2:       5      -6      2
3:      14     -28     20      -5
4:      42    -120    135     -70      14
5:     132    -495    770    -616     252     -42
6:     429   -2002   4004   -4368    2730    -924    132
7:    1430   -8008  19656  -27300   23100  -11880   3432    -429
8:    4862  -31824  92820 -157080  168300 -116688  51051  -12870  1430
9:   16796 -125970 426360 -852720 1108536 -969969 570570 -217360 48620 -4862
... formatted by _Wolfdieter Lang_, Jan 20 2020
N(2; 2, x)= 5 - 6*x + 2*x^2.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
C. A. Francisco, J. Mermin, and J. Schweig, Catalan numbers, binary trees, and pointed pseudotriangulations, 2013.
V. E. Hoggatt Jr. and Marjorie Bicknell-Johnson, Numerator Polynomial Coefficient Arrays for Catalan and Related Sequence Convolution Triangles, The Fibonacci Quarterly 15 (1977) 30-34. [On p. 31, in the line n = 1, 14 is missing in S_1^4. - _Wolfdieter Lang_, Jan 20 2020 ]
Joseph T. Iosue, Adam Ehrenberg, Dominik Hangleiter, Abhinav Deshpande, and Alexey V. Gorshkov, Page curves and typical entanglement in linear optics, arXiv:2209.06838 [quant-ph], 2022.
A. Lakshminarayan, Z. Puchala, and K. Zyczkowski, Diagonal unitary entangling gates and contradiagonal quantum states, arXiv preprint arXiv:1407.1169 [quant-ph], 2014.
Jian Zhou, On Some Mathematics Related to the Interpolating Statistics, arXiv:2108.10514 [math-ph], 2021.

Cf. A000108, A002694, A009766, A033184, A062992, A084938, A089434, A094385.
For an unsigned version see Borel's triangle, A234950.
Sums include: A000012 (row), A000079 (diagonal), A064062 (signed row), A071356 (signed diagonal).

A062991:= func< n,k | (-1)^k*Binomial(2*n+2,n-k)*Binomial(n+k,k)/(n+1) >;
[A062991(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 27 2024

T[n_, k_] := 2 (-1)^k Binomial[2n+1, n] (n-k+1) Binomial[n+1, k]/((k+n+1)(k+n+2));
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 19 2018 *)

def A062991(n,k): return (-1)^k*binomial(2*n+2,n-k)*binomial(n+k,k)/(n+1)
flatten([[A062991(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 27 2024

T(n, k) = [x^k] N(2; n, x) with N(2; n, x) = (N(2; n-1, x) - A000108(n)*(1-x)^(n+1))/x, N(2; 0, x) = 1.
T(n, k) = T(n-1, k+1) + (-1)^k*binomial(n+1, k+1)*binomial(2*n+1, n)/(2*n+1) if k=0, .., (n-2); T(n, k) = (-1)^k*binomial(n+1, k+1)*binomial(2*n+1, n)/(2*n+1) if k=(n-1) or n; else 0.
O.g.f. (with offset 1) is the series reversion w.r.t. x of x*(1+x*t)/(1+x)^2. If R(n,t) denotes the n-th row polynomial of this triangle then R(n,1-t) is the n-th row polynomial of A009766. Cf. A089434. - Peter Bala, Jul 15 2012
From G. C. Greubel, Sep 27 2024: (Start)
Sum_{k=0..n} T(n, k) = A000012(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A064062(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A000079(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A071356(n). (End)

A025235 a(n) = (1/2)*s(n+2), where s = A014431.

Original entry on oeis.org

1, 1, 3, 7, 21, 61, 191, 603, 1961, 6457, 21595, 72975, 249085, 857013, 2970007, 10356323, 36311633, 127937649, 452738867, 1608426647, 5734534629, 20511509549, 73583105007, 264687136235, 954482676217, 3449853902761, 12495597328011, 45349353908383

Offset: 0

Clark Kimberling

nonn

Number of lattice paths in the first quadrant from (0,0) to (n,0) using only steps H=(1,0), U=(1,1) and D=(1,-1), where the U steps come in two colors: red (R) and green (G) (i.e., Motzkin paths with the up steps in two colors). E.g., a(3)=7 because we have HHH, HRD, HGD, RDH, GDH, RHD and GHD. - Emeric Deutsch, Dec 25 2003
Equals inverse binomial transform of A071356: (1, 2, 6, 20, 72, ...). - Gary W. Adamson, Sep 03 2010
a(n) is the number of increasing unary-binary trees with associated permutation that avoids 231. For more information about increasing unary-binary trees with an associated permutation, see A245888. - Manda Riehl, Aug 07 2014

			x + x^2 + 3*x^3 + 7*x^4 + 21*x^5 + 61*x^6 + 191*x^7 + 603*x^8 + 1961*x^9 + ...
a(4) = 21 since the top row of M^4 = (21, 11, 7, 1, 1)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Paul Barry, Generalized Catalan recurrences, Riordan arrays, elliptic curves, and orthogonal polynomials, arXiv:1910.00875 [math.CO], 2019.
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
Paul Barry, On Motzkin-Schröder Paths, Riordan Arrays, and Somos-4 Sequences, J. Int. Seq. (2023) Vol. 26, Art. 23.4.7.
Stefano Capparelli and Alberto Del Fra, Dyck Paths, Motzkin Paths, and the Binomial Transform, Journal of Integer Sequences, 18 (2015), #15.8.5.
Xiang-Ke Chang, X.-B. Hu, H. Lei, and Y.-N. Yeh, Combinatorial proofs of addition formulas, The Electronic Journal of Combinatorics, 23(1) (2016), #P1.8.
Serkan Demiriz, Adem Şahin, and Sezer Erdem, Some topological and geometric properties of novel generalized Motzkin sequence spaces, Rendiconti Circ. Mat. Palermo Ser. 2 (2025) Vol. 74, No. 136. See p. 4.
Maciej Dziemiańczuk, Counting Lattice Paths With Four Types of Steps, Graphs and Combinatorics, September 2013, DOI 10.1007/s00373-013-1357-1.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Louis W. Shapiro and Carol J. Wang, A bijection between 3-Motzkin paths and Schroder paths with no peak at odd height, JIS 12 (2009) 09.3.2.

Cf. A071356, A001003, A068764, A217275.

Join[{1}, Table[Sum[2^(k - 1)*Binomial[n + 1, k]*Binomial[n - k + 1, k - 1]/(n + 1), {k,0,n}], {n,0,50}]] (* G. C. Greubel, Jan 27 2017 *)
a[n_] := Hypergeometric2F1[1/2 - n/2, -n/2, 2, 8];
Table[a[n], {n, 0, 27}] (* Peter Luschny, Mar 18 2018 *)

{a(n) = if( n<0, 0, polcoeff( serreverse( x / (1 + x + 2*x^2 + x * O(x^n))), n+1))} /* Michael Somos, Jul 12 2003 */

{a(n) = if( n<0, 0, polcoeff( (1 - x - sqrt(1 - 2*x -7*x^2 + x^3 * O(x^n)) ) / 4, n+2))} /* Michael Somos, Mar 31 2007 */

{a(n) = local(A); if( n<0, 0, A = x * O(x^n); n! * simplify( polcoeff( exp(x + A) * besseli(1, 2*x * quadgen(8) + A), n)))} /* Michael Somos, Mar 31 2007 */

a(n) = Sum_{k=0..n} 2^(k-1)*binomial(n+1, k)*binomial(n-k+1, k-1)/(n+1 ). - Len Smiley
G.f.: (1 - x - sqrt(1 - 2*x - 7*x^2)) / (4*x^2). - Michael Somos, Jun 08 2000
G.f. (for offset 1) is series reversion of x / (1 + x + 2*x^2). - Michael Somos, Jul 12 2003
a(n) = Sum_{k=0..n} binomial(n, k)*2^(k/2)*C(k/2)*(1+(-1)^k)/2, where C(n)=A000108(n). - Paul Barry, Dec 22 2003
E.g.f.: exp(x)*BesselI(1, 2*sqrt(2)*x)/(sqrt(2)*x). - Vladeta Jovovic, Mar 31 2004
From Gary W. Adamson, Feb 21 2012: (Start)
a(n) is the leftmost term in the top row of M^n, M is an infinite square production matrix as follows:
  1, 1, 0, 0, 0, 0, ...
  2, 0, 1, 0, 0, 0, ...
  2, 2, 0, 1, 0, 0, ...
  2, 2, 2, 0, 1, 0, ...
  2, 2, 2, 2, 0, 1, ...
  2, 2, 2, 2, 2, 0, ...
  2, 2, 2, 2, 2, 2, ...
  ... (End)
From Vaclav Kotesovec, Sep 29 2012: (Start)
a(n) ~ (1+2*sqrt(2))^(n+3/2)/(2*sqrt(Pi)*2^(3/4)*n^(3/2)).
Recurrence: (n+2)*a(n) = (2*n+1)*a(n-1) + 7*(n-1)*a(n-2). (End)
a(n) = hypergeom([-n/2, (1-n)/2], [2], 8). - Peter Luschny, May 28 2014
G.f.: 1/(1 - x - 2*x^2/(1 - x - 2*x^2/(1 - x - 2*x^2/(1 - x - 2*x^2/(1 - ....))))), a continued fraction. - Ilya Gutkovskiy, May 26 2017

A107267 A square array of Motzkin related transforms, read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 6, 3, 1, 0, 9, 20, 12, 4, 1, 0, 21, 72, 54, 20, 5, 1, 0, 51, 272, 261, 112, 30, 6, 1, 0, 127, 1064, 1323, 672, 200, 42, 7, 1, 0, 323, 4272, 6939, 4224, 1425, 324, 56, 8, 1, 0, 835, 17504, 37341, 27456, 10625, 2664, 490, 72, 9, 1

Offset: 0

Paul Barry, May 15 2005

Rows are transforms of k^n, k>=0, under the matrix A107131. As a number triangle, with T(n,k)=if(k<=n,sum{j=0..n-k, (1/(j+1))C(j+1,n-k-j+1)C(n-k,j)k^j},0), row sums are A107268 and diagonal sums are A107269. Rows are series reversions of x/(1+kx+kx^2), k>=0. Conjecture: rows count weighted Motzkin paths.

Row k counts colored Motzkin paths, where H(1,0) and U(1,1) each have k colors and D(1,-1) one color. - Paul Barry, May 16 2005

			Array begins
  1, 0,  0,   0,    0,     0,      0, ...
  1, 1,  2,   4,    9,    21,     51, ...
  1, 2,  6,  20,   72,   272,   1064, ...
  1, 3, 12,  54,  261,  1323,   6939, ...
  1, 4, 20, 112,  672,  4224,  27456, ...
  1, 5, 30, 200, 1425, 10625,  81875, ...
  1, 6, 42, 324, 2664, 22896, 203256, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Rows n=0..6 give A000007, A001006, A071356, A107264, A003645, A107265, A107266.
Main diagonal gives A292716.
Cf. A000108.

Number array T(n,k) = Sum_{j=0..k} n^j * binomial(k,j) * binomial(j+1,k-j+1)/(j+1).
G.f. of row k: 1/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
From Seiichi Manyama, May 05 2019: (Start)
T(n,k) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * binomial(2*j,j)/(j+1) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * A000108(j).
(k+2) * T(n,k) = n * (2*k+1) * T(n,k-1) - n * (n-4) * (k-1) * T(n,k-2). (End)

A071943 Triangle T(n,k) (n>=0, 0 <= k <= n) read by rows giving number of underdiagonal lattice paths from (0,0) to (n,k) using only steps R=(1,0), V=(0,1) and D=(1,2).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 7, 9, 1, 4, 12, 24, 31, 1, 5, 18, 46, 89, 113, 1, 6, 25, 76, 183, 342, 431, 1, 7, 33, 115, 323, 741, 1355, 1697, 1, 8, 42, 164, 520, 1376, 3054, 5492, 6847, 1, 9, 52, 224, 786, 2326, 5900, 12768, 22669, 28161, 1, 10, 63, 296, 1134, 3684, 10370

Offset: 0

N. J. A. Sloane, Jun 15 2002

For another interpretation of this array see the Example section.

			T(3,2)=7 because we have RRRVV, RRVRV, RRVVR, RVRRV, RVRVR, RRD and RDR.
Array begins:
1,
1, 1,
1, 2, 3,
1, 3, 7, 9,
1, 4, 12, 24, 31,
1, 5, 18, 46, 89, 113,
1, 6, 25, 76, 183, 342, 431,
1, 7, 33, 115, 323, 741, 1355, 1697,
...
Equivalently, let U(n,k) (for n >= 0, 0 <= k <= n) be the number of walks from (0,0) to (n,k) using steps (1,1), (1,-1) and (0,-1). Then U(n,n-k) = T(n,k). The U(n,k) array begins:
4:  0  0  0  0  1  5 ...
3:  0  0  0  1  4 18 ...
2:  0  0  1  3 12 46 ...
1:  0  1  2  7 24 89 ...
0:  1  1  3  9 31 113 ...
-------------------------
k/n:0  1  2  3  4  5 ...
The recurrence for this version is: U(0,0)=1, U(n,k)=0 for k>n or k<0; U(n,k) = U(n,k+1) + U(n-1,k+1) + U(n,k-1). E.g. 46 = 18 + 4 + 24. Also U(n,0) = A052709(n-1).

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
James East, Nicholas Ham, Lattice paths and submonoids of Z^2, arXiv:1811.05735 [math.CO], 2018.
D. Merlini, D. G. Rogers, R. Sprugnoli and M. C. Verri, On some alternative characterizations of Riordan arrays, Canad J. Math., 49 (1997), 301-320.
N. J. A. Sloane, Rows 0 through 100
N. J. A. Sloane, Illustration of the initial terms of the U(n,k) array

Diagonal entries yield A052709. Row sums are A071356.

Related arrays: A071944, A071945, A071946.

U:=proc(n,k) option remember;
if (n < 0) then RETURN(0);
elif (n=0) then
   if (k=0) then RETURN(1); else RETURN(0); fi;
elif (k>n or k<0) then RETURN(0);
else RETURN(U(n,k+1)+U(n-1,k+1)+U(n-1,k-1));
fi;
end;
for n from 0 to 20 do
lprint( [seq(U(n,n-i),i=0..n)] );
od:

t[0, 0] = 1; t[n_, k_] /; k<0 || k>n = 0; t[n_, k_] := t[n, k] = t[n, k-1] + t[n-1, k] + t[n-1, k-2]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 07 2014, after N. J. A. Sloane *)

G.f.=(1-q)/[z(2t+2t^2z-1+q)], where q=sqrt(1-4tz-4t^2z^2).

Define T(0,0)=1 and T(n,k)=0 for k<0 and k >n. Then the array is generated by the recurrence T(n,k) = T(n,k-1) + T(n-1,k) + T(n-1,k-2). For example, T(5,3) = 46 = T(5,2) + T(4,3) + T(4,1) = 18 + 24 + 4. - N. J. A. Sloane, Mar 28 2013

Edited by Emeric Deutsch, Dec 21 2003

Edited by N. J. A. Sloane, Mar 28 2013

A369126 Expansion of (1/x) * Series_Reversion( x / ((1+x)^4+x^4) ).

Original entry on oeis.org

1, 4, 22, 140, 970, 7104, 54096, 424008, 3398224, 27721024, 229410328, 1921308272, 16253502512, 138683973120, 1192142838656, 10314377770720, 89749921081280, 784913791336192, 6895599255571840, 60825440855493376, 538507243041624864, 4783482648574893056

Offset: 0

Seiichi Manyama, Jan 13 2024

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for reversions of series

Cf. A071356, A192132, A369128.

Cf. A127902.

my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)^4+x^4))/x)

a(n) = sum(k=0, n\4, binomial(n+1, k)*binomial(4*n-4*k+4, n-4*k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} binomial(n+1,k) * binomial(4*n-4*k+4,n-4*k).

D-finite with recurrence -3*(3*n+2)*(3*n+4)*(1746*n-6043)*(n+1)*a(n) +4*(519282*n^4 -1632448*n^3 +319539*n^2 +77803*n-72516)*a(n-1) +16*(-1055610*n^4 +5245655*n^3 -8423433*n^2 +5306215*n-1129842)*a(n-2) +96*(n-2) *(150552*n^3 -673240*n^2 +868987*n -301954)*a(n-3) -64*(n-2) *(n-3) *(174726*n^2 -528221*n +220460)*a(n-4) -512*(7353*n-3733)*(n-2)*(n-3)*(n-4)*a(n-5)=0. - R. J. Mathar, Jan 24 2024

A336707 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} 2^(n-j) * binomial(n,j) * binomial(n+(k-1)*j,j-1) for n > 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 11, 20, 1, 1, 5, 19, 45, 72, 1, 1, 6, 30, 100, 197, 272, 1, 1, 7, 44, 201, 562, 903, 1064, 1, 1, 8, 61, 364, 1445, 3304, 4279, 4272, 1, 1, 9, 81, 605, 3249, 10900, 20071, 20793, 17504, 1, 1, 10, 104, 940, 6502, 30526, 85128, 124996, 103049, 72896

Offset: 0

Seiichi Manyama, Aug 01 2020

			Square array begins:
    1,   1,    1,     1,     1,     1,      1, ...
    1,   1,    1,     1,     1,     1,      1, ...
    2,   3,    4,     5,     6,     7,      8, ...
    6,  11,   19,    30,    44,    61,     81, ...
   20,  45,  100,   201,   364,   605,    940, ...
   72, 197,  562,  1445,  3249,  6502,  11857, ...
  272, 903, 3304, 10900, 30526, 73723, 158034, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0-3 give: A071356(n-1), A001003, A007564, A118346.
Main diagonal gives A336712.
Cf. A336575, A336706, A336708, A336709.

T[0, k_] := 1; T[n_, k_] := Sum[2^(n - j) * Binomial[n, j] * Binomial[n + (k - 1)*j, j - 1], {j, 1, n}] / n; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 01 2020 *)

{T(n, k) = if(n==0, 1, sum(j=1, n, 2^(n-j)*binomial(n, j)*binomial(n+(k-1)*j, j-1))/n)}

{T(n, k) = local(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^k/(1-2*x*A)); polcoef(A, n)}

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x)^k / (1 - 2 * x * A_k(x)).

A374497 Expansion of 1/(1 - 4x - 4x^2)^(3/2).

Original entry on oeis.org

1, 6, 36, 200, 1080, 5712, 29792, 153792, 787680, 4009280, 20304768, 102405888, 514678528, 2579028480, 12890311680, 64283809792, 319954540032, 1589720712192, 7886437652480, 39069462835200, 193307835764736, 955361266917376, 4716674314223616, 23264437702656000

Offset: 0

Seiichi Manyama, Jul 09 2024

Cf. A001788, A002457, A025163.
Cf. A006139, A374511, A374513.
Cf. A071356.

a[n_]:= Sum[(2*k+1)*Binomial[2*k,k]*Binomial[k,n-k],{k,0,n}]; Array[a,24,0] (* Stefano Spezia, May 08 2025 *)

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

a(0) = 1, a(1) = 6; a(n) = (2*(2*n+1)*a(n-1) + 4*(n+1)*a(n-2))/n.
a(n) = binomial(n+2,2) * A071356(n).
a(n) = Sum_{k=0..n} (2*k+1) * binomial(2*k,k) * binomial(k,n-k). - Seiichi Manyama, Oct 19 2024
a(n) = ((n+2)/2) * Sum_{k=0..floor(n/2)} 2^(n-k) * binomial(n+1,n-2*k) * binomial(2*k+1,k). - Seiichi Manyama, Aug 20 2025

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A071943 Triangle T(n,k) (n>=0, 0 <= k <= n) read by rows giving number of underdiagonal lattice paths from (0,0) to (n,k) using only steps R=(1,0), V=(0,1) and D=(1,2).

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