A168491 -id:A168491 - cp's OEIS frontend

A171567 Riordan array (f(x), x*f(x)) where f(x) is the g.f. of A168491.

1, -1, 1, 2, -2, 1, -5, 5, -3, 1, 14, -14, 9, -4, 1, -42, 42, -28, 14, -5, 1, 132, -132, 90, -48, 20, -6, 1, -429, 429, -297, 165, -75, 27, -7, 1, 1430, -1430, 1001, -572, 275, -110, 35, -8, 1, -4862, 4862, -3432, 2002, -1001, 429, -154, 44, -9, 1

Offset: 0

Philippe Deléham, Dec 11 2009

Equal to B^(-2)*A054336, B = A007318. T(n,0)=(-1)^n*A000108(n). Unsigned version in A033184.

			Triangle begins : 1 ; -1,1 ; 2,-2,1 ; -5,5,-3,1 ; 14,-14,9,-4,1 ;...

Cf. A033184, A171488, A054336, A154930

T(n,k) = (-1)^(n-k)*(k+1)*binomial(2n-k, n-k)/(n+1), 0<=k<=n; else 0.

T(n,k) = T(n-1,k-1) - T(n-1,k) + sum_{i, i>=0} T(n-1,k+1+i)*(-1)^i. - Philippe Deléham, Feb 23 2012

A001405 a(n) = binomial(n, floor(n/2)).

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, 924, 1716, 3432, 6435, 12870, 24310, 48620, 92378, 184756, 352716, 705432, 1352078, 2704156, 5200300, 10400600, 20058300, 40116600, 77558760, 155117520, 300540195, 601080390, 1166803110

Offset: 0

N. J. A. Sloane

Sperner's theorem says that this is the maximal number of subsets of an n-set such that no one contains another.
When computed from index -1, [seq(binomial(n,floor(n/2)), n = -1..30)]; -> [1,1,1,2,3,6,10,20,35,70,126,...] and convolved with aerated Catalan numbers [seq(((n+1) mod 2)*binomial(n,n/2)/((n/2)+1), n = 0..30)]; -> [1,0,1,0,2,0,5,0,14,0,42,0,132,0,...] shifts left by one: [1,1,2,3,6,10,20,35,70,126,252,...] and if again convolved with aerated Catalan numbers, gives A037952 apart from the initial term. - Antti Karttunen, Jun 05 2001 [This is correct because the g.f.'s satisfy (1+x*g001405(x))*g126120(x) = g001405(x) and g001405(x)*g126120(x) = g037952(x)/x. - R. J. Mathar, Sep 23 2021]
Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Gives for n >= 1 the maximum absolute column sum norm of the inverse of the Vandermonde matrix (a_ij) i=0..n-1, j=0..n-1 with a_00=1 and a_ij=i^j for (i,j) != (0,0). - Torsten Muetze, Feb 06 2004
Image of Catalan numbers A000108 under the Riordan array (1/(1-2x),-x/(1-2x)) or A065109. - Paul Barry, Jan 27 2005
Number of left factors of Dyck paths, consisting of n steps. Example: a(4)=6 because we have UDUD, UDUU, UUDD, UUDU, UUUD and UUUU, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Apr 23 2005
Number of dispersed Dyck paths of length n; they are defined as concatenations of Dyck paths and (1,0)-steps on the x-axis; equivalently, Motzkin paths with no (1,0)-steps at positive height. Example: a(4)=6 because we have HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD, where U=(1,1), H=(1,0), and D=(1,-1). - Emeric Deutsch, Jun 04 2011
a(n) is odd iff n=2^k-1. - Jon Perry, May 05 2005
An inverse Chebyshev transform of binomial(1,n)=(1,1,0,0,0,...) where g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), with c(x) the g.f. of A000108. - Paul Barry, May 13 2005
In a random walk on the number line, starting at 0 and with 0 absorbing after the first step, number of ways of ending up at a positive integer after n steps. - Joshua Zucker, Jul 31 2005
Maximum number of sums of the form Sum_{i=1..n} e(i)*a(i) that are congruent to 0 mod q, where e_i=0 or 1 and gcd(a_i,q)=1, provided that q > ceiling(n/2). - Ralf Stephan, Apr 27 2003
Also the number of standard tableaux of height <= 2. - Mike Zabrocki, Mar 24 2007
Hankel transform of this sequence forms A000012 = [1,1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
A001263 * [1, -2, 3, -4, 5, ...] = [1, -1, -2, 3, 6, -10, -20, 35, 70, -126, ...]. - Gary W. Adamson, Jan 02 2008
Equals right border of triangle A153585. - Gary W. Adamson, Dec 28 2008
Second binomial transform of A168491. - Philippe Deléham, Nov 27 2009
a(n) is also the number of distinct strings of length n, each of which is a prefix of a string of balanced parentheses; see example. - Lee A. Newberg, Apr 26 2010
Number of symmetric balanced strings of n pairs of parentheses; see example. - Joerg Arndt, Jul 25 2011
a(n) is the number of permutation patterns modulo 2. - Olivier Gérard, Feb 25 2011
For n >= 2, a(n-1) is the number of incongruent two-color bracelets of 2*n-1 beads, n of which are black (A007123), having a diameter of symmetry. - Vladimir Shevelev, May 03 2011
The number of permutations of n elements where p(k-2) < p(k) for all k. - Joerg Arndt, Jul 23 2011
Also size of the equivalence class of S_{n+1} containing the identity permutation under transformations of positionally adjacent elements of the form abc <--> cba where a < b < c, cf. A210668. - Tom Roby, May 15 2012
a(n) is the number of symmetric Dyck paths of length 2n. - Matt Watson, Sep 26 2012
a(n) is divisible by A000108(floor(n/2)) = abs(A129996(n-2)). - Paul Curtz, Oct 23 2012
a(n) is the number of permutations of length n avoiding both 213 and 231 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Number of symmetric standard Young tableaux of shape (n,n). - Ran Pan, Apr 10 2015
From Luciano Ancora, May 09 2015: (Start)
Also "stepped path" in the array formed by partial sums of the all 1's sequence (or a Pascal's triangle displayed as a square). Example:
  [1], [1],  1,    1,    1,     1,    1, ... A000012
   1,  [2], [3],   4,    5,     6,    7, ...
   1,   3,  [6], [10],  15,    21,   28, ...
   1,   4,  10,  [20], [35],   56,   84, ...
   1,   5,  15,   35,  [70], [126], 210, ...
Sequences in second formula are the mixed diagonals shown in this array. (End)
a(n) = A265848(n,n). - Reinhard Zumkeller, Dec 24 2015
The constant Sum_{n >= 0} a(n)/n! is 1 + A130820. - Peter Bala, Jul 02 2016
Number of meanders (walks starting at the origin and ending at any altitude >= 0 that may touch but never go below the x-axis) with n steps from {-1,1}. - David Nguyen, Dec 20 2016
a(n) is also the number of paths of n steps (either up or down by 1) that end at the maximal value achieved along the path. - Winston Luo, Jun 01 2017
Number of binary n-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions. - Juan A. Olmos, Dec 21 2017
Equivalently, a(n) is the number of subsets of {1,...,n} containing as many even numbers as odd numbers. - Gus Wiseman, Mar 17 2018
a(n) is the number of Dyck paths with semilength = n+1, returns to the x-axis = floor((n+3)/2) and up movements in odd positions = floor((n+3)/2).  Example: a(4)=6, U=up movement in odd position, u=up movement in even position, d=down movement, -=return to x-axis: Uududd-Ud-Ud-, Ud-Uudd-Uudd-, Uudd-Uudd-Ud-, Ud-Ud-Uududd-, Uudd-Ud-Uudd-, Ud-Uududd-Ud-. - Roger Ford, Dec 29 2017
Let C_n(R, H) denote the transition matrix from the ribbon basis to the homogeneous basis of the graded component of the algebra of noncommutative symmetric functions of order n. Letting I(2^(n-1)) denote the identity matrix of order 2^(n-1), it has been conjectured that the dimension of the kernel of C_n(R, H) - I(2^(n-1)) is always equal to a(n-1). - John M. Campbell, Mar 30 2018
The number of U-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are U-equivalent iff the positions of pattern U are identical in these paths. - Sergey Kirgizov, Apr 08 2018
All binary self-dual codes of length 2n, for n > 0, must contain at least a(n) codewords of weight n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 2n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself n times. A permutation equivalent code can be constructed by augmenting two identity matrices of length n together. - Nathan J. Russell, Nov 25 2018
Closed under addition. - Torlach Rush, Apr 18 2019
The sequence starting (1, 2, 3, 6, ...) is the invert transform of A097331: (1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, ...). - Gary W. Adamson, Feb 22 2020
From Gary W. Adamson, Feb 24 2020: (Start)
The sequence is the culminating limit of an infinite set of sequences with convergents of 2*cos(Pi/N), N = (3, 5, 7, 9, ...).
The first few such sequences are:
N = 3: (1,  1,  1,  1,  1,  1,  1,  1, ...)
N = 5: (1,  1,  2,  3,  5,  8, 13, 21, ...) = A000045
N = 7: (1,  1,  2,  3,  6, 10, 19, 33, ...) = A028495, a(n)/a(n-1) tends to 1.801937...
N = 9  (1,  1,  2,  3,  6, 10, 20, 35, ...) = A061551, a(n)/a(n_1) tends to 1.879385...
...
In the limit one gets the current sequence with ratio 2. (End)
a(n) is also the number of monotone lattice paths from (0,0) to (floor(n/2),ceiling(n/2)). These are the number of Grand Dyck paths when n is even. - Nachum Dershowitz, Aug 12 2020
The maximum number of preimages that a permutation of length n+1 can have under the consecutive-132-avoiding stack-sorting map. - Colin Defant, Aug 28 2020
Counts faro permutations of length n. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one. - Sergey Kirgizov, Jan 12 2021
Per "Sperner's Theorem", the largest possible familes of finite sets none of which contain any other sets in the family. - Renzo Benedetti, May 26 2021
a(n-1) are the incomplete, primitive Dyck paths of n steps without a first return: paths of U and D steps starting at the origin, never touching the horizontal axis later on, and ending above the horizontal axis. n=1: {U}, n=2: {UU}, n=3: {UUU, UUD}, n=4: {UUUU, UUUD, UUDU}, n=5: {UUUUU, UUUUD, UUUDD, UUDUU, UUUDU, UUDUD}. For comparison: A037952 counts incomplete Dyck paths with n steps with any number of intermediate returns to the horizontal axis, ending above the horizontal axis. - R. J. Mathar, Sep 24 2021
a(n) is the number of noncrossing partitions of [n] whose nontrivial blocks are of type {a,b}, with a <= n/2, b > n/2. - Francesca Aicardi, May 29 2022
Maximal coefficient of (1+x)^n. - Vaclav Kotesovec, Dec 30 2022
Sums of lower-left-to-upper-right diagonals of the Catalan Triangle A001263. - Howard A. Landman, Sep 16 2024

			For n = 4, the a(4) = 6 distinct strings of length 4, each of which is a prefix of a string of balanced parentheses, are ((((, (((), (()(, ()((, ()(), and (()). - _Lee A. Newberg_, Apr 26 2010
There are a(5)=10 symmetric balanced strings of 5 pairs of parentheses:
[ 1] ((((()))))
[ 2] (((()())))
[ 3] ((()()()))
[ 4] ((())(()))
[ 5] (()()()())
[ 6] (()(())())
[ 7] (())()(())
[ 8] ()()()()()
[ 9] ()((()))()
[10] ()(()())() - _Joerg Arndt_, Jul 25 2011
G.f. = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 20*x^6 + 35*x^7 + 70*x^8 + ...
The a(4)=6 binary 4-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions are 0000, 1100, 1001, 0110, 0011, 1111. - _Juan A. Olmos_, Dec 21 2017

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Index entries for sequences of k-nomial coefficients
Index entries for "core" sequences.

Row sums of Catalan triangle A053121 and of symmetric Dyck paths A088855.
Enumerates the structures encoded by A061854 and A061855.
First differences are in A037952.
Apparently a(n) = lim_{k->infinity} A094718(k, n).
Partial sums are in A036256. Column k=2 of A182172. Column k=1 of A335570.
Bisections: A000984 (even part), A001700 (odd part).
Cf. A000712, A001006, A001700, A005773, A005817, A007578, A007579, A022916, A022917 (permutation patterns mod k), A049401, A051920, A063886, A130820, A132815, A153585, A239241, A265848.
Cf. A097331.
Cf. A000045, A028495, A061551
Cf. A107373, A340567, A340568, A340569 (popularity of certain patterns in faro permutations).

List([0..40],n->Binomial(n,Int(n/2))); # Muniru A Asiru, Apr 08 2018

a001405 n = a007318_row n !! (n `div` 2) -- Reinhard Zumkeller, Nov 09 2011

[Binomial(n, Floor(n/2)): n in [0..40]]; // Vincenzo Librandi, Nov 16 2014

A001405 := n->binomial(n, floor(n/2)): seq(A001405(n), n=0..33);

Table[Binomial[n, Floor[n/2]], {n, 0, 40}] (* Stefan Steinerberger, Apr 08 2006 *)
Table[DifferenceRoot[Function[{a,n},{-4 n a[n]-2 a[1+n]+(2+n) a[2+n] == 0,a[1] == 1,a[2] == 1}]][n], {n, 30}] (* Luciano Ancora, Jul 08 2015 *)
Array[Binomial[#,Floor[#/2]]&,40,0] (* Harvey P. Dale, Mar 05 2018 *)

A001405(n):=binomial(n,floor(n/2))$
makelist(A001405(n),n,0,30); /* Martin Ettl, Nov 01 2012 */

PARI
```
a(n) = binomial(n, n\2);
```

first(n) = x='x+O('x^n); Vec((-1+2*x+sqrt(1-4*x^2))/(2*x-4*x^2)) \\ Iain Fox, Dec 20 2017 (edited by Iain Fox, May 07 2018)

from math import comb
def A001405(n): return comb(n,n//2) # Chai Wah Wu, Jun 07 2022

a(n) = max_{k=0..n} binomial(n, k).
a(2*n) = A000984(n), a(2*n+1) = A001700(n).
By symmetry, a(n) = binomial(n, ceiling(n/2)). - Labos Elemer, Mar 20 2003
P-recursive with recurrence: a(0) = 1, a(1) = 1, and for n >= 2, (n+1)*a(n) = 2*a(n-1) + 4*(n-1)*a(n-2). - Peter Bala, Feb 28 2011
G.f.: (1+x*c(x^2))/sqrt(1-4*x^2) = 1/(1 - x - x^2*c(x^2)); where c(x) = g.f. for Catalan numbers A000108.
G.f.: (-1 + 2*x + sqrt(1-4*x^2))/(2*x - 4*x^2). - Lee A. Newberg, Apr 26 2010
G.f.: 1/(1 - x - x^2/(1 - x^2/(1 - x^2/(1 - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 12 2009
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = Sum_{k = 0..2*m} (-1)^k*a(k)*a(2*m-k). - Len Smiley, Dec 09 2001
G.f.: (sqrt((1+2*x)/(1-2*x)) - 1)/(2*x). - Vladeta Jovovic, Apr 28 2003
The o.g.f. A(x) satisfies A(x) + x*A^2(x) = 1/(1-2*x). - Peter Bala, Feb 28 2011
E.g.f.: BesselI(0, 2*x) + BesselI(1, 2*x). - Vladeta Jovovic, Apr 28 2003
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = 2*a(2*m) - c(m), where c(m)=A000108(m) are the Catalan numbers. - Christopher Hanusa (chanusa(AT)washington.edu), Nov 25 2003
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*A000108(k). - Paul Barry, Jan 27 2005
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(1, n-2*k). - Paul Barry, May 13 2005
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..floor((n+1)/2)} (binomial(n+1, k)*(cos((n-2*k+1)*Pi/2) + sin((n-2*k+1)*Pi/2))).
a(n) = Sum_{k=0..n+1}, (binomial(n+1, (n-k+1)/2)*(1-(-1)^(n-k))*(cos(k*Pi/2) + sin(k*Pi))/2). (End)
a(n) = Sum_{k=floor(n/2)..n} (binomial(n,n-k) - binomial(n,n-k-1)). - Paul Barry, Sep 06 2007
Inverse binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double inverse binomial transform of A001700. Row sums of triangle A132815. - Gary W. Adamson, Aug 31 2007
a(n) = Sum_{k=0..n} A120730(n,k). - Philippe Deléham, Oct 16 2008
a(n) = Sum_{k = 0..floor(n/2)} (binomial(n,k) - binomial(n,k-1)). - Nishant Doshi (doshinikki2004(AT)gmail.com), Apr 06 2009
Sum_{n>=0} a(n)/10^(n+1) = 0.1123724... = (sqrt(3)-sqrt(2))/(2*sqrt(2)); Sum_{n>=0} a(n)/100^(n+1) = 0.0101020306102035... = (sqrt(51)-sqrt(49))/(2*sqrt(49)). - Mark Dols, Jul 15 2010
Conjectured: a(n) = 2^n*2F1(1/2,-n;2;2), useful for number of paths in 1-d for which the coordinate is never negative. - Benjamin Phillabaum, Feb 20 2011
a(2*m+1) = (2*m+1)*a(2*m)/(m+1), e.g., a(7) = (7/4)*a(6) = (7/4)*20 = 35. - Jon Perry, Jan 20 2011
From Peter Bala, Feb 28 2011: (Start)
Let F(x) be the logarithmic derivative of the o.g.f. A(x). Then 1+x*F(x) is the o.g.f. for A027306.
Let G(x) be the logarithmic derivative of 1+x*A(x). Then x*G(x) is the o.g.f. for A058622. (End)
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal; and V = the vector [1,0,0,0,...]. a(n) = M^n*V, leftmost term. - Gary W. Adamson, Jun 13 2011
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal. a(n) = M^n_{1,1}. - Corrected by Gary W. Adamson, Jan 30 2012
a(n) = A007318(n, floor(n/2)). - Reinhard Zumkeller, Nov 09 2011
a(n+1) = Sum_{k=0..n} a(n-k)*A097331(k) = a(n) + Sum_{k=0..(n-1)/2} A000108(k)*a(n-2*k-1). - Philippe Deléham, Nov 27 2011
a(n) = A214282(n) - A214283(n), for n > 0. - Reinhard Zumkeller, Jul 14 2012
a(n) = Sum_{k=0..n} A168511(n,k)*(-1)^(n-k). - Philippe Deléham, Mar 19 2013
a(n+2*p-2) = Sum_{k=0..floor(n/2)} A009766(n-k+p-1, k+p-1) + binomial(n+2*p-2, p-2), for p >= 1. - Johannes W. Meijer, Aug 02 2013
O.g.f.: (1-x*c(x^2))/(1-2*x), with the o.g.f. c(x) of Catalan numbers A000108. See the rewritten formula given by Lee A. Newberg above. This is the o.g.f. for the row sums the Riordan triangle A053121. - Wolfdieter Lang, Sep 22 2013
a(n) ~ 2^n / sqrt(Pi * n/2). - Charles R Greathouse IV, Oct 23 2015
a(n) = 2^n*hypergeom([1/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
a(2*k) = Sum_{i=0..k} binomial(k, i)*binomial(k, i), a(2*k+1) = Sum_{i=0..k} binomial(k+1, i)*binomial(k, i). - Juan A. Olmos, Dec 21 2017
a(0) = 1, a(n) = 2 * a(n-1) for even n, a(n) = (2*n/(n+1)) * a(n-1) for odd n. - James East, Sep 25 2019
a(n) = A037952(n) + A000108(n/2) where A(.)=0 for non-integer argument. - R. J. Mathar, Sep 23 2021
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=0} 1/a(n) = 2*Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 - 2*Pi/(9*sqrt(3)). (End)
For k>2, Sum_{n>=0} a(n)/k^n = (sqrt((k+2)/(k-2)) - 1)*k/2. - Vaclav Kotesovec, May 13 2022
From Peter Bala, Mar 24 2023: (Start)
a(n) = Sum_{k = 0..n+1} (-1)^(k+binomial(n+2,2)) * k/(n+1) * binomial(n+1,k)^2.
(n + 1)*(2*n - 1)*a(n) = (-1)^(n+1)*2*a(n-1) + 4*(n - 1)*(2*n + 1)*a(n-2) with a(0) = a(1) = 1. (End)
a(n) = Integral_{x=-2..2} x^n*W(x)*dx, n>=0, where W(x) = sqrt((2+x)/(2-x))/(2*Pi) is a positive function on x=(-2,2) and is singular at x = 2. Therefore a(n) is a positive definite sequence. - Karol A. Penson, May 12 2025

A091867 Triangle read by rows: T(n,k) = number of Dyck paths of semilength n having k peaks at odd height.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 3, 4, 6, 0, 1, 6, 15, 10, 10, 0, 1, 15, 36, 45, 20, 15, 0, 1, 36, 105, 126, 105, 35, 21, 0, 1, 91, 288, 420, 336, 210, 56, 28, 0, 1, 232, 819, 1296, 1260, 756, 378, 84, 36, 0, 1, 603, 2320, 4095, 4320, 3150, 1512, 630, 120, 45, 0, 1, 1585, 6633, 12760, 15015, 11880, 6930, 2772, 990, 165, 55, 0, 1

Offset: 0

Emeric Deutsch, Mar 10 2004

Number of ordered trees with n edges having k leaves at odd height. Row sums are the Catalan numbers (A000108). T(n,0)=A005043(n). Sum_{k=0..n} k*T(n,k) = binomial(2n-2,n-1).
T(n,k)=number of Dyck paths of semilength n and having k ascents of length 1 (an ascent is a maximal string of consecutive up steps). Example: T(4,2)=6 because we have UdUduud, UduuddUd, uuddUdUd, uudUdUdd, UduudUdd and uudUddUd (the ascents of length 1 are indicated by U instead of u).
T(n,k) is the number of Łukasiewicz paths of length n having k level steps (i.e., (1,0)). A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(4,1)=4 because we have HU(2)DD, U(2)HDD, U(2)DHD and U(2)DDH, where H=(1,0), U(1,1), U(2)=(1,2) and D=(1,-1). - Emeric Deutsch, Jan 06 2005
T(n,k) = number of noncrossing partitions of [n] containing k singleton blocks. Also, T(n,k) = number of noncrossing partitions of [n] containing k adjacencies. An adjacency is an occurrence of 2 consecutive integers in the same block (here 1 and n are considered consecutive). In fact, the statistics # singletons and # adjacencies have a symmetric joint distribution.
Exponential Riordan array [e^x*(Bessel_I(0,2x)-Bessel_I(1,2x)),x]. - Paul Barry, Mar 03 2011
T(n,k) is the number of ordered trees having n edges and exactly k nodes with one child. - Geoffrey Critzer, Feb 25 2013
From Tom Copeland, Nov 04 2014: (Start)
Summing the coeff. of the partitions in A134264 for a Lagrange inversion formula (see also A249548) containing (h_1)^k = (1')^k gives this triangle, so this array's o.g.f. H(x,t) = x + t * x^2 + (1 + t^2) * x^3 ... is the inverse of the o.g.f. of A104597 with a sign change, i.e., H^(-1)(x,t) = (x-x^2) / [1 + (t-1)(x-x^2)] = Cinv(x)/[1 + (t-1)Cinv(x)] = P[Cinv(x),t-1] where Cinv(x)= x * (1-x) is the inverse of C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the Catalan numbers A000108, and P(x,t) = x/(1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x/(1-t*x). Therefore,
O.g.f.: H(x,t) = C[Pinv(x,t-1)] = C[P(x,1-t)] = C[x/(1-(t-1)x)] = {1-sqrt[1-4*x/(1-(t-1)x)]}/2 (for A091867). Reprising,
Inverse O.g.f.: H^(-1)(x,t) = x*(1-x) / [1 + (t-1)x(1-x)] = P[Cinv(x),t-1].
From general arguments in A134264, the row polynomials are an Appell sequence with lowering operator d/dt, having the umbral property (p(.,t)+a)^n=p(n,t+a) with e.g.f. = e^(x*t)/w(x), where 1/w(x)= e.g.f. of first column for the Motzkin numbers in A005043. (Mislabeled argument corrected on Jan 31 2016.)
Cf. A124644 (t-shifted polynomials), A026378 (t=-4), A001700 (t=-3), A005773 (t=-2), A126930 (t=-1) and A210736 (t=-1, a(0)=0, unsigned), A005043 (t=0), A000108 (t=1), A007317 (t=2), A064613 (t=3), A104455 (t=4), A030528 (for inverses).
(End)
The sequence of binomial transforms A126930, A005043, A000108, ... in the above comment appears in A126930 and the link therein to a paper by F. Fite et al. on page 42. - Tom Copeland, Jul 23 2016

			T(4,2)=6 because we have (ud)uu(ud)dd, uu(ud)dd(ud), uu(ud)(ud)dd, (ud)(ud)uudd, (ud)uudd(ud) and uudd(ud)(ud) (here u=(1,1), d=(1,-1) and the peaks at odd height are shown between parentheses).
Triangle begins:
   1;
   0,   1;
   1,   0,   1;
   1,   3,   0,   1;
   3,   4,   6,   0,  1;
   6,  15,  10,  10,  0,  1;
  15,  36,  45,  20, 15,  0, 1;
  36, 105, 126, 105, 35, 21, 0, 1;
  ...

R. Sedgewick and P. Flajolet, Analysis of Algorithms, Addison and Wesley, 1996, page 254 (first edition)

Alois P. Heinz, Rows n = 0..200, flattened
David Callan, On conjugates for set partitions and integer compositions , arXiv:math/0508052 [math.CO], 2005.
A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
F. Yano and H. Yoshida, Some set partition statistics in non-crossing partitions and generating functions, Discr. Math., 307 (2007), 3147-3160.

Cf. A000108, A001700, A005043, A005773, A007317, A026378, A030528, A064613, A104455, A104597, A124644, A126930, A210736.

T := proc(n,k) if k>n then 0 elif k=n then 1 else (binomial(n+1,k)/(n+1))*sum(binomial(n+1-k,j)*binomial(n-k-j-1,j-1),j=1..floor((n-k)/2)) fi end: seq(seq(T(n,k),k=0..n),n=0..12);
T := (n,k) -> (-1)^(n+k)*binomial(n,k)*hypergeom([-n+k,1/2],[2],4): seq(seq(simplify(T(n, k)), k=0..n), n=0..10); # Peter Luschny, Jul 27 2016
# alternative Maple program:
b:= proc(x, y, t) option remember; expand(`if`(x=0, 1,
      `if`(y>0, b(x-1, y-1, 0)*z^irem(t*y, 2), 0)+
      `if`(y (p-> seq(coeff(p, z, i), i=0..n))(b(2*n, 0$2)):
seq(T(n), n=0..16);  # Alois P. Heinz, May 12 2017

nn=10;cy = ( 1 + x - x y - ( -4x(1+x-x y) + (-1 -x + x y)^2)^(1/2))/(2(1+x-x y)); Drop[CoefficientList[Series[cy,{x,0,nn}],{x,y}],1]//Grid  (* Geoffrey Critzer, Feb 25 2013 *)
Table[Which[k == n, 1, k > n, 0, True, (Binomial[n + 1, k]/(n + 1)) Sum[Binomial[n + 1 - k, j] Binomial[n - k - j - 1, j - 1], {j, Floor[(n - k)/2]}]], {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 25 2016 *)

T(n, k) = [binomial(n+1, k)/(n+1)]*Sum_{j=1..floor((n-k)/2)} binomial(n+1-k, j)*binomial(n-k-j-1, j-1) for kn. G.f.=G=G(t, z) satisfies z(1+z-tz)G^2-(1+z-tz)G+1=0. T(n, k)=r(n-k)*binomial(n, k), where r(n)=A005043(n) are the Riordan numbers.
G.f.: 1/(1-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-... (continued fraction). - Paul Barry, Aug 03 2009
Sum_{k=0..n} T(n,k)*x^k = A126930(n), A005043(n), A000108(n), A007317(n), A064613(n), A104455(n) for x = -1,0,1,2,3,4 respectively. - Philippe Deléham, Dec 03 2009
Sum_{k=0..n} (-1)^(n-k)*T(n,k)*x^k = A168491(n), A099323(n+1), A001405(n), A005773(n+1), A001700(n), A026378(n+1), A005573(n), A122898(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively. - Philippe Deléham, Dec 03 2009
E.g.f.: e^(x+xy)*(Bessel_I(0,2x)-Bessel_I(1,2x)). - Paul Barry, Mar 10 2010
From Tom Copeland, Nov 06 2014: (Start)
O.g.f.: H(x,t) = {1-sqrt[1-4x/(1-(t-1)x)]}/2 (shifted index, as given in Copeland's comment, see comp. inverse there).
H(x,t)= x / [1-(C.+(t-1))x] = Sum_{n>=1} (C.+ (t-1))^(n-1)*x^n umbrally, e.g., (a.+b.)^2 = a_0*b_2 + 2 a_1*b1_+ a_0*b_2, where (C.)^n = C_n are the Catalan numbers (1,1,2,5,14,..) of A000108.
This shows directly that the lowering operator for the polynomials is D=d/dt, i.e., D p(n,t)= D(C. + (t-1))^n = n * (C. + (t-1))^(n-1) = n*p(n-1,t), so that the polynomials form an Appell sequence, and that p(n,0) gives a Motzkin sum, or Riordan, number A005043.
(End)
T(n,k) = (-1)^(n+k)*binomial(n,k)*hypergeom([k-n,1/2],[2],4). - Peter Luschny, Jul 27 2016

A085880 Triangle T(n,k) read by rows: multiply row n of Pascal's triangle (A007318) by the n-th Catalan number (A000108).

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 5, 15, 15, 5, 14, 56, 84, 56, 14, 42, 210, 420, 420, 210, 42, 132, 792, 1980, 2640, 1980, 792, 132, 429, 3003, 9009, 15015, 15015, 9009, 3003, 429, 1430, 11440, 40040, 80080, 100100, 80080, 40040, 11440, 1430, 4862, 43758, 175032, 408408, 612612, 612612, 408408, 175032, 43758, 4862

Offset: 0

N. J. A. Sloane, Aug 17 2003

Coefficients of terms in the series reversion of (1-k*x-(k+1)*x^2)/(1+x). - Paul Barry, May 21 2005
Equals A131427 * A007318 as infinite lower triangular matrices. [Philippe Deléham, Sep 15 2008]
Sum_{k=0..n} T(n,k)*x^k = A168491(n), A000007(n), A000108(n), A151374(n), A005159(n), A151403(n), A156058(n), A156128(n), A156266(n), A156270(n), A156273(n), A156275(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Nov 15 2013
Diagonal sums are A052709(n+1). - Philippe Deléham, Nov 15 2013

			Triangle starts:
[ 1]     1;
[ 2]     1,     1;
[ 3]     2,     4,      2;
[ 4]     5,    15,     15,      5;
[ 5]    14,    56,     84,     56,     14;
[ 6]    42,   210,    420,    420,    210,     42;
[ 7]   132,   792,   1980,   2640,   1980,    792,    132;
[ 8]   429,  3003,   9009,  15015,  15015,   9009,   3003,    429;
[ 9]  1430, 11440,  40040,  80080, 100100,  80080,  40040,  11440,  1430;
[10]  4862, 43758, 175032, 408408, 612612, 612612, 408408, 175032, 43758, 4862;
...

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.

Flat(List([0..10], n-> List([0..n], k-> Binomial(n,k)*Binomial(2*n,n)/( n+1) ))); # G. C. Greubel, Feb 07 2020

[Binomial(n,k)*Catalan(n): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 07 2020

seq(seq(binomial(n, k)*binomial(2*n, n)/(n+1), k = 0..n), n = 0..10); # G. C. Greubel, Feb 07 2020

Table[Binomial[n, k]*CatalanNumber[n], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 07 2020 *)

tabl(nn) = {for (n=0, nn, c =  binomial(2*n,n)/(n+1); for (k=0, n, print1(c*binomial(n, k), ", ");); print(););} \\ Michel Marcus, Apr 09 2015

[[binomial(n,k)*catalan_number(n) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 07 2020

Triangle given by [1, 1, 1, 1, 1, 1, ...] DELTA [1, 1, 1, 1, 1, 1, ...] where DELTA is Deléham's operator defined in A084938.
Sum_{k>=0} T(n, k) = A151374(n) (row sums). - Philippe Deléham, Aug 11 2005
G.f.: (1-sqrt(1-4*(x+y)))/(2*(x+y)). - Vladimir Kruchinin, Apr 09 2015

A343439 G.f.: 1 + 2^0x/(1 + 2^1x/(1 + 2^2x/(1 + 2^3x/(1 + 2^4*x/(1 + ...))))).

Original entry on oeis.org

1, 1, -2, 12, -136, 2736, -99616, 6810816, -900563072, 234247256832, -120883821425152, 124271556482829312, -255006726559759042560, 1045529090595650037657600, -8569159507007490469146992640, 140431398588497630920722150113280, -4602217897540461023955069241211781120

Offset: 0

Seiichi Manyama, Apr 15 2021

sign

Seiichi Manyama, Table of n, a(n) for n = 0..81

Cf. A015083, A168491, A216966.

a(n) = my(A=1+O(x)); for(i=1, n, A=1+2^(n-i)*x/A); polcoef(A, n);

a(n) = if(n<2, 1, -sum(k=1, n-1, 2^k*a(k)*a(n-k)));

G.f. satisfies: A(x) = 1 + x/A(2*x).
G.f.: 1/(Sum_{k>=0} A015083(k) * (-x)^k).
a(0) = a(1) = 1 and a(n) = -Sum_{k=1..n-1} 2^k*a(k)*a(n-k) for n > 1.
a(n) = (-2)^(n-1) * A015083(n-1) for n > 0.

A122898 Expansion of (sqrt(21x^2 - 10x + 1) + 7x - 1) / (2x(1 - 7x)).

Original entry on oeis.org

1, 6, 37, 233, 1491, 9660, 63195, 416610, 2763595, 18426026, 123375927, 829053197, 5588050069, 37764371676, 255800207277, 1736181639585, 11804962371795, 80394249836010, 548283258074895, 3744067955618403, 25596986050620681

Offset: 0

Paul Barry, Sep 18 2006

3rd binomial transform of C(2n+1,n+1) (A001700); 5th binomial transform of C(n,floor(n/2)) (A001405); 7th binomial transform of (-1)^n*A000108(n) = A168491(n). Hankel transform is (1,1,1,.....). Row sums of Riordan array (1/(1+5x+x^2), x/(1+5x+x^2))^(-1). Counts Motzkin paths with 5 colors for horizontal steps. [Corrected by Philippe Deléham, Nov 29 2009]

Binomial transform of A005573. 7th binomial transform of A168491. - Philippe Deléham, Nov 28 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Isaac DeJager, Madeleine Naquin, Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.

a := n -> (-1)^n*simplify(GegenbauerC(n,-n,5/2) - GegenbauerC(n-1,-n,5/2)):
seq(a(n), n=0..21); # Peter Luschny, May 13 2016

CoefficientList[Series[(Sqrt[21*x^2-10*x+1]+7*x-1)/(2*x*(1-7*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

a(n) = Sum{k=0..n, 3^(n-k)C(n,k)C(2k+1,k+1)}.
a(n) = Sum{k=0..n, 5^(n-k)C(n,k)C(k,floor(k/2))}.
a(n) = Sum{k=0..n, 7^(n-k)C(n,k)*(-1)^k*C(k)} where C(n)=A000108(n).
a(n) = Sum{k=0..n, sum{j=0..n, 3^(n-j)*C(n,k)*C(n-k,j-k)*C(j+1,k+1)}}.
G.f.: 1/(1-6x-x^2/(1-5x-x^2/(1-5x-x^2/(1-5x-x^2/(1-...(continued fraction). - Philippe Deléham, Nov 28 2009
D-finite with recurrence: (n+1)*a(n) = 2*(5*n+1)*a(n-1) - 21*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 7^(n+1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 19 2012
G.f.: G(0)/(2*x) - 1/(2*x), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1-3*x) - 2*x*(1-3*x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1-3*x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
a(n) = (-1)^n*(GegenbauerC(n,-n,5/2) - GegenbauerC(n-1,-n,5/2)). - Peter Luschny, May 13 2016
E.g.f.: exp(5*x)*(BesselI(0,2*x) + BesselI(1,2*x)). - Ilya Gutkovskiy, Sep 20 2017

A171651 Triangle T, read by rows : T(n,k) = A007318(n,k)*A005773(n+1-k).

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 13, 15, 6, 1, 35, 52, 30, 8, 1, 96, 175, 130, 50, 10, 1, 267, 576, 525, 260, 75, 12, 1, 750, 1869, 2016, 1225, 455, 105, 14, 1, 2123, 6000, 7476, 5376, 2450, 728, 140, 16, 1, 6046, 19107, 27000, 22428, 12096, 4410, 1092, 180, 18, 1

Offset: 0

Philippe Deléham, Dec 14 2009

			Triangle begins:
   1;
   2,   1;
   5,   4,  1;
  13,  15,  6, 1;
  35,  52, 30, 8, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A097692.
Cf. A007318, A005773.
Cf. A168491, A099323, A001405, A005773, A001700, A026378, A005573, A122898.

b:= proc(u, d, t) option remember; `if`(u=0 and d=0, 1/2,
      expand(`if`(u=0, 0, b(u-1, d, 2)*`if`(t=3, x, 1))
      +`if`(d=0, 0, b(u, d-1, `if`(t=2, 3, 1)))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n+1$2, 1)):
seq(T(n), n=0..12);  # Alois P. Heinz, Apr 29 2015
# second program:
A171651:= (n, k)-> binomial(n,k)*add((-1)^(n-k-j)*binomial(n-k,j)*binomial(2*j+1,j+1),j=0..n-k): seq(print(seq(A171651(n, k), k=0..n)), n=0..9);  # Mélika Tebni, Dec 16 2023

b[u_, d_, t_] := b[u, d, t] = If[u == 0 && d == 0, 1/2, Expand[If[u == 0, 0, b[u-1, d, 2]*If[t == 3, x, 1]] + If[d == 0, 0, b[u, d-1, If[t == 2, 3, 1]]]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n}]][b[n+1, n+1, 1] ];
Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, May 21 2016, after Alois P. Heinz *)

Sum_{k, 0<=k<=n} T(n,k)*x^k = A168491(n), A099323(n), A001405(n), A005773(n+1), A001700(n), A026378(n+1), A005573(n), A122898(n) for x = -3, -2, -1, 0, 1, 2, 3, 4 respectively.

E.g.f. of column k: exp(x)*(BesselI(0,2*x)+BesselI(1,2*x))*x^k / k!. - Mélika Tebni, Dec 16 2023

Corrected by Philippe Deléham, Dec 18 2009

A327682 Expansion of Product_{k>0} (-1+sqrt(1+4x^k))/(2x^k).

Original entry on oeis.org

1, -1, 1, -5, 14, -40, 122, -404, 1362, -4608, 15881, -55709, 197402, -705114, 2539282, -9210196, 33605471, -123262137, 454268676, -1681305246, 6246544735, -23288217459, 87096982499, -326680267261, 1228547420236, -4631474743422, 17499462106763, -66257720483935, 251356773101419

Offset: 0

Seiichi Manyama, Sep 22 2019

sign

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

Cf. A000108, A081362, A168491, A309867, A322204, A327683.

m = 28; CoefficientList[Series[Product[(-1 + Sqrt[1 + 4*x^k])/(2*x^k), {k, 1, m}], {x, 0, m}], x] (* Amiram Eldar, May 06 2021 *)

N=66; x='x+O('x^N); Vec(prod(k=1, N, (-1+sqrt(1+4*x^k))/(2*x^k)))

N=66; x='x+O('x^N); Vec(prod(i=1, N, sum(j=0, N\i, (-1)^j*binomial(2*j, j)*x^(i*j)/(j+1))))

a(n) ~ (-1)^n * c * 4^n / n^(3/2), where c = 1/(2*sqrt(Pi)) * Product_{k>=1} (-1 + sqrt(1 + 4*(-1/4)^k)) / (2*(-1/4)^k) = 0.5396673413761086071059510679780476790558662471136055... - Vaclav Kotesovec, May 06 2021

A104726 Triangle generated as the matrix product of A026729 and A000012 (triangular views), read by rows.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 3, 3, 1, 5, 5, 5, 4, 1, 8, 8, 8, 8, 5, 1, 13, 13, 13, 13, 12, 6, 1, 21, 21, 21, 21, 21, 17, 7, 1, 34, 34, 34, 34, 34, 33, 23, 8, 1, 55, 55, 55, 55, 55, 55, 50, 30, 9, 1, 89, 89, 89, 89, 89, 89, 88, 73, 38

Offset: 0

Gary W. Adamson, Mar 20 2005

If the triangular factors A026729 and A000012 are commuted in the product, A004070 results.

Riordan array (1/(1-x-x^2), x*(1+x)). - Philippe Deléham, Mar 06 2013

			First few rows of the triangle are
1;
1, 1;
2, 2, 1;
3, 3, 3, 1;
5, 5, 5, 4, 1;
8, 8, 8, 8, 5, 1;
13, 13, 13, 13, 12, 6, 1;
21, 21, 21, 21, 21, 17, 7, 1;
...
Production array begins
1, 1
1, 1, 1
-1, -1, 1, 1
2, 2, -1, 1, 1
-5, -5, 2, -1, 1, 1
14, 14, -5, 2, -1, 1, 1
-42, -42, 14, -5, 2, -1, 1, 1
132, 132, -42, 14, -5, 2, -1, 1, 1
-429, -429, 132, -42, 14, -5, 2, -1, 1, 1
... which is based on A000108 or A168491. - _Philippe Deléham_, Mar 06 2013

Cf. A001629 (row sums), A026729, A004070, A000071.

A104726 := proc(n,k)
        add( binomial(j,n-j),j=k..n) ;
end proc:
seq(seq(A104726(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Oct 30 2011

T(n,k) = sum_{j=k..n} binomial(j,n-j). - R. J. Mathar, Oct 30 2011
T(n,0) = T(n-1,0) + T(n-2,0), T(n,k) = T(n-1,k-1) + T(n-2,k-1) for k>0. - Philippe Deléham, Mar 06 2013
T(2*n,n) = A000045(2n+1) = A001519(n+1) = A122367(n). - Philippe Deléham, Mar 06 2013

A378783 Triangular array T(n,k) read by rows: T(n, k) = c_k(n+1). The sequence c_k(m) has the ordinary generating function C_k(x) which satisfies C_k(x) = 1/(1+C_k(x)*Sum_{t=0..k} x^(t+1)).

Original entry on oeis.org

-1, 2, 1, -5, -1, -2, 14, 1, 5, 4, -42, -1, -12, -8, -9, 132, 1, 29, 18, 22, 21, -429, -1, -73, -43, -54, -50, -51, 1430, 1, 190, 105, 135, 124, 128, 127, -4862, -1, -505, -262, -345, -315, -326, -322, -323, 16796, 1, 1363, 666, 896, 813, 843, 832, 836, 835

Offset: 0

Thomas Scheuerle, Dec 07 2024

			Triangle begins:
  [0]    -1
  [1]     2,  1
  [2]    -5, -1,   -2
  [3]    14,  1,    5,    4
  [4]   -42, -1,  -12,   -8,   -9
  [5]   132,  1,   29,   18,   22,   21
  [6]  -429, -1,  -73,  -43,  -54,  -50,  -51
  [7]  1430,  1,  190,  105,  135,  124,  128,  127
  [8] -4862, -1, -505, -262, -345, -315, -326, -322, -323
.

Cf. A001006, A000108, A152171, A166587, A168491, A378816.

T[n_,k_]:=SeriesCoefficient[(2 / (Sqrt[1+4*Sum[x^(t+1),{t,0,k}]] + 1) - 1)/x,{x,0,n}];Table[T[n,k],{n,0,9},{k,0,n}]//Flatten (* Stefano Spezia, Dec 08 2024 *)

column(n, max_n) = { my(s = 1,x = 'x,cu); for(k = 0, max_n-1, cu = cu+polcoeff(1/s+O(x^(k+1)), k, x); cu = cu-polcoeff(1/s+O(x^(k+1)), k-1-n, x); s = s+cu*x^(k+1)); Vec(1/s+O(x^max_n)) };
T(n, k) = column(k, n+2)[n+2]
T(n, k) = polcoeff(2 / (sqrt(1+4*x*sum(t=0, k, x^t)) + 1) + O(x^(n+2)), n+1, x)

G.f. column k: (2 / (sqrt(1+4*Sum_{t=0..k}x^(t+1)) + 1) - 1)/x.
T(n, 0) = (-1)^(n+1)*Catalan(n+1) = A168491(n+1).
T(n, 2) = (-1)^(n+1)*A152171(n+1).
T(n, n) = (-1)^(n+1)*A001006(n) = -A166587(n+1).
A378816(n) = Limit_{k->oo} (T(k, k-n) - T(k, k-n-1)).

cp's OEIS Frontend

A171567 Riordan array (f(x), x*f(x)) where f(x) is the g.f. of A168491.

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A001405 a(n) = binomial(n, floor(n/2)).

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A085880 Triangle T(n,k) read by rows: multiply row n of Pascal's triangle (A007318) by the n-th Catalan number (A000108).

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A343439 G.f.: 1 + 2^0*x/(1 + 2^1*x/(1 + 2^2*x/(1 + 2^3*x/(1 + 2^4*x/(1 + ...))))).

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A122898 Expansion of (sqrt(21*x^2 - 10*x + 1) + 7*x - 1) / (2*x*(1 - 7*x)).

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A171651 Triangle T, read by rows : T(n,k) = A007318(n,k)*A005773(n+1-k).

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A343439 G.f.: 1 + 2^0x/(1 + 2^1x/(1 + 2^2x/(1 + 2^3x/(1 + 2^4*x/(1 + ...))))).

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A327682 Expansion of Product_{k>0} (-1+sqrt(1+4x^k))/(2x^k).