A094527 -id:A094527 - cp's OEIS frontend

A088218 Total number of leaves in all rooted ordered trees with n edges.

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052

Offset: 0

Michael Somos, Sep 24 2003

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
  ()  (11)  (22)    (33)
            (121)   (132)
            (1111)  (231)
                    (1122)
                    (1221)
                    (2112)
                    (2211)
                    (11121)
                    (12111)
                    (111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....

L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Anwar Al Ghabra, K. Gopala Krishna, Patrick Labelle, and Vasilisa Shramchenko, Enumeration of multi-rooted plane trees, arXiv:2301.09765 [math.CO], 2023.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
B. Dasarathy and C. Yang, A transformation on ordered trees, Comput. J. 23 (1980) 161-164. - _Nachum Dershowitz_, Sep 01 2016
Toufik Mansour and Mark Shattuck, A statistic on n-color compositions and related sequences, Proc. Indian Acad. Sci. (Math. Sci.) Vol. 124, No. 2, May 2014, pp. 127-140.
Anthony Mansuy, Preordered forests, packed words and contraction algebras, J. Algebra 411 (2014) 259-311.
Mircea Merca, A Note on Cosine Power Sums, J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
A. Vogt, Resummation of small-x double logarithms in QCD: semi-inclusive electron-positron annihilation, arXiv preprint arXiv:1108.2993 [hep-ph], 2011.

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218 (this sequence), ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218 (this sequence), ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000027, A000070, A000097, A000108, A001622, A006232, A008965, A039599, A045992, A058696, A094527, A097070, A110162, A110555, A180662, A238279, A239830, A325534, A325535, A333213, A344607, A344611, A344617.

[Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014

seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014

a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)

{a(n) = sum( i=0, n, binomial(n+i-2,i))};

{a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};

{a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};

{a(n) = if( n<0, 0, binomial( 2*n - 1, n))};

{a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};

def A088218(n):
    return rising_factorial(n,n)/falling_factorial(n,n)
[A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1)  ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

A005810 a(n) = binomial(4n,n).

Original entry on oeis.org

1, 4, 28, 220, 1820, 15504, 134596, 1184040, 10518300, 94143280, 847660528, 7669339132, 69668534468, 635013559600, 5804731963800, 53194089192720, 488526937079580, 4495151581425648, 41432089765583440, 382460951663844400

Offset: 0

N. J. A. Sloane

Start off with 0 balls in a box. Find the number of ways you can throw 3 balls back out. Then continue to throw 4 balls into the box after each stage. (I.e., the first stage is 0. Then at the next stage there are 4 ways to throw 3 balls back out.) - Ruppi Rana (ruppirana007(AT)hotmail.com), Mar 03 2004
Central coefficients of A094527. - Paul Barry, Mar 08 2011
This is the case m = 2n in Catalan's formula (2m)!*(2n)!/(m!*(m+n)!*n!) - see Umberto Scarpis in References. - Bruno Berselli, Apr 27 2012
A generating function in terms of a (labyrinthine) solution to a depressed quartic equation is given in the Copeland link for signed A005810. - Tom Copeland, Oct 10 2012
Conjecture: a(n) == 4 (mod n^3) iff n is prime. - Gary Detlefs, Apr 03 2013
For prime p, the congruence a(p) = binomial(4*p,p) = 4 (mod p^3)  is a known generalization of Wolstenholme's theorem. See Mestrovic, Section 6, equation 35. - Peter Bala, Dec 28 2014

			G.f. = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + 15504*x^5 + 134596*x^6 + ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
Umberto Scarpis, Sui numeri primi e sui problemi dell'analisi indeterminata in Questioni riguardanti le matematiche elementari, Nicola Zanichelli Editore (1924-1927, third Edition), page 11.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000 (terms 0..100 from T. D. Noe, terms 101..213 from Muniru A Asiru)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Tom Copeland, Discriminating Deltas, Depressed Equations, and Generalized Catalan Numbers, 2012, pp. 5-6.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Ruppi Rana, Title? [Broken link]

Row 4 of A060539.
binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).
Cf. A007318, A182400, A262261, A378806, A378807.

List([0..20],n->Binomial(4*n,n)); # Muniru A Asiru, Mar 19 2018

a005810 n = a007318 (4*n) n  -- Reinhard Zumkeller, Mar 04 2012

[ Binomial(4*n,n): n in [0..100] ]; // Vincenzo Librandi, Apr 13 2011

seq(binomial(4*n,n),n=0..20); # Muniru A Asiru, Mar 19 2018

Table[Binomial[4n,n],{n,0,19}] (* Geoffrey Critzer, Sep 15 2013 *)

a(n) = binomial(4*n, n); \\ Altug Alkan, Mar 19 2018

from math import comb
def A005810(n): return comb(n<<2,n) # Chai Wah Wu, Aug 01 2023

a(n) is asymptotic to c*(256/27)^n/sqrt(n) with c = sqrt(2 / (3 Pi)) = 0.460658865961780639... - Benoit Cloitre, Jan 26 2003; corrected by Charles R Greathouse IV, Dec 14 2006
a(n) = Sum_{k=0..2n} binomial(2n,k) * binomial(2n,k-n). - Paul Barry, Mar 08 2011
G.f.: g/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
D-finite with recurrence: 3*n*(3*n-1)*(3*n-2)*a(n) - 8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1) = 0. - R. J. Mathar, Dec 02 2012
a(n) = binomial(4*n,n-1)*(3*n+1)/n. - Gary Detlefs, Apr 03 2013
a(n) = C(4*n-1,n-1)*C(16*n^2,2)/(3*n*C(4*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
a(n) = Sum_{i,j,k = 0..n} binomial(n,i)*binomial(n,j)*binomial(n,k)* binomial(n,i+j+k). - Peter Bala, Dec 28 2014
a(n) = GegenbauerC(n, -2*n, -1). - Peter Luschny, May 07 2016
From Ilya Gutkovskiy, Nov 22 2016: (Start)
O.g.f.: 3F2(1/4,1/2,3/4; 1/3,2/3; 256*x/27).
E.g.f.: 3F3(1/4,1/2,3/4; 1/3,2/3,1; 256*x/27). (End)
a(n) = hypergeom([-3*n, -1*n], [1], 1). - Peter Luschny, Mar 19 2018
RHS of the identity Sum_{k = 0..2*n} (-1)^(n+k)*binomial(4*n, k)* binomial(4*n, 2*n-k) = binomial(4*n,n). - Peter Bala, Oct 07 2021
From Peter Bala, Feb 20 2022: (Start)
The o.g.f. A(x) satisfies the differential equation
(-256*x^3 + 27*x^2)*A(x)''' + (-1152*x^2 + 54*x)*A(x)'' + (-816*x + 6)*A(x)' - 24*A(x) = 0 with A(0) = 1, A'(0) = 4 and A''(0) = 56.
Algebraic equation: (1 - A(x))*(1 + 3*A(x))^3 +  256*x*A(x)^4 = 0.
Sum_{n >= 1} a(n)*( x*(3*x + 4)^3/(256*(1 + x)^4) )^n = x. (End)
From Amiram Eldar, Dec 07 2024: (Start)
Sum_{n>=1} 1/a(n) = A378806.
Sum_{n>=1} (-1)^n/a(n) = A378807. (End)
From Peter Bala, Jun 29 2025: (Start)
a(n) = (1/8)^n * Sum_{k = n..4*n} binomial(k, n) * binomial(4*n, k).
Sum_{n >= 0 } a(n)*(1/128)^n = (1/5)*(sqrt(2) + sqrt(7 + 5*sqrt(2))). (End)
From Seiichi Manyama, Aug 16 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(4*n+1,k).
G.f.: 1/(1 - 4*x*g^3) where g = 1+x*g^4 is the g.f. of A002293. (End)

More terms from Henry Bottomley, Oct 06 2000

Corrected by T. D. Noe, Jan 16 2007

A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 14, 14, 6, 1, 42, 48, 27, 8, 1, 132, 165, 110, 44, 10, 1, 429, 572, 429, 208, 65, 12, 1, 1430, 2002, 1638, 910, 350, 90, 14, 1, 4862, 7072, 6188, 3808, 1700, 544, 119, 16, 1, 16796, 25194, 23256, 15504, 7752, 2907, 798, 152, 18, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005
Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005
T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005
Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007
Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007
Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007
This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008
T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008
The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012
O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:
  ((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012
The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)
The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). -  Tom Copeland, Dec 11 2015 (revised Dec 28 2015)
T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017
Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

			Triangle T(n,k) starts:
n\k     0      1      2      3      4     5    6    7   8  9 10
0:      1
1:      2      1
2:      5      4      1
3:     14     14      6      1
4:     42     48     27      8      1
5:    132    165    110     44     10     1
6:    429    572    429    208     65    12    1
7:   1430   2002   1638    910    350    90   14    1
8:   4862   7072   6188   3808   1700   544  119   16   1
9:  16796  25194  23256  15504   7752  2907  798  152  18  1
10: 58786  90440  87210  62016  33915 14364 4655 1120 189 20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012.
Production matrix begins:
2, 1
1, 2, 1
0, 1, 2, 1
0, 0, 1, 2, 1
0, 0, 0, 1, 2, 1
0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 0, 1, 2, 1
- _Philippe Deléham_, Nov 07 2011
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: T(5,1) = 165 = 1*42 + 2*48 +1*27. The Riordan A-sequence is [1,2,1].
Recurrence from Riordan Z-sequence [2,1]: T(5,0) = 132 = 2*42 + 1*48. (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
  Example for rho(N) = 2*cos(Pi/N) powers:
  n=2: rho(N)^5 = 5*R(N, 2) + 4*R(N, 4) + 1*R(N, 6) = 5*S(1, rho(N)) + 4*S(3, rho(N)) + 1*S(5, rho(N)), identical in N >= 1. For N=5 (the pentagon with only one distinct diagonal) the degree delta(5) = 2, hence R(5, 4) and R(5, 6) can be reduced, namely to R(5, 1) = 1 and R(5, 6) = -R(5,1) = -1, respectively. Thus rho(5)^5 = 5*R(N, 2) + 4*1  + 1*(-1) = 3 + 5*R(N, 2) = 3 + 5*rho(5), with the golden section rho(5). (End)

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Mirror image of A050166. Row sums are A001700.

Cf. A000108, A008313, A039599, A183134, A094527, A033184, A035324, A053122, A172417.

/* As triangle: */ [[Binomial(2*n,n-k) - Binomial(2*n,n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2015

T:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); # N. J. A. Sloane, Aug 26 2013
# Uses function PMatrix from A357368. Adds row and column above and to the left.
PMatrix(10, n -> binomial(2*n, n) / (n + 1)); # Peter Luschny, Oct 07 2022

Flatten[Table[Binomial[2n, n-k] - Binomial[2n, n-k-2], {n,0,9}, {k,0,n}]] (* Jean-François Alcover, May 03 2011 *)

T(n,k)=binomial(2*n,n-k) - binomial(2*n,n-k-2) \\ Charles R Greathouse IV, Nov 07 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle.
def A039598_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1
    for i in range(2*n) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
        if b : print([D[z] for z in (1..h-1) ])
A039598_triangle(10)  # Peter Luschny, May 01 2012

Row n: C(2n, n-k) - C(2n, n-k-2).
a(n, k) = C(2n+1, n-k)*2*(k+1)/(n+k+2) = A050166(n, n-k) = a(n-1, k-1) + 2*a(n-1, k)+ a (n-1, k+1) [with a(0, 0) = 1 and a(n, k) = 0 if n<0 or nHenry Bottomley, Sep 24 2001
From Philippe Deléham, Feb 14 2004: (Start)
T(n, 0) = A000108(n+1), T(n, k) = 0 if n0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+2) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
Sum_{k>=0} T(m, k)*T(n, k) = A000108(m+n+1). (End)
T(n, k) = A009766(n+k+1, n-k) = A033184(n+k+2, 2k+2). - Philippe Deléham, Feb 14 2004
Sum_{j>=0} T(k, j)*A039599(n-k, j) = A028364(n, k). - Philippe Deléham, Mar 04 2004
Antidiagonal Sum_{k=0..n} T(n-k, k) = A000957(n+3). - Gerald McGarvey, Jun 05 2005
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super- and subdiagonals and [2,2,2,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
G.f.: G(t,x) = C^2/(1-txC^2), where C = (1-sqrt(1-4x))/(2x) is the Catalan function. From here G(-1,x)=C, i.e., the alternating row sums are the Catalan numbers (A000108). - Emeric Deutsch, Jan 20 2007
Sum_{k=0..n} T(n,k)*x^k = A000957(n+1), A000108(n), A000108(n+1), A001700(n), A049027(n+1), A076025(n+1), A076026(n+1) for x=-2,-1,0,1,2,3,4 respectively (see square array in A067345). - Philippe Deléham, Mar 21 2007, Nov 04 2011
Sum_{k=0..n} T(n,k)*(k+1) = 4^n. - Philippe Deléham, Mar 30 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A035324(n,k), A035324 with offset 0 (0 <= k <= n). - Philippe Deléham, Mar 30 2007
T(n,k) = A053121(2*n+1,2*k+1). - Philippe Deléham, Apr 16 2007, Apr 18 2007
T(n,k) = A039599(n,k) + A039599(n,k+1). - Philippe Deléham, Sep 11 2007
Sum_{k=0..n+1} T(n+1,k)*k^2 = A029760(n). - Philippe Deléham, Dec 16 2007
Sum_{k=0..n} T(n,k)*A059841(k) = A000984(n). - Philippe Deléham, Nov 12 2008
G.f.: 1/(1-xy-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-.... (continued fraction).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A001700(n), A194723(n+1), A194724(n+1), A194725(n+1), A194726(n+1), A194727(n+1), A194728(n+1), A194729(n+1), A194730(n+1) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Nov 03 2011
From Peter Bala, Dec 21 2014: (Start)
This triangle factorizes in the Riordan group as ( C(x), x*C(x) ) * ( 1/(1 - x), x/(1 - x) ) = A033184 * A007318, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Let U denote the lower unit triangular array with 1's on or below the main diagonal and zeros elsewhere. For k = 0,1,2,... define U(k) to be the lower unit triangular block array
/I_k 0\
\ 0  U/ having the k X k identity matrix I_k as the upper left block; in particular, U(0) = U. Then this array equals the bi-infinite product (...*U(2)*U(1)*U(0))*(U(0)*U(1)*U(2)*...). (End)
From Peter Bala, Jul 21 2015: (Start)
O.g.f. G(x,t) = (1/x) * series reversion of ( x/f(x,t) ), where f(x,t) = ( 1 + (1 + t)*x )^2/( 1 + t*x ).
1 + x*d/dx(G(x,t))/G(x,t) = 1 + (2 + t)*x + (6 + 4*t + t^2)*x^2 + ... is the o.g.f for A094527. (End)
Conjecture: Sum_{k=0..n} T(n,k)/(k+1)^2 = H(n+1)*A000108(n)*(2*n+1)/(n+1), where H(n+1) = Sum_{k=0..n} 1/(k+1). - Werner Schulte, Jul 23 2015
From Werner Schulte, Jul 25 2015: (Start)
Sum_{k=0..n} T(n,k)*(k+1)^2 = (2*n+1)*binomial(2*n,n). (A002457)
Sum_{k=0..n} T(n,k)*(k+1)^3 = 4^n*(3*n+2)/2.
Sum_{k=0..n} T(n,k)*(k+1)^4 = (2*n+1)^2*binomial(2*n,n).
Sum_{k=0..n} T(n,k)*(k+1)^5 = 4^n*(15*n^2+15*n+4)/4. (End)
The o.g.f. G(x,t) is such that G(x,t+1) is the o.g.f. for A035324, but with an offset of 0, and G(x,t-1) is the o.g.f. for A033184, again with an offset of 0. - Peter Bala, Sep 20 2015
Denote this lower triangular array by L; then L * transpose(L) is the Cholesky factorization of the Hankel matrix ( 1/(i+j)*binomial(2*i+2*j-2, i+j-1) )A172417%20read%20as%20a%20square%20array.%20See%20Chamberland,%20p.%201669.%20-%20_Peter%20Bala">i,j >= 1 = A172417 read as a square array. See Chamberland, p. 1669. - _Peter Bala, Oct 15 2023

Typo in one entry corrected by Philippe Deléham, Dec 16 2007

A032443 a(n) = Sum_{i=0..n} binomial(2*n, i).

Original entry on oeis.org

1, 3, 11, 42, 163, 638, 2510, 9908, 39203, 155382, 616666, 2449868, 9740686, 38754732, 154276028, 614429672, 2448023843, 9756737702, 38897306018, 155111585372, 618679078298, 2468152192772, 9848142504068, 39301087452632, 156861290196878, 626155256640188

Offset: 0

J. H. Conway, Dec 11 1999

Array interpretation: first row is filled with 1's, first column with powers of 2, b(i,j) = b(i-1,j) + b(i,j-1); then a(n) = b(n,n). - Benoit Cloitre, Apr 01 2002
   1   1   1   1   1   1   1 ...
   2   3   4   5   6   7   8 ...
   4   7  11  16  22 ...
   8  15  26  42  64 ...
  16  31  ..  99 163 ...
From Gary W. Adamson, Dec 27 2008: (Start)
This is an analog of the Catalan sequence: Let M denote an infinite Cartan matrix (-1's in the super and subdiagonals and (2,2,2,...) in the main diagonal which we modify to (1,2,2,2,...)). Then A000108 can be generated by accessing the leftmost term in M^n * [1,0,0,0,...]. Change the operation to M^n * [1,2,3,...] to get this sequence. Or, take iterates M * [1,2,3,...] -> M * ANS, -> M * ANS,...; accessing the leftmost term. (End)
Convolved with the Catalan sequence, A000108: (1, 1, 2, 5, 14, ...) = powers of 4, A000302: (1, 4, 16, 64, ...). - Gary W. Adamson, May 15 2009
Row sums of A094527. - Paul Barry, Sep 07 2009
Hankel transform of the aeration of this sequence is C(n+2, 2). - Paul Barry, Sep 26 2009
Number of 4-ary words of length n in which the number of 1's does not exceed the number of 0's. - David Scambler, Aug 14 2012
Number of options available to a voter who has up to n (0-n) votes to allot among 2n candidates. - Lorraine Lee, Apr 27 2019
2*a(n-1) is the number of all triangulations that can be obtained from a Möbius strip with n chosen points on its edge. See Bazier-Matte et al. - Michel Marcus, Sep 15 2020

			G.f. = 1 + 3*x + 11*x^2 + 42*x^3 + 163*x^4 + 638*x^5 + 2510*x^6 + 9908*x^7 + ...
According to the second formula, we see the fourth row of Pascal's triangle has the terms 1,4,6,4,1 and the partial sums are 1,5,11,15,16. Using these we get 1*1 + 4*5 + 6*11 + 4*15*1*16 = 1 + 20 + 66 + 60 + 16 = 163 = a(4). - _J. M. Bergot_, Apr 29 2014

D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Véronique Bazier-Matte, Ruiyan Huang, and Hanyi Luo, Number of Triangulations of a Möbius Strip, arXiv:2009.05785 [math.CO], 2020.
A. Bernini, F. Disanto, R. Pinzani and S. Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007) # 07.9.7.
Celeste Damiani, Paul Martin, and Eric C. Rowell, Generalisations of Hecke algebras from Loop Braid Groups, arXiv:2008.04840 [math.GT], 2020.
M. Klazar, Twelve countings with rooted plane trees, European Journal of Combinatorics 18 (1997), 195-210; Addendum, 18 (1997), 739-740.
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.

Binomial transform of A027914. Hankel transform is {1, 2, 3, 4, ..., n, ...}.

Cf. A000108, A000302, A110166.

[(4^n+Binomial(2*n, n))/2:n in [0.. 30]]; // Vincenzo Librandi, Oct 18 2014

A032443:=n->(4^n + binomial(2*n, n))/2; seq(A032443(n), n=0..30); # Wesley Ivan Hurt, Apr 15 2014

a[ n_] := If[ n<-3, 0, (4^n + Binomial[2 n, n]) / 2]; Table[a[n], {n, 0, 30}] (* Michael Somos, Jan 25 2014 *)

A032443 = n->sum(i=0,n,binomial(2*n,i)) \\ M. F. Hasler, Jan 02 2014

A032443 = n->binomial(2*n-1,n)+4^n\2 \\ M. F. Hasler, Jan 02 2014

{a(n) = if( n<-3, 0, (4^n + binomial(2*n, n)) / 2)} /* Michael Somos, Jan 25 2014 */

a(n) = (4^n+binomial(2*n, n))/2. - David W. Wilson
a(n) = Sum_{0 <= i_1 <= i_2 <= n} binomial(n, i_2) * binomial(n, i_1+i_2). - Benoit Cloitre, Oct 14 2004
Sequence with interpolated zeros has a(n) = Sum_{k=0..floor(n/2)} (if((n-2k) mod 2)=0, C(n, k), 0). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=0..n} C(n+k-1, k)*2^(n-k). - Paul Barry, Sep 28 2007
E.g.f.: exp(2*x)*(exp(2*x) + BesselI(0,2*x))/2. For BesselI see Abramowitz-Stegun (reference and link under A008277), p. 375, eq. 9.6.10. See also A000984 for its e.g.f. - Wolfdieter Lang, Jan 16 2012
From Sergei N. Gladkovskii, Aug 13 2012: (Start)
G.f.: (1/sqrt(1-4*x) + 1/(1-4*x))/2 = G(0)/2 where G(k) = 1 + ((2*k)!)/(k!)^2/(4^k - 4*x*(16^k)/( 4*x*(4^k) + ((2*k)!)/(k!)^2/G(k+1))); (continued fraction).
E.g.f.: G(0)/2 where G(k) = 1 + ((2*k)!)/(k!)^2/(4^k - 4*x*(16^k)/( 4*x*(4^k) + (k+1)*((2*k)!)/(k!)^2/G(k+1))); (continued fraction).
(End)
O.g.f.: (1 - x*(2 + c(x)))/(1 - 4*x)^(3/2), with c the o.g.f. of A000108 (Catalan). - Wolfdieter Lang, Nov 22 2012
D-finite with recurrence: n*a(n) + 2*(-4*n+3)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Dec 04 2012
a(n) = binomial(2*n-1,n) + floor(4^n/2), or a(n+1) = A001700(n) + A004171(n), for all n >= 0. See A000346 for the difference. - M. F. Hasler, Jan 02 2014
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n > -4. - Michael Somos, Jan 25 2014
a(n) = coefficient of x^n in (4*x + 1 / (1 + x))^n. - Michael Somos, Jan 25 2014
Binomial transform is A110166. - Michael Somos, Jan 25 2014
Asymptotics: a(n) ~ 2^(2*n-1)*(1+1/sqrt(Pi*n)). - Fung Lam, Apr 13 2014
Self-convolution is A240879. - Fung Lam, Apr 13 2014
a(0) = 1, a(n+1) = A001700(n) + 2^(2n+1). - Philippe Deléham, Oct 11 2014
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 10*x^2 + 35*x^3 + 126*x^4 + ... is the o.g.f. for A001700. - Peter Bala, Jul 21 2015
a(n) = 4*a(n-1) - A000108(n-1). - Bob Selcoe, Apr 02 2017
a(n) = [x^n] 1/((1 - x)^n*(1 - 2*x)). - Ilya Gutkovskiy, Oct 12 2017
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 08 2022
O.g.f.: ( hypergeom([1/4, 3/4], [1/2], 4*x) )^2. - Peter Bala, Mar 04 2022
a(n) = binomial(2*n, n) * hypergeom([1, -n], [n + 1], -1). - Peter Luschny, Oct 06 2023

A100100 Triangle T(n,k) = binomial(2*n-k-1, n-k) read by rows.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 10, 6, 3, 1, 35, 20, 10, 4, 1, 126, 70, 35, 15, 5, 1, 462, 252, 126, 56, 21, 6, 1, 1716, 924, 462, 210, 84, 28, 7, 1, 6435, 3432, 1716, 792, 330, 120, 36, 8, 1, 24310, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 92378, 48620, 24310, 11440, 5005, 2002

Offset: 0

Paul Barry, Nov 08 2004

Sometimes called a Catalan triangle, although there are many other triangles that carry that name - see A009766, A008315, A028364, A033184, A053121, A059365, A062103.
Number of nodes of outdegree k in all ordered trees with n edges. Equivalently, number of ascents of length k in all Dyck paths of semilength n. Example: T(3,2) = 3 because the Dyck paths of semilength 3 are UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and UUUDDD, where U = (1,1), D = (1,-1), the ascents of length 2 being shown between parentheses. - Emeric Deutsch, Nov 19 2006
Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A088218 and g(x) is the g.f. of A000108. - Philippe Deléham, Jan 23 2010
T(n,k) is the number of nonnegative paths of upsteps U = (1,1) and downsteps D = (1,-1) of length 2*n with k returns to  ground level, the horizontal line through the initial vertex. Example: T(2,1) = 2 counts UDUU, UUDD. Also, T(n,k) = number of these paths whose last descent has length k, that is, k downsteps follow the last upstep. Example: T(2,1) = 2 counts UUUD, UDUD. - David Callan, Nov 21 2011
Belongs to the hitting-time subgroup of the Riordan group. Multiplying this triangle by the square Pascal matrix gives A092392 read as a square array. See the example below. - Peter Bala, Nov 03 2015

			From _Paul Barry_, Mar 15 2010: (Start)
Triangle begins in row n=0 with columns 0<=k<=n as:
    1;
    1,   1;
    3,   2,   1;
   10,   6,   3,  1;
   35,  20,  10,  4,  1;
  126,  70,  35, 15,  5, 1;
  462, 252, 126, 56, 21, 6, 1;
Production matrix begins
  1, 1;
  2, 1, 1;
  3, 1, 1, 1;
  4, 1, 1, 1, 1;
  5, 1, 1, 1, 1, 1;
  6, 1, 1, 1, 1, 1, 1;
  7, 1, 1, 1, 1, 1, 1, 1;
(End)
A092392 as a square array = A100100 * square Pascal matrix:
/1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
|2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
|6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
|20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
|70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, An algebraic structure for Faber polynomials, Lin. Alg. Applic. 433 (2010) 1170-1179.
Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009), no. 12, 3962-3974.

Row sums are A000984. Equivalent to A092392, to which A088218 has been added as a first column. Columns include A088218, A000984, A001700, A001791, A002054, A002694. Diagonal sums are A100217. Matrix inverse is A100218.

Cf. A059481 (mirrored). Cf. A033184, A094527, A113955.

a100100 n k = a100100_tabl !! n !! n
a100100_row n = a100100_tabl !! n
a100100_tabl = [1] : f a092392_tabl where
   f (us : wss'@(vs : wss)) = (vs !! 1 : us) : f wss'
-- Reinhard Zumkeller, Jan 15 2014

/* As triangle */ [[Binomial(2*n - k - 1, n - k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 21 2018

A100100 := proc(n,k)
    binomial(2*n-k-1,n-1) ;
end proc:
seq(seq(A100100(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Flatten[Table[Binomial[2 n - k - 1, n - k], {n, 0, 11}, {k, 0, n}]] (* Vincenzo Librandi, Nov 21 2018 *)

T(n,k)=binomial(2*n-k-1,n-k) \\ Charles R Greathouse IV, Jan 16 2012

From Peter Bala, Sep 06 2015: (Start)
Matrix product A094527 * P^(-1) = A113955 * P^(-2), where P denotes Pascal's triangle A007318.
Essentially, the logarithmic derivative of A033184. (End)
Let column(k) = [T(n, k), n >= k], then the generating function for column(k) is (2/(sqrt(1-4*x)+1))^(k-1)/sqrt(1-4*x). - Peter Luschny, Mar 19 2021
O.g.f. row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, i.e. o.g.f. of the triangle, G(z,x) = 1/((2 - c(z))*(1 - x*z*c(z))), with c the o.g.f. of A000108 (Catalan). See the Riordan coomment by Philippe Deléham above. - Wolfdieter Lang, Apr 06 2021

A110162 Riordan array ((1-x)/(1+x), x/(1+x)^2).

Original entry on oeis.org

1, -2, 1, 2, -4, 1, -2, 9, -6, 1, 2, -16, 20, -8, 1, -2, 25, -50, 35, -10, 1, 2, -36, 105, -112, 54, -12, 1, -2, 49, -196, 294, -210, 77, -14, 1, 2, -64, 336, -672, 660, -352, 104, -16, 1, -2, 81, -540, 1386, -1782, 1287, -546, 135, -18, 1, 2, -100, 825, -2640, 4290, -4004, 2275, -800, 170, -20, 1

Offset: 0

Paul Barry, Jul 14 2005

Inverse of Riordan array A094527. Rows sums are A099837. Diagonal sums are A110164. Product of Riordan array A102587 and inverse binomial transform (1/(1+x), x/(1+x)).
Coefficients of polynomials related to Cartan matrices of types C_n and B_n: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2), with p(x,0) = 1; p(x,1) = 2-x; p(x,2) = x^2-4*x-2. - Roger L. Bagula, Apr 12 2008
From Wolfdieter Lang, Nov 16 2012: (Start)
The alternating row sums are given in A219233.
For n >= 1 the row polynomials in the variable x^2 are R(2*n,x):=2*T(2*n,x/2) with Chebyshev's T-polynomials. See A127672 and also the triangle A127677.
(End)
From Peter Bala, Jun 29 2015: (Start)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = x/(1 + x)^2 and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 - 2*x + sqrt(1 - 4*x))/2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)

			Triangle T(n,k) begins:
m\k  0    1    2     3     4     5     6    7    8   9 10 ...
0:   1
1:  -2    1
2:   2   -4    1
3:  -2    9   -6     1
4:   2  -16   20    -8     1
5:  -2   25  -50    35   -10     1
6:   2  -36  105  -112    54   -12     1
7:  -2   49 -196   294  -210    77   -14    1
8:   2  -64  336  -672   660  -352   104  -16    1
9:  -2   81 -540  1386 -1782  1287  -546  135  -18   1
10:  2 -100  825 -2640  4290 -4004  2275 -800  170 -20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 16 2012
Row polynomial n=2: P(2,x) = 2 - 4*x + x^2. R(4,x):= 2*T(4,x/2) = 2 - 4*x^2 + x^4. For P and R see a comment above. - _Wolfdieter Lang_, Nov 16 2012.

G. C. Greubel, Rows n=0..100 of triangle, flattened
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
T. M. Richardson, The Reciprocal Pascal Matrix, arXiv:1405.6315 [math.CO], 2014.

Cf. A128411. See A127677 for an almost identical triangle.

Cf. A136674, A053122.

/* As triangle */ [[(-1)^(n-k)*(Binomial(n+k,n-k) + Binomial(n+k-1,n-k-1)): k in [0..n]]: n in [0.. 12]]; // Vincenzo Librandi, Jun 30 2015

Table[If[n==0 && k==0, 1, (-1)^(n-k)*(Binomial[n+k, n-k] + Binomial[n+k-1, n-k-1])], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 16 2018 *)

{T(n,k) = (-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1))};
for(n=0, 12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 16 2018

[[(-1)^(n-k)*(binomial(n+k,n-k) + binomial(n+k-1,n-k-1)) for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 16 2018

T(n,k) = (-1)^(n-k)*(C(n+k,n-k) + C(n+k-1,n-k-1)), with T(0,0) = 1. - Paul Barry, Mar 22 2007
From Wolfdieter Lang, Nov 16 2012: (Start)
O.g.f. row polynomials P(n,x) := Sum(T(n,k)*x^k, k=0..n): (1-z^2)/(1+(x-2)*z+z^2) (from the Riordan property).
O.g.f. column No. k: ((1-x)/(1+x))*(x/(1+x)^2)^k, k >= 0.
T(0,0) = 1, T(n,k) = (-1)^(n-k)*(2*n/(n+k))*binomial(n+k,n-k), n>=1, and T(n,k) = 0 if n < k. (From the Chebyshev T-polynomial formula due to Waring's formula.)
(End)
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 29 2013

A094531 Array read by rows: right-hand side of triangle A027907 of trinomial coefficients.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 7, 6, 3, 1, 19, 16, 10, 4, 1, 51, 45, 30, 15, 5, 1, 141, 126, 90, 50, 21, 6, 1, 393, 357, 266, 161, 77, 28, 7, 1, 1107, 1016, 784, 504, 266, 112, 36, 8, 1, 3139, 2907, 2304, 1554, 882, 414, 156, 45, 9, 1, 8953, 8350, 6765, 4740, 2850, 1452, 615, 210, 55

Offset: 0

Paul Barry, May 07 2004

Sometimes called a Motzkin triangle, although that name is usually reserved for A026300.
Expand (1+x+x^2)^n and take last (nonzero) coefficient of first row, last two coefficients of second row, etc.
Equals A094531*(1,xc(-x^2)) where c(x) is the g.f. of A000108. - Paul Barry, May 12 2009
Coefficients of Faber polynomials for (1/x+1+x): Fa(n,x) = Sum_{k=0..n} T(n,k)*x^k, g.f.: -log((sqrt(-3*t^2-2*t+1)-t+1)/2-t*x) = Sum_{n>0} Fa(n,x)*t^n/n. - Vladimir Kruchinin, Jul 01 2013

			Triangle begins:
    1;
    1,   1;
    3,   2,   1;
    7,   6,   3,   1;
   19,  16,  10,   4,   1;
   51,  45,  30,  15,   5,   1;
  141, 126,  90,  50,  21,   6,   1;
  393, 357, 266, 161,  77,  28,   7,   1;
  ...

Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. (2009) Vol 309, No. 12, 3962-3974.
Ana Luzón, Donatella Merlini, Manuel A. Morón, and Renzo Sprugnoli, Complementary Riordan arrays, Discrete Applied Mathematics, 172 (2014) 75-87.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012.
Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 23.
Paul Peart and Wen-jin Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
Louis W. Shapiro, Seyoum Getu, Wen-Jin Woan, and Leon C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.
Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, Some matrix identities on colored Motzkin paths, Discrete Mathematics 340.12 (2017): 3081-3091.

Binomial transform is triangle A094527. Row sums are A027914.

Cf. A111808 (row reversed).

T := (n, k) -> simplify(GegenbauerC(n-k, -n, -1/2)):
for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, May 12 2016

max = 10; se = Series[ -Log[ (Sqrt[-3*t^2 - 2*t + 1] - t + 1)/2 - t*x], {t, 0, max + 1}, {x, 0, max}]; a[n_, k_] := SeriesCoefficient[se, {t, 0, n}, {x, 0, k}]*n; a[0, 0] = 1; Table[a[n, k], {n, 0, max }, {k, 0, n}] // Flatten  (* Jean-François Alcover, Jul 02 2013, after Vladimir Kruchinin *)
Table[Binomial[n, k] Hypergeometric2F1[(k - n)/2, (k - n + 1)/2, k + 1, 4], {n, 0, 9}, {k, 0, n}] // Flatten (* or *)
Table[If[n == 0, 1, GegenbauerC[n - k, -n, -1/2]], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, May 12 2016 *)

Riordan array ( 1/sqrt(1-2*x-3*x^2), (1-x-sqrt(1-2*x-3*x^2))/(2*x) ). - N. J. A. Sloane, Jun 02 2005
Product of Riordan arrays (1/(1-x), x/(1-x)) (Pascal's triangle, A007318) and (1/sqrt(1-4x^2), (1-sqrt(1-4*x^2))/(2*x)) (A108044). Inverse is A102587. - Paul Barry, Jul 14 2005
Column k has e.g.f. exp(x)*Bessel_I(k, 2x). - Paul Barry, Jul 14 2005
T(n, k) = Sum_{i=0..n} C(n-k-i, i)*C(n, k+i). - Paul Barry, Nov 04 2005
T(n, k) = Sum_{j=0..n} C(n,j)*C(j,n-k-j). - Paul Barry, Oct 25 2006
From Paul Barry, May 12 2009: (Start)
Production matrix is
  1, 1;
  2, 1, 1;
  0, 1, 1, 1;
  0, 0, 1, 1, 1;
  0, 0, 0, 1, 1, 1; (End)
From Peter Bala, Jun 29 2015: (Start)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = (1 - x -sqrt(1 - 2*x - 3*x^2))/(2*x) and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan, Example 1.1).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = 1 + x + x^2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)
From Peter Luschny, May 12 2016: (Start)
T(n,k) = binomial(n, k)*hypergeom([(k-n)/2, (k-n+1)/2], [k+1], 4):
T(n,k) = GegenbauerC(n-k, -n, -1/2). (End)

cp's OEIS Frontend

A088218 Total number of leaves in all rooted ordered trees with n edges.

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A005810 a(n) = binomial(4n,n).

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A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

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A032443 a(n) = Sum_{i=0..n} binomial(2*n, i).

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A100100 Triangle T(n,k) = binomial(2*n-k-1, n-k) read by rows.

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A128417 Number triangle T(n,k) = 2^(n-k)C(2n,n-k).

A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2x + 1)^n + (x^2 - 1)(x + 1)^n.

A321127 Irregular triangle read by rows: row n gives the coefficients in the expansion of ((x + 1)^(2n) + (x^2 - 1)(2*(x + 1)^n - 1))/x.