A037965 -id:A037965 - cp's OEIS frontend

A033876 Expansion of 1/(2x) (1/(1-4*x)^(3/2)-1).

3, 15, 70, 315, 1386, 6006, 25740, 109395, 461890, 1939938, 8112468, 33801950, 140408100, 581690700, 2404321560, 9917826435, 40838108850, 167890003050, 689232644100, 2825853840810, 11572544300460, 47342226683700, 193485622098600, 790066290235950, 3223470464162676

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

a(n) is the trace of the zigzag matrix Z(n+1) (see A088961). - Paul Boddington, Nov 03 2003
The number of edges in the odd graph O_k (for k >= 2) can be computed as 0.5*(2k-1)*C(2k-2,k-1). This sequence gives the number of edges in O_k for integer values of k from k=2. - K.V.Iyer, Mar 04 2009
Apparently the number of peaks in all symmetric Dyck paths with semilength 2n+2. - David Scambler, Apr 29 2013
For n > 0, also the number of maximal and maximum cliques in the (n+2)-odd graph. - Eric W. Weisstein, Nov 30 2017

			G.f. = 3 + 15*x + 70*x^2 + 315*x^3 + 1386*x^4 + 6006*x^5 + 25740*x^6 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..250
Eric Weisstein's World of Mathematics, Maximal Clique.
Eric Weisstein's World of Mathematics, Maximum Clique.
Eric Weisstein's World of Mathematics, Odd Graph.

Cf. A000984, A001622, A001700, A001803, A002457, A037965, A088961.

a033876 n = sum $ zipWith (!!) zss [0..n] where
   zss = take (n+1) $ g (take (n+1) (1 : [0,0..])) where
       g us = (take (n+1) $ g' us) : g (0 : init us)
       g' vs = last $ take (2 * n + 3) $
                      map snd $ iterate h (0, vs ++ reverse vs)
   h (p,ws) = (1 - p, drop p $ zipWith (+) ([0] ++ ws) (ws ++ [0]))
-- Reinhard Zumkeller, Oct 25 2013

[(2*n+3)*Binomial(2*n+1, n) : n in [0..40]]; // Wesley Ivan Hurt, Nov 30 2017

[seq((n+2)*binomial(2*(n+2),n+2)/4, n=0..22)]; # Zerinvary Lajos, Jan 04 2007

Table[nn = 2 n + 1; (2 n + 1)! Coefficient[Series[Exp[x] (x^n/n!)^2/2, {x, 0, nn}], x^(2 n + 1)], {n, 30}] (* Geoffrey Critzer, Apr 19 2017 *)
Table[n Binomial[2 n, n]/4, {n, 2, 20}] (* Eric W. Weisstein, Nov 30 2017 *)
Table[(4^n Gamma[n + 3/2])/(Sqrt[Pi] Gamma[n + 1]), {n, 20}] (* Eric W. Weisstein, Nov 30 2017 *)
CoefficientList[Series[((1 - 4 x)^(-3/2) - 1)/(2 x), {x, 0, 20}], x] (* Eric W. Weisstein, Nov 30 2017 *)

x='x+O('x^66); Vec( 1/(2*x) * (1/(1-4*x)^(3/2)-1) ) \\ Joerg Arndt, May 01 2013

a(n) = (2*n+3)*binomial(2*n+1, n). - Paul Boddington, Nov 03 2003
Equals n*A000984/4, n >= 2. - Zerinvary Lajos, Jan 04 2007
For n >= 1, 1/a(n-1) = Sum_{k>=0} binomial(2*k,k)/(4^(n+k)*(n+k+1)) = int(4*t^n/sqrt(1-4*t), t=0..1/4). - Groux Roland, Jan 17 2011
G.f.: - 1/(2*x) + G(0)/(4*x), where G(k)= 1 + 1/(1 - 2*x*(2*k+3)/(2*x*(2*k+3) + (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 18 2013
a(n) = 2^(2*n+1)*binomial(n+3/2, 1/2). - Peter Luschny, May 06 2014
0 = a(n)*(16*a(n+1) - 2*a(n+2)) + a(n+1)*(-6*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
a(n-2) = n*binomial(2*n, n)/4 for n > 1. - Eric W. Weisstein, Nov 30 2017
G.f.: ((1 - 4*x)^(-3/2) - 1)/2 (by definition). - Eric W. Weisstein, Nov 30 2017
D-finite with recurrence: (n+1)*a(n) +2*(-2*n-3)*a(n-1)=0. - R. J. Mathar, Jan 28 2020
G.f.: (1F0(3/2;;4*x)-1)/(2*x). - R. J. Mathar, Jan 28 2020
From Amiram Eldar, Mar 04 2023: (Start)
Sum_{n>=0} 1/a(n) = 4*Pi/(3*sqrt(3)) - 2.
Sum_{n>=0} (-1)^n/a(n) = 2 - 8*log(phi)/sqrt(5), where phi is the golden ratio (A001622). (End)
From Mélika Tebni, Sep 04 2024: (Start)
a(n) = A037965(n+2) - A001700(n).
E.g.f.: exp(2*x)*((3+8*x)*BesselI(0, 2*x) + (1+8*x)*BesselI(1, 2*x)). (End)
a(n) = 2^n*JacobiP(n+1, 1/2, -n-1, 3). - Peter Luschny, Jan 22 2025

A037966 a(n) = n^2binomial(2n-2, n-1).

Original entry on oeis.org

0, 1, 8, 54, 320, 1750, 9072, 45276, 219648, 1042470, 4862000, 22355476, 101582208, 457002364, 2038517600, 9026235000, 39710085120, 173712232710, 756088415280, 3276123843300, 14138105520000, 60790319209620, 260516811228960, 1113068351807880, 4742456099097600, 20154752301937500, 85453569951920352

Offset: 0

N. J. A. Sloane

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Nikita Gogin and Mika Hirvensalo, On the Moments of Squared Binomial Coefficients, (2020).
Han Mao Kiah, Alexander Vardy, and Hanwen Yao, Efficient Algorithms for the Bee-Identification Problem, arXiv:2212.09952 [cs.IT], 2022.

Cf. A000108, A000984, A037965, A336214.

[0] cat [n^3*Catalan(n-1): n in [1..30]]; // G. C. Greubel, Jun 19 2022

Array[#^2*Binomial[2#-2, #-1] &, 27, 0] (* Michael De Vlieger, Jul 15 2020 *)

{a(n) = n^2*binomial(2*n-2, n-1)} \\ Seiichi Manyama, Jul 15 2020

[n^3*catalan_number(n-1) for n in (0..30)] # G. C. Greubel, Jun 19 2022

a(n) = Sum_{k=0..n} k^2*binomial(n,k)^2. - Paul Barry, Mar 04 2003
a(n) = n^2*A000984(n-1). - Zerinvary Lajos, Jan 18 2007, corrected Jul 26 2015
a(n) = n*A037965(n). - Zerinvary Lajos, Jan 18 2007, corrected Jul 26 2015
(n-1)^3*a(n) = 2*n^2*(2*n-3)*a(n-1). - R. J. Mathar, Jul 26 2015
E.g.f.: x*exp(2*x)*((1 + 2*x)*BesselI(0,2*x) + 2*x*BesselI(1,2*x)). - Ilya Gutkovskiy, Mar 04 2021

More terms from Seiichi Manyama, Jul 15 2020

A037972 a(n) = n^2(n+1)binomial(2*n-2, n-1)/2.

Original entry on oeis.org

0, 1, 12, 108, 800, 5250, 31752, 181104, 988416, 5212350, 26741000, 134132856, 660284352, 3199016548, 15288882000, 72209880000, 337535723520, 1563410094390, 7182839945160, 32761238433000, 148450107960000, 668693511305820, 2995943329133040, 13356820221694560

Offset: 0

N. J. A. Sloane

nonn

Identity (3.78), S_{3}, in H. W. Gould, Combinatorial Identities, Morgantown, 1972, page 31.

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A000108, A000217, A037965, A037966, A074334, A329913, A329444.

[n^2*(n+1)*Binomial(2*n-2, n-1)/2: n in [0..30]]; // G. C. Greubel, Jun 22 2022

Table[n^2*Binomial[n+1,2]*CatalanNumber[n-1], {n,0,30}] (* G. C. Greubel, Jun 22 2022 *)

{a(n) = n^2*(n+1)*binomial(2*n-2, n-1)/2} \\ Seiichi Manyama, Aug 09 2020

[n^2*binomial(n+1,2)*catalan_number(n-1) for n in (0..30)] # G. C. Greubel, Jun 22 2022

a(n) = Sum_{k=0..n} k^3*(C(n,k))^2. [heruneedollar (heruneedollar(AT)gmail.com), Mar 20 2010]
a(n) = A000217(n)*A037965(n). - R. J. Mathar, Jul 26 2015
(n-1)^2*a(n) = 2*(11*n-16)*a(n-1) + 8*n*(2*n-5)*a(n-2). - R. J. Mathar, Oct 20 2015
(n-1)^3*a(n) = 2*n*(n+1)*(2*n-3)*a(n-1). - R. J. Mathar, Oct 20 2015
G.f.: x * (1 - 2*x + 10*x^2 - 12*x^3) / (1 - 4*x)^(7/2). - Ilya Gutkovskiy, Nov 17 2021

A093375 Array T(m,n) read by ascending antidiagonals: T(m,n) = m*binomial(n+m-2, n-1) for m, n >= 1.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 18, 8, 1, 6, 25, 40, 30, 10, 1, 7, 36, 75, 80, 45, 12, 1, 8, 49, 126, 175, 140, 63, 14, 1, 9, 64, 196, 336, 350, 224, 84, 16, 1, 10, 81, 288, 588, 756, 630, 336, 108, 18, 1, 11, 100, 405, 960, 1470, 1512, 1050, 480, 135, 20, 1, 12

Offset: 1

Ralf Stephan, Apr 28 2004

Number of n-long m-ary words avoiding the pattern 1-1'2'.
T(n,n+1) = Sum_{i=1..n} T(n,i).
Exponential Riordan array [(1+x)e^x, x] as a number triangle. - Paul Barry, Feb 17 2009
From Peter Bala, Jul 22 2014: (Start)
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A059298. (End)

			Array T(m,n) (with rows m >= 1 and columns n >= 1) begins as follows:
   1   1   1   1   1   1 ...
   2   4   6   8  10  12 ...
   3   9  18  30  45  63 ...
   4  16  40  80 140 224 ...
   5  25  75 175 350 630 ...
   ...
Triangle S(n,k) = T(n-k+1, k+1) begins
.n\k.|....0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|....2....1
..2..|....3....4....1
..3..|....4....9....6....1
..4..|....5...16...18....8....1
..5..|....6...25...40...30...10....1
..6..|....7...36...75...80...45...12....1
...

Muniru A Asiru, Antidiagonals, n=1..100 flattened
Yasemin Alp and E. Gokcen Kocer, Exponential Almost-Riordan Arrays, Results Math 79, 173 (2024). See page 7.
Sergey Kitaev and Toufik Mansour, Partially ordered generalized patterns and k-ary words, arXiv:math/0210023 [math.CO], 2002.
Wikipedia, Sheffer sequence.

Rows include A045943. Columns include A002411, A027810.
Main diagonal is A037965. Subdiagonals include A002457.
Antidiagonal sums are A001792.
See A103283 for a signed version.
Cf. A103406, A059298, A073107 (unsigned inverse).

nmax:=14;; T:=List([1..nmax],n->List([1..nmax],k->k*Binomial(n+k-2,n-1)));;
b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][1]][b[i][j][2]]))); # Muniru A Asiru, Aug 07 2018

nmax = 10;
T = Transpose[CoefficientList[# + O[z]^(nmax+1), z]& /@ CoefficientList[(1 - x z)/(1 - z - x z)^2 + O[x]^(nmax+1), x]];
row[n_] := T[[n+1, 1 ;; n+1]];
Table[row[n], {n, 0, nmax}] // Flatten (* Jean-François Alcover, Aug 07 2018 *)

# uses[riordan_array from A256893]
riordan_array((1+x)*exp(x), x, 8, exp=true) # Peter Luschny, Nov 02 2019

Triangle = P*M, the binomial transform of the infinite bidiagonal matrix M with (1,1,1,...) in the main diagonal and (1,2,3,...) in the subdiagonal, and zeros elsewhere. P = Pascal's triangle as an infinite lower triangular matrix. - Gary W. Adamson, Nov 05 2006
From Peter Bala, Sep 20 2012: (Start)
E.g.f. for triangle: (1 + z)*exp((1 + x)*z) = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2/2! + ....
O.g.f. for triangle: (1 - x*z)/(1 - z - x*z)^2 = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2 + ....
The n-th row polynomial R(n,x) of the triangle equals (1+x)^n + n*(1+x)^(n-1) for n >= 0 and satisfies d/dx(R(n,x)) = n*R(n-1,x), as well as R(n,x+y) = Sum_{k = 0..n} binomial(n,k)*R(k,x)*y^(n-k). The row polynomials are a Sheffer sequence of Appell type.
Matrix inverse of the triangle is a signed version of A073107. (End)
From Tom Copeland, Oct 20 2015: (Start)
With offset 0 and D = d/dx, the raising operator for the signed row polynomials P(n,x) is RP = x - d{log[e^D/(1-D)]}/dD = x - 1 - 1/(1-D) =  x - 2 - D - D^2 + ..., i.e., RP P(n,x) = P(n+1,x).
The e.g.f. for the signed array is (1-t) * e^(-t) * e^(x*t).
From the Appell formalism, the row polynomials PI(n,x) of A073107 are the umbral inverse of this entry's row polynomials; that is, P(n,PI(.,x)) = x^n = PI(n,P(.,x)) under umbral composition. (End)
From Petros Hadjicostas, Nov 01 2019: (Start)
As a triangle, we let S(n,k) = T(n-k+1, k+1) = (n-k+1)*binomial(n, k) for n >= 0 and 0 <= k <= n. See the example below.
As stated above by Peter Bala, Sum_{n,k >= 0} S(n,k)*z^n*x^k = (1 - x*z)/(1 - z -x*z)^2.
Also, Sum_{n, k >= 0} S(n,k)*z^n*x^k/n! = (1+z)*exp((1+x)*z).
As he also states, the n-th row polynomial is R(n,x) = Sum_{k = 0..n} S(n, k)*x^k = (1 + x)^n + n*(1 + x)^(n-1).
If we define the signed triangle S*(n,k) = (-1)^(n+k) * S(n,k) = (-1)^(n+k) * T(n-k+1, k+1), as Tom Copeland states, Sum_{n,k >= 0} S^*(n,k)*t^n*x^k/n! = (1-t)*exp((1-x)*(-t)) = (1-t) * e^(-t) * e^(x*t).
Apparently, S*(n,k) = A103283(n,k).
As he says above, the signed n-th row polynomial is P(n,x) =  (-1)^n*R(n,-x) = (x - 1)^n - n*(x - 1)^(n-1).
According to Gary W. Adamson, P(n,x) is "the monic characteristic polynomial of the n X n matrix with 2's on the diagonal and 1's elsewhere." (End)

A097070 Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.

Original entry on oeis.org

1, 2, 9, 40, 175, 756, 3234, 13728, 57915, 243100, 1016158, 4232592, 17577014, 72804200, 300874500, 1240940160, 5109183315, 21002455980, 86213785350, 353452638000, 1447388552610, 5920836618840, 24197138082780, 98801168731200, 403095046038750, 1643337883690776, 6694900194799404

Offset: 1

Amy J. Kolan, Sep 15 2004

Number of compositions of n into n parts, allowing zeros = binomial(2*n-1,n) = A088218 = essentially A001700.

			The compositions for n=2 are 20, 02, 11. There are two 1's in these so a(2) = 2.
From _Robert G. Wilson v_, Sep 16 2004: (Start)
The case n = 5:
A. There are 5 combinations associated with the numbers 50000: 50000, 05000, 00500, 00050, 00005.
B. There are 20 combinations associated with the numbers 41000.
C. There are 20 combinations associated with 32000.
D. There are 30 combinations associated with 31100.
E. There are 30 combinations associated with 22100.
F. There are 20 combinations associated with 21110.
G. There is one combinations associated with 11111.
The number of 1's associated with A is 0, with B 20, with C 0, with D 60, with E 30, with F 60 and with G 5. 0 + 20 + 0 + 60 + 30 + 60 + 5 = 175.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
G.-S. Cheon, H. Kim, and L. W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.

Cf. A037965.
Cf. A000984, A001622, A088218.
cf. A005430, A002740, A002457, A000917, A110609, A097613, A001791, A110609.

List([1..30], n-> n*Binomial(2*n-3, n-1)); # G. C. Greubel, Jul 27 2019

[n*Binomial(2*n-3, n-1): n in [1..30]]; // Vincenzo Librandi, Jul 13 2019

A097070 := n -> ifelse(n=1, 1, 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3)):
seq(simplify(A097070(n)), n = 1..28);  # Peter Luschny, Jan 22 2025

Table[n*Binomial[2n-3, n-1], {n, 30}] (* Robert G. Wilson v, Sep 17 2004 *)

a(n) = n*binomial(2*n-3, n-1); \\ Joerg Arndt, Feb 17 2015

[n*binomial(2*n-3, n-1) for n in (1..30)] # G. C. Greubel, Jul 27 2019

a(n) = n*binomial(2*n-3, n-1).
More generally, total number of k's (k>=0) in all ordered partitions of n into n parts, allowing zeros, is n*binomial(2*n-k-2, n-2) if n >= k, 0 otherwise.
Total number of 0's is given by A005430.
From Vladeta Jovovic, Sep 17 2004: (Start)
a(n) = Sum_{k=0..n} k*binomial(n, k)*binomial(n-2, k-2).
G.f.: x*(1 -2*x +(1-4*x)^(3/2))/(2*(1-4*x)^(3/2)).
E.g.f.: (x/2)*(exp(2*x)*BesselI(0, 2*x)+1). (End)
a(n) = A014107(n)*A000108(n-2). - Philippe Deléham, Apr 12 2007
a(n) = n*A088218(n-1) for n > 0. - Werner Schulte, Jan 22 2017
From Bruce J. Nicholson, Jul 11 2019: (Start)
a(n) = A002740(n) + A097613(n).
a(n) = A110609(n-1) - A002457(n-2) + A097613(n).
a(n) = A005430(n-1) - A000917(n-3) for n > 1.
a(n) = A002457(n-1) - A037965(n) - A000917(n-3) for n > 1.
a(n) = A037965(n)/2.
a(n) = A001700(n-2)*n.
a(n) = A001791(n-2)*n + A000984(n-2)*n for n > 1. (End)
From Amiram Eldar, May 16 2022: (Start)
Sum_{n>=1} 1/a(n) = 4*Pi/(3*sqrt(3)) - Pi^2/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*log(phi)/sqrt(5) - 4*log(phi)^2, where phi is the golden ratio (A001622). (End)
a(n) = 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3) for n > 1. - Peter Luschny, Jan 22 2025

Formula, more terms and comments from Vladeta Jovovic, Sep 15 2004

A123160 Triangle read by rows: T(n,k) = n!(n+k-1)!/((n-k)!(n-1)!*(k!)^2) for 0 <= k <= n, with T(0,0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 9, 18, 10, 1, 16, 60, 80, 35, 1, 25, 150, 350, 350, 126, 1, 36, 315, 1120, 1890, 1512, 462, 1, 49, 588, 2940, 7350, 9702, 6468, 1716, 1, 64, 1008, 6720, 23100, 44352, 48048, 27456, 6435, 1, 81, 1620, 13860, 62370, 162162, 252252, 231660, 115830, 24310

Offset: 0

Roger L. Bagula, Oct 02 2006

T(n,k) is also the number of order-preserving partial transformations (of an n-element chain) of width k (width(alpha) = |Dom(alpha)|). - Abdullahi Umar, Aug 25 2008

			Triangle begins:
  1;
  1,  1;
  1,  4,   3;
  1,  9,  18,  10;
  1, 16,  60,  80,  35;
  1, 25, 150, 350, 350, 126;
  ...

Frederick T. Wall, Chemical Thermodynamics, W. H. Freeman, San Francisco, 1965 pages 296 and 305

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.

Cf. A005773, A037965, A059481, A085373, A088218, A088617, A122899, A123164, A178301.

[Binomial(n,k)*Binomial(n+k-1,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jun 19 2022

T:=proc(n,k) if k=0 and n=0 then 1 elif k<=n then n!*(n+k-1)!/(n-k)!/(n-1)!/(k!)^2 else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, m_]= If [n==m==0, 1, n!*(n+m-1)!/((n-m)!*(n-1)!(m!)^2)];
Table[T[n, m], {n,0,10}, {m,0,n}]//Flatten
max = 9; s = (x+1)/(2*Sqrt[(1-x)^2-4*y])+1/2 + O[x]^(max+2) + O[y]^(max+2); T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]; Table[T[n-k, k], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 18 2015, after Vladimir Kruchinin *)

def A123160(n,k): return binomial(n, k)*binomial(n+k-1, k)
flatten([[A123160(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 19 2022

T(n, m) = n!*(n + m - 1)!/((n - m)!*(n - 1)!(m!)^2), with T(0, 0) = 1.
T(n, k) = binomial(n,k)*binomial(n+k-1,k). The row polynomials (except the first) are (1+x)*P(n,0,1,2x+1), where P(n,a,b,x) denotes the Jacobi polynomial. The columns of this triangle give the diagonals of A122899. - Peter Bala, Jan 24 2008
T(n, k) = binomial(n,k)*(n+k-1)!/((n-1)!*k!).
T(n, k)= binomial(n,k)*binomial(n+k-1,n-1). - Abdullahi Umar, Aug 25 2008
G.f.: (x+1)/(2*sqrt((1-x)^2-4*y)) + 1/2. - Vladimir Kruchinin, Jun 16 2015
From _Peter Bala, Jul 20 2015: (Start)
O.g.f. (1 + x)/( 2*sqrt((1 - x)^2 - 4*x*y) ) + 1/2 = 1 + (1 + y)*x + (1 + 4*y + 3*y^2)*x^2 + ....
For n >= 1, the n-th row polynomial R(n,y) = (1 + y)*r(n-1,y), where r(n,y) is the n-th row polynomial of A178301.
exp( Sum_{n >= 1} R(n,y)*x^n/n ) = 1 + (1 + y)*x + (1 + 3*y + 2*y^2)*x^2 + ... is the o.g.f for A088617. (End)
From G. C. Greubel, Jun 19 2022: (Start)
T(n, n) = A088218(n).
T(n, n-1) = A037965(n).
T(n, n-2) = A085373(n-2).
Sum_{k=0..n} T(n, k) = A123164(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A005773(n). (End)

Edited by N. J. A. Sloane, Oct 26 2006 and Jul 03 2008

A189911 Row sums of the extended Catalan triangle A189231.

Original entry on oeis.org

1, 2, 4, 9, 18, 40, 80, 175, 350, 756, 1512, 3234, 6468, 13728, 27456, 57915, 115830, 243100, 486200, 1016158, 2032316, 4232592, 8465184, 17577014, 35154028, 72804200, 145608400, 300874500, 601749000, 1240940160, 2481880320, 5109183315, 10218366630

Offset: 0

Peter Luschny, May 01 2011

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Luschny, The lost Catalan numbers.

Cf. A037965, A097070, A056040, A212303.

A189911 := proc(n) local a,b,d; if n = 0 then 1 else
a := GAMMA(n-floor(n/2)); b := GAMMA(floor(n/2+3/2));
d := GAMMA(floor(n/2+1))^2; GAMMA(n+1)*(a*b+d)/(a*b*d) fi end: seq(A189911(n),n=0..32);
A189911 := proc(n) h:=irem(n,2); g:=iquo(n,2); (g+h+1)*binomial(2*g+h,g+h) end; # Peter Luschny, Oct 24 2013

a[n_] := Module[{q, r}, {q, r} = QuotientRemainder[n, 2]; (q+r+1)*Pochhammer[q+1, q+r]/(q+r)!]; Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Jan 09 2014 *)

def A189911():
    r, n = 1, 1
    while True:
        yield r
        h = n//2
        r *= 2 if is_even(n) else (h+2)*(2*h+1)/(h+1)^2
        n += 1
a = A189911(); [next(a) for i in range(16)]  # Peter Luschny, Oct 24 2013

Let a = Gamma(n-floor(n/2)), b = Gamma(floor(n/2+3/2)), d = Gamma( floor(n/2+1))^2, c = Gamma(n+1). Then a(n) = c*(a*b+d)/(a*b*d).
a(n) = A162246(n,n) + A162246(n,n+1) for n > 0.
From Peter Luschny, Oct 24 2013 : (Start)
E.g.f.: (x+1)*(BesselI(0, 2*x)+BesselI(1, 2*x)).
O.g.f.: I*(2*x^2-1)/(2*sqrt(2*x+1)*x*(2*x-1)^(3/2))-1/(2*x).
Recurrence: a(0) = 1; a(n) = a(n-1)*2 if n is even else ([n/2]+2)*(2*[n/2]+1)/([n/2]+1)^2. ([.] the floor brackets.)
a(n) = A056040(n) + A212303(n) = n$*(1+[(n+1)/2]^((-1)^n)), where n$ is the swinging factorial.
a(2*n) = (n+1)*C(2*n, n) (A037965);
a(2*n+1) = (n+2)*C(2*n+1, n+1) (A097070). (End)
Sum_{n>=0} 1/a(n) = 4*Pi/sqrt(3) - Pi^2/3 - 2. - Amiram Eldar, Aug 20 2022
D-finite with recurrence: (n-2)*(n+1)^2*a(n) - (2*(n-2)^2+2*n-12)*a(n-1) - 4*(n+2)*(n-1)^2*a(n-2) = 0. - Georg Fischer, Nov 25 2022

A134757 A007318 * A100071.

Original entry on oeis.org

1, 3, 11, 37, 123, 401, 1293, 4131, 13107, 41353, 129873, 406319, 1267093, 3940431, 12224579, 37845117, 116944371, 360771417, 1111332129, 3418840431, 10504903809, 32242682787, 98863833159, 302863592073, 927025884477, 2835306153351, 8665554849903

Offset: 1

Gary W. Adamson, Nov 08 2007

nonn

Also A007318^(-1) * A037965. - Gary W. Adamson, Nov 10 2007

			a(3) = 11 = (1, 2, 1) dot (1, 2, 6) = (1 + 4 + 6), where A100071 = (1, 2, 6, 12, 30, ...).
a(3) = 11 = (1, -2, 1) dot (1, 4, 18) = (1 - 8 + 18). - _Gary W. Adamson_, Nov 10 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A037965, A100071, A107965.

A134757:= func< n | (&+[(-1)^(n-k-1)*(k+1)^2*Binomial(n-1,k)*Catalan(k) : k in [0..n-1]]) >;
[A134757(n): n in [1..40]]; // G. C. Greubel, May 28 2024

a[n_]:= a[n]= Sum[(-1)^(n-k-1)*Binomial[n-1,k]*(k+1)*Binomial[2*k, k], {k,0,n-1}];
Table[a[n], {n,40}] (* G. C. Greubel, May 28 2024 *)

def A134757(n): return sum((-1)^(n-k-1)*(k+1)*binomial(n-1,k)*binomial( 2*k, k) for k in range(n))
[A134757(n) for n in range(1,41)] # G. C. Greubel, May 28 2024

Binomial transform of A100071 starting [1, 2, 6, 12, 30, ...].
Inverse binomial transform of A037965 starting [1, 4, 18, 80, 350, ...].
a(n) = (n-1)! * [x^(n-1)] exp(x)*((1 + 2*x)*BesselI(0, 2*x) + 2*x*BesselI(1, 2*x)) for n>0, a(0) = 0. - Peter Luschny, Aug 26 2012
D-finite with recurrence (n-1)*a(n) = 3*(n-1)*a(n-1) +(n+1)*a(n-2) -3*(n-3)*a(n-3). - R. J. Mathar, Nov 09 2021
G.f.: x*(1-x)/((1-3*x)*sqrt((1+x)*(1-3*x))). - G. C. Greubel, May 28 2024

A141692 Triangle read by rows: T(n,k) = n*(binomial(n - 1, k - 1) - binomial(n - 1, k)), 0 <= k <= n.

Original entry on oeis.org

0, -1, 1, -2, 0, 2, -3, -3, 3, 3, -4, -8, 0, 8, 4, -5, -15, -10, 10, 15, 5, -6, -24, -30, 0, 30, 24, 6, -7, -35, -63, -35, 35, 63, 35, 7, -8, -48, -112, -112, 0, 112, 112, 48, 8, -9, -63, -180, -252, -126, 126, 252, 180, 63, 9, -10, -80, -270, -480, -420, 0, 420, 480, 270, 80, 10

Offset: 0

Roger L. Bagula, Sep 09 2008

The row sums are zero.

Row n consists of the coefficients in the expansion of n*(x - 1)*(x + 1)^(n - 1). - Franck Maminirina Ramaharo, Oct 02 2018

			Triangle begins:
    0;
   -1,   1;
   -2,   0,    2;
   -3,  -3,    3,    3;
   -4,  -8,    0,    8,    4;
   -5, -15,  -10,   10,   15,   5;
   -6, -24,  -30,    0,   30,  24,   6;
   -7, -35,  -63,  -35,   35,  63,  35,   7;
   -8, -48, -112, -112,    0, 112, 112,  48,   8;
   -9, -63, -180, -252, -126, 126, 252, 180,  63,  9;
  -10, -80, -270, -480, -420,   0, 420, 480, 270, 80, 10;
  ...

G. C. Greubel, Table of n, a(n) for n = 0..5150 (Rows n=1..100 of triangle, flattened; offset adapted by _Georg Fischer_, Jan 31 2019)
Rida T. Farouki, The Bernstein polynomial basis: A centennial retrospective, Computer Aided Geometric Design Vol. 29 (2012), 379-419.
Eric Weisstein's World of Mathematics, Bernstein Polynomial
Wikipedia, Bernstein polynomial

Cf. A007318, A112467, A128433, A128434.

a:=proc(n,k) n*(binomial(n-1,k-1)-binomial(n-1,k)); end proc: seq(seq(a(n,k),k=0..n),n=0..10); # Muniru A Asiru, Oct 03 2018

Table[Table[n*(Binomial[n - 1, k - 1] - Binomial[n - 1, k]),{k, 0, n}],{n, 0, 12}]//Flatten

T(n, k) := n*(binomial(n - 1, k - 1) - binomial(n - 1, k))$
tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, n))$ /* Franck Maminirina Ramaharo, Oct 02 2018 */

T(n,k) = n*(B(1/2;n-1,k-1) - B(1/2;n-1,k))*2^(n - 1), where B(t;n,k) = binomial(n,k)*t^k*(1 - t)^(n - k) denotes the k-th Benstein basis polynomial of degree n.
T(n,k) = n*A112467(n,k).
From Franck Maminirina Ramaharo, Oct 02 2018: (Start)
T(n,k) = -T(n,n-k)
T(n,0) = -n.
T(n,1) = -A067998(n)
E.g.f.: (x*y - y)/(x*y + y - 1)^2.
Sum_{k=0..n} abs(T(n,k)) = 2*A100071(n).
Sum_{k=0..n} T(n,k)^2 = 2*A037965(n).
Sum_{k=0..n} k*T(n,k) = A001787(n).
Sum_{k=0..n} k^2*T(n,k) = A014477(n-1). (End)

Edited, new name and offset corrected by Franck Maminirina Ramaharo, Oct 02 2018

A296771 Row sums of A050157.

Original entry on oeis.org

1, 3, 13, 58, 257, 1126, 4882, 20980, 89497, 379438, 1600406, 6720748, 28117498, 117254268, 487589572, 2022568168, 8371423177, 34581780478, 142605399982, 587138954428, 2413944555742, 9911778919348, 40650232625212, 166534680737368, 681576405563722

Offset: 0

Peter Luschny, Dec 21 2017

nonn

Robert Israel, Table of n, a(n) for n = 0..1657

Cf. A000346, A037965, A050157, A296770.

A296771 := n -> add(binomial(2*n, n) - binomial(2*n, n+k+1), k=0..n):
seq(A296771(n), n=0..24);

a[n_] := 4^n ((n - 1/2)! (2 n + 3)/(2 Sqrt[Pi] n!) - 1/2);
Table[a[n], {n, 0, 24}]

a(n) = sum(k=0, n, binomial(2*n, n) - binomial(2*n, n+k+1)) \\ Iain Fox, Dec 21 2017

a(n) = Sum_{k=0..n} (binomial(2*n, n) - binomial(2*n, n+k+1)).
a(n) = 2^(2*n-1)*(((n-1/2)!*(2*n+3))/(sqrt(Pi)*n!) - 1).
a(n) ~ 4^n*(sqrt(n/Pi) - 1/2).
a(n) = A037965(n+1) - A000346(n-1) for n >= 1.
From Robert Israel, Dec 21 2017: (Start)
a(n) = (n+3/2)*binomial(2*n,n) - 2^(2*n-1).
G.f.: (3/2-4*x)*(1-4*x)^(-3/2) - (1/2)*(1-4*x)^(-1).
64*(n+1)*(2*n+1)*a(n)-8*(2*n+3)*(5*n+4)*a(n+1)+2*(n+2)*(8*n+11)*a(n+2)-(n+3)*(n+2)*a(n+3)=0. (End)

cp's OEIS Frontend

A033876 Expansion of 1/(2*x) * (1/(1-4*x)^(3/2)-1).

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A037966 a(n) = n^2*binomial(2*n-2, n-1).

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A037972 a(n) = n^2*(n+1)*binomial(2*n-2, n-1)/2.

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A093375 Array T(m,n) read by ascending antidiagonals: T(m,n) = m*binomial(n+m-2, n-1) for m, n >= 1.

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A097070 Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.

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A123160 Triangle read by rows: T(n,k) = n!*(n+k-1)!/((n-k)!*(n-1)!*(k!)^2) for 0 <= k <= n, with T(0,0) = 1.

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A033876 Expansion of 1/(2x) (1/(1-4*x)^(3/2)-1).

A037966 a(n) = n^2binomial(2n-2, n-1).

A037972 a(n) = n^2(n+1)binomial(2*n-2, n-1)/2.

A123160 Triangle read by rows: T(n,k) = n!(n+k-1)!/((n-k)!(n-1)!*(k!)^2) for 0 <= k <= n, with T(0,0) = 1.