A026641 -id:A026641 - cp's OEIS frontend

A109244 A tree-node counting triangle.

1, 1, 1, 4, 2, 1, 13, 7, 3, 1, 46, 24, 11, 4, 1, 166, 86, 40, 16, 5, 1, 610, 314, 148, 62, 22, 6, 1, 2269, 1163, 553, 239, 91, 29, 7, 1, 8518, 4352, 2083, 920, 367, 128, 37, 8, 1, 32206, 16414, 7896, 3544, 1461, 541, 174, 46, 9, 1, 122464, 62292, 30086, 13672, 5776, 2232

Offset: 0

Paul Barry, Jun 23 2005

Columns include A026641,A014300,A014301. Inverse matrix is A109246. Row sums are A014300. Diagonal sums are A109245.

			Rows begin:
  1;
  1,1;
  4,2,1;
  13,7,3,1;
  46,24,11,4,1;
  166,86,40,16,5,1;

G. C. Greubel, Rows n=0..100 of triangle, flattened

Flat(List([0..12], n-> List([0..n], k-> Sum([0..n], j-> (-1)^(n-j)*Binomial(n+j-k, j-k) )))); # G. C. Greubel, Feb 19 2019

[[(&+[(-1)^(n-j)*Binomial(n+j-k, j-k): j in [0..n]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 19 2019

Table[Sum[(-1)^(n-j)*Binomial[n+j-k, j-k], {j,0,n}], {n,0,12}, {k,0,n}] //Flatten  (* G. C. Greubel, Feb 19 2019 *)

{T(n,k) = sum(j=0,n, (-1)^(n-j)*binomial(n+j-k, j-k))};
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 19 2019

[[sum((-1)^(n-j)*binomial(n+j-k, j-k) for j in (0..n)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 19 2019

Number triangle T(n, k) = Sum_{i=0..n} (-1)^(n-i)*binomial(n+i-k, i-k).
Riordan array (1/(1-x*c(x)-2*x^2*c(x)^2), x*c(x)) where c(x)=g.f. of A000108.
The production matrix M (discarding the zeros) is:
  1, 1;
  3, 1, 1;
  3, 1, 1, 1;
  3, 1, 1, 1, 1;
... such that the n-th row of the triangle is the top row of M^n. - Gary W. Adamson, Feb 16 2012

A192368 Number of lattice paths from (0,0) to (n,n) using steps (1,0), (2,0), (0,2), (1,1).

Original entry on oeis.org

1, 1, 6, 19, 94, 396, 1870, 8541, 40284, 189274, 899260, 4281168, 20487156, 98299384, 473118174, 2282322211, 11034087438, 53443135944, 259283934816, 1259795078566, 6129223177272, 29856164309124, 145592506783224, 710686739172096, 3472285996766556, 16979257639328076

Offset: 0

Joerg Arndt, Jul 01 2011

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A001850, A026641, A036355, A137644, A192364, A192365, A192369.

s := RootOf( 16*x*(3*s+1)*s+(s^2-18*s+1)*(s-1), s):
ogf := -16*(3*s+1)*s^(3/2)/(3*s^4+2*s^3-76*s^2+6*s+1):
series(ogf, x=0, 20); # Mark van Hoeij, Apr 16 2013
# second Maple program:
b:= proc(x, y) option remember;
      `if`(min(x, y)<0, 0, `if`(max(x, y)=0, 1,
       b(x-1, y)+b(x-2, y)+b(x, y-2)+b(x-1, y-1)))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..35);  # Alois P. Heinz, May 16 2017

a[0, 0] = 1; a[n_, k_] /; n >= 0 && k >= 0 := a[n, k] = a[n, k - 1] + a[n, k - 2] + a[n - 1, k - 1] + a[n - 2, k]; a[, ] = 0;
a[n_] := a[n, n];
a /@ Range[0, 25] (* Jean-François Alcover, Oct 14 2019 *)

/* same as in A092566 but use */
steps=[[1,0], [2,0], [0,2], [1,1]];
/* Joerg Arndt, Jun 30 2011 */

G.f. -16*(3*s+1)*s^(3/2)/(3*s^4+2*s^3-76*s^2+6*s+1) where s satisfies 16*x*(3*s+1)*s+(s^2-18*s+1)*(s-1) = 0. - Mark van Hoeij, Apr 16 2013

A220074 Triangle read by rows giving coefficients T(n,k) of [x^(n-k)] in Sum_{i=0..n} (x-1)^i, 0 <= n <= k.

Original entry on oeis.org

1, 1, 0, 1, -1, 1, 1, -2, 2, 0, 1, -3, 4, -2, 1, 1, -4, 7, -6, 3, 0, 1, -5, 11, -13, 9, -3, 1, 1, -6, 16, -24, 22, -12, 4, 0, 1, -7, 22, -40, 46, -34, 16, -4, 1, 1, -8, 29, -62, 86, -80, 50, -20, 5, 0, 1, -9, 37, -91, 148, -166, 130, -70, 25, -5, 1

Offset: 0

Mokhtar Mohamed, Dec 03 2012

If the triangle is viewed as a square array S(m, k) = T(m+k, k), 0 <= m, 0 <= k, its first row is (1,0,1,0,1,...) with e.g.f. cosh(x), g.f. 1/(1-x^2) and subsequent rows have g.f. 1/((1+x)^n*(1-x^2)) (substitute x for -x in g.f. for A059259).
By column, S(m, k) is the coefficient of [x^m] in the generating function Sum_{i=0..k} (-1)^i/(1-x)^(i+1).
This is a rational generating function down column k with a power of (1-x) in the denominator; therefore column k is a polynomial in m respectively n. - Mathew Englander, May 14 2014
Column k multiplied by k! seems to correspond to row k of A054651, considered as a polynomial and then evaluated on the negative integers. For example, row 5 of A054651 represents the polynomial x^5 - 5*x^4 + 25*x^3 + 5*x^2 + 94*x + 120. Evaluating that for x = -1, x = -2, x = -3, ... gives (0, -360, -1440, -4080, -9600, -19920, -37680, ...) which is 5! times column 5 of this triangle. - Mathew Englander, May 23 2014
This triangle provides a solution to a question in the mathematics of gambling. For 0 < p < 1 and positive integers N and G with N < G, suppose you begin with N dollars and make repeated wagers, each time winning 1 dollar with probability p and losing 1 dollar with probability 1-p. You continue betting 1 dollar at a time until you have either G dollars (your Goal) or 0 (bankrupt). What is the probability of reaching your Goal before going bankrupt, as a function of p, N, and G? (This is a type of one-dimensional random walk.) Answer: Let Q_m_(x) be the polynomial whose coefficients are given by row m-1 of the triangle (e.g., Q_6_(x) = 1 - 4x + 7x^2 - 6x^3 + 3x^4). Then, the probability of reaching G dollars before going bankrupt is p^(G-N)*Q_N_(p)/Q_G_(p). - Mathew Englander, May 23 2014
From Paul Curtz, Mar 17 2017: (Start)
Consider the triangle Ja(n+1,k) (here, but generally Ja(n,k)) composed of the triangle a(n) prepended with a column of 0's, i.e.,
  0;
  0,   1;
  0,   1,   0;
  0,   1,  -1,   1;
  0,   1,  -2,   2,   0;
  0,   1,  -3,   4,  -2,   1;
  0,   1,  -4,   7,  -6,   3,   0;
  0,   1,  -5,  11, -13,   9,  -3,   1;
  ... .
The row sums are 0, 1, 1, ... = A057427(n), the most elementary autosequence of the first kind (a sequence of the first kind has 0's as main diagonal of its array of successive differences).
The row sums of the absolute values are A001045(n).
Ja applied to a sequence written in its reluctant form yields an autosequence of the first kind. Example: the reluctant form of A001045(n) is 0, 0, 1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 1, 3, 5, ... = Jl.
Jl multiplied by Ja gives the triangle Jal:
  0;
  0,   1;
  0,   1,   0;
  0,   1,  -1,   3;
  0,   1,  -2,   6,   0;
  0,   1,  -3,  12, -10,  11;
  0,   1,  -4,  21, -30,  33,   0;
  0,   1,  -5,  33, -65,  99, -63,  43;
  ... .
The row sums are A001045(n). (End)

			Triangle begins:
  1;
  1,   0;
  1,  -1,   1;
  1,  -2,   2,    0;
  1,  -3,   4,   -2,    1;
  1,  -4,   7,   -6,    3,    0;
  1,  -5,  11,  -13,    9,   -3,    1;
  1,  -6,  16,  -24,   22,  -12,    4,    0;
  1,  -7,  22,  -40,   46,  -34,   16,   -4,   1;
  1,  -8,  29,  -62,   86,  -80,   50,  -20,   5,   0;
  1,  -9,  37,  -91,  148, -166,  130,  -70,  25,  -5, 1;
  1, -10,  46, -128,  239, -314,  296, -200,  95, -30, 6, 0;
  ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Dmitry Efimov, Hafnian of two-parameter matrices, arXiv:2101.09722 [math.CO], 2021.
Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Ângela Mestre, José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
OEIS Wiki, Autosequence

Similar to the triangles A080242, A108561, A112555, A071920.
Cf. A000124 (column 2), A003600 (column 3), A223718 (column 4, conjectured), A257890 (column 5).
Cf. A026641, A054651, A059259.

Flat(List([0..12], n-> List([0..n], k-> Sum([0..k], j-> (-1)^j*Binomial(n-k+j, j))))); # G. C. Greubel, Feb 18 2019

[[(&+[(-1)^j*Binomial(n-k+j, j): j in [0..k]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 18 2019

A059259A := proc(n,k)
    1/(1+y)/(1-x-y) ;
    coeftayl(%,x=0,n) ;
    coeftayl(%,y=0,k) ;
end proc:
A059259 := proc(n,k)
    A059259A(n-k,k) ;
end proc:
A220074 := proc(i,j)
    (-1)^j*A059259(i,j) ;
end proc: # R. J. Mathar, May 14 2014

Table[Sum[(-1)^i*Binomial[n-k+i,i], {i, 0, k}], {n, 0, 12}, {k, 0, n} ]//Flatten (* Michael De Vlieger, Jan 27 2016 *)

{T(n,k) = sum(j=0,k, (-1)^j*binomial(n-k+j,j))};
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 18 2019

[[sum((-1)^j*binomial(n-k+j,j) for j in (0..k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 18 2019

Sum_{k=0..n} T(n,k) = 1.
T(n,k) = Sum_{i=0..k} (-1)^i*binomial(n-k+i, i).
T(2*n,n) = (-1)^n*A026641(n).
T(n,k) = (-1)^k*A059259(n,k).
T(n,0) = 1, T(n,n) = (1+(-1)^n)/2, and T(n,k) = T(n-1,k) - T(n-1,k-1) for 0 < k < n. - Mathew Englander, May 24 2014

Definition and comments clarified by Li-yao Xia, May 15 2014

A387085 a(n) = Sum_{k=0..n} (-3)^(n-k) * binomial(2*n+1,k).

Original entry on oeis.org

1, 0, 4, 8, 36, 120, 456, 1680, 6340, 23960, 91224, 348656, 1337896, 5149872, 19877904, 76907808, 298176516, 1158168792, 4505865144, 17555689008, 68490100536, 267518448912, 1046041377264, 4094231982048, 16039426479336, 62887835652720, 246761907761776, 968943740083040

Offset: 0

Seiichi Manyama, Aug 16 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A000302, A000984, A001700, A026641, A141223, A377011, A386957.
Cf. A226733, A226705, A226751.
Cf. A005810, A079589, A183160.

[&+[(-3)^(n-k) * Binomial(2*n+1,k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 31 2025

Table[Sum[(-3)^(n-k)*Binomial[2*n+1,k],{k,0,n}],{n,0,25}] (* Vincenzo Librandi, Aug 31 2025 *)

a(n) = sum(k=0, n, (-3)^(n-k)*binomial(2*n+1, k));

a(n) = [x^n] (1+x)^(2*n+1)/(1+3*x).
a(n) = [x^n] 1/((1-x)^(n+1) * (1+2*x)).
a(n) = Sum_{k=0..n} (-2)^k * 3^(n-k) * binomial(2*n+1,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} (-2)^k * binomial(2*n-k,n-k).
G.f.: 1/( 4*x - 1 + 2*sqrt(1 - 4*x) ).
G.f.: 1/(1 - 4*x*(-1+g)) where g = 1+x*g^2 is the g.f. of A000108.
G.f.: g^2/((-2+3*g) * (2-g)) where g = 1+x*g^2 is the g.f. of A000108.
G.f.: B(x)^2/(1 + 2*(B(x)-1)), where B(x) is the g.f. of A000984.
D-finite with recurrence 3*n*a(n) +2*(-4*n+3)*a(n-1) +8*(-2*n+1)*a(n-2)=0. - R. J. Mathar, Aug 19 2025

A109078 Number of symmetric Dyck paths of semilength n and having no hills (i.e., no peaks at level 1).

Original entry on oeis.org

1, 0, 1, 2, 4, 6, 13, 22, 46, 80, 166, 296, 610, 1106, 2269, 4166, 8518, 15792, 32206, 60172, 122464, 230252, 467842, 884236, 1794196, 3406104, 6903352, 13154948, 26635774, 50922986, 103020253, 197519942, 399300166, 767502944, 1550554582

Offset: 0

Emeric Deutsch, Jun 17 2005

nonn

Column 0 of A109077.

			a(4)=4 because we have uudduudd, uudududd, uuududdd and uuuudddd, where u=(1,1), d=(1,-1).

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A109077.

Bisections are A026641 and A072547.

R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( 2*(1-x-x*Sqrt(1-4*x^2)+2*x^2 +Sqrt(1-4*x^2))/(1+Sqrt(1-4*x^2)-2*x)/(1+Sqrt(1-4*x^2)+2*x^2) )); // G. C. Greubel, Apr 29 2019

g:=2*(1-z-z*sqrt(1-4*z^2)+2*z^2+sqrt(1-4*z^2))/(1+sqrt(1-4*z^2)-2*z)/(1+sqrt(1-4*z^2)+2*z^2): gser:=series(g,z=0,39): 1, seq(coeff(gser,z^n),n=1..36);

CoefficientList[Series[2*(1-x-x*Sqrt[1-4*x^2]+2*x^2+Sqrt[1-4*x^2])/(1+ Sqrt[1-4*x^2]-2*x)/(1+Sqrt[1-4*x^2]+2*x^2), {x, 0, 40}], x] (* Vaclav Kotesovec, Mar 20 2014 *)

my(x='x+O('x^40)); Vec(2*(1-x-x*sqrt(1-4*x^2)+2*x^2 +sqrt(1-4*x^2))/(1+sqrt(1-4*x^2)-2*x)/(1+sqrt(1-4*x^2)+2*x^2)) \\ G. C. Greubel, Mar 16 2017

(2*(1-x-x*sqrt(1-4*x^2)+2*x^2 +sqrt(1-4*x^2))/(1+sqrt(1-4*x^2)-2*x)/(1+sqrt(1-4*x^2)+2*x^2)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 29 2019

G.f.: 2*(1 -z +2*z^2 +(1-z)*q)/((1-2*z+q)*(1+2*z^2+q)), where q = sqrt(1-4*z^2).
a(n) ~ 2^(n+3/2)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 20 2014
D-finite with recurrence 4*(n+1)*a(n) +2*(-n-3)*a(n-1) +2*(-7*n+11)*a(n-2) +(7*n-27)*a(n-3) +2*(-4*n+5)*a(n-4) +4*(n-3)*a(n-5)=0. - R. J. Mathar, Jul 26 2022

A110541 A number triangle of sums of binomial products.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 5, 7, 4, 1, 1, 1, 8, 19, 13, 5, 1, 1, 1, 13, 51, 46, 21, 6, 1, 1, 1, 21, 141, 166, 89, 31, 7, 1, 1, 1, 34, 393, 610, 393, 151, 43, 8, 1, 1, 1, 55, 1107, 2269, 1761, 776, 235, 57, 9, 1, 1, 1, 89, 3139, 8518, 7985, 4056, 1363, 344, 73, 10, 1, 1

Offset: 0

Paul Barry, Jul 25 2005

Columns include A000045, A002426, A026641. Rows include A000012, A000027, A002061(n+1). Row sums are A110542.

			Rows begin
  1;
  1,   1;
  1,   1,   1;
  1,   2,   1,   1;
  1,   3,   3,   1,   1;
  1,   5,   7,   4,   1,   1;
  1,   8,  19,  13,   5,   1,   1;
  1,  13,  51,  46,  21,   6,   1,   1;
  1,  21, 141, 166,  89,  31,   7,   1,   1;
As a number square read by antidiagonals, rows begin
  1,   1,   1,   1,   1,   1, ...
  1,   1,   1,   1,   1,   1, ...
  1,   2,   3,   4,   5,   6, ...
  1,   3,   7,  13,  21,  31, ...
  1,   5,  19,  46,  89, 151, ...
  1,   8,  51, 166, 393, 776, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Flat(List([0..12], n-> List([0..n], k-> Sum([0..n-k], j-> Binomial((k-1)*(n-k)-(k-2)*j, j)*Binomial(j, n-k-j) )))); # G. C. Greubel, Feb 19 2019

[[(&+[Binomial((k-1)*(n-k)-(k-2)*j, j)*Binomial(j, n-k-j): j in [0..n-k]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 19 2019

T[n_, k_] := Sum[Binomial[(k-1)*(n-k) - (k-2)*j, j]*Binomial[j, n-k-j], {j, 0, n-k}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 31 2017 *)

for(n=0,20, for(k=0,n, print1(sum(j=0,n-k, binomial((k-1)*(n-k) -(k-2)*j, j)*binomial(j, n-k-j)), ", "))) \\ G. C. Greubel, Aug 31 2017

[[sum(binomial((k-1)*(n-k) -(k-2)*j, j)*binomial(j, n-k-j) for j in (0..n-k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 19 2019

T(n,k) = Sum_{j=0..n-k} C((k-1)*(n-k)-(k-2)*j, j)*C(j, n-k-j).

A128316 Triangle read by rows: A000012 * A128315 as infinite lower triangular matrices.

Original entry on oeis.org

1, 1, 1, 3, -1, 1, 2, 3, -2, 1, 4, -1, 4, -3, 1, 4, 3, -5, 7, -4, 1, 6, -3, 10, -13, 11, -5, 1, 4, 8, -14, 23, -24, 16, -6, 1, 7, -2, 15, -33, 46, -40, 22, -7, 1, 7, 4, -15, 47, -79, 86, -62, 29, -8, 1, 9, -6, 30, -73, 131, -166, 148, -91, 37, -9, 1, 7, 12, -37, 103, -204, 297, -314, 239, -128, 46, -10, 1

Offset: 1

Gary W. Adamson, Feb 25 2007

A128316 * [1,2,3...] = A000034: [1,2,1,2,...].

			First few rows of the triangle:
  1;
  1,  1;
  3, -1,   1;
  2,  3   -2,   1;
  4, -1,   4,  -3,   1;
  4,  3,  -5,   7,  -4,  1;
  6, -3,  10, -13,  11, -5,  1;
  4,  8, -14,  23, -24, 16, -6, 1;
  ...

G. C. Greubel, Rows n = 1..100 of the triangle, flattened

Cf. A000012, A000034, A014300, A026641, A059851, A128315.

Sums include: A000027 (row), A032766, A047215, A344817 (alternating sign).

A128316:= func< n,k | (&+[(-1)^(j+k)*Floor(n/j)*Binomial(j-1,k-1): j in [k..n]]) >;
[A128316(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jun 23 2024

T[n_, k_]:= Sum[(-1)^(j+k)*Floor[n/j]*Binomial[j-1,k-1], {j,k,n}];
Table[T[n,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jun 23 2024 *)

def A128316(n,k): return sum((-1)^(j+k)*int(n//j)*binomial(j-1,k-1) for j in range(k,n+1))
flatten([[A128316(n,k) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Jun 23 2024

Sum_{k=1..n} T(n, k) = A000027(n) (row sums).
T(n, 1) = A059851(n).
From G. C. Greubel, Jun 23 2024: (Start)
T(n, k) = A010766(n,k) * AA130595(n-1, k-1) as infinite lower triangular matrices.
T(n, k) = Sum_{j=k..n} (-1)^(j+k) * floor(n/j) * binomial(j-1, k-1).
T(2*n-1, n) = (-1)^(n-1)*A026641(n).
T(2*n-2, n-1) = (-1)^n*A014300(n-1), for n >= 2.
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = A344817(n).
Sum_{k=1..n} k*T(n, k) = A032766(n-1).
Sum_{k=1..n} (k+1)*T(n, k) = A047215(n). (End)

a(28) = 1 inserted and more terms from Georg Fischer, Jun 06 2023

A181933 a(n) = Sum_{k=0..n} binomial(n+k,k)sin(Pi(n+k)/2).

Original entry on oeis.org

0, 1, -3, 9, -30, 106, -385, 1421, -5304, 19966, -75658, 288222, -1102790, 4234868, -16312773, 63003869, -243896960, 946066678, -3676303578, 14308370014, -55768166380, 217640082188, -850345208538, 3325907590274, -13020993588680

Offset: 0

Robert G. Wilson v, Apr 02 2012

sign

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A026641, A176332.

f[n_] := Sum[ Binomial[n + k, k] Sin[Pi (n + k)/2], {k, 0, n}]; Array[f, 25, 0]

makelist(coeff(taylor(1/2*(sqrt(4*x+1)*(1+x)-3*x-1)/(sqrt(4*x+1)*(x^2+3*x+1)-4*x^2-5*x-1),x,0,20),x,n),n,0,20); /* Vladimir Kruchinin, Mar 28 2016 */

x='x+O('x^50); concat([0], Vec((1/2)*(sqrt(4*x+1)*(1+x)-3*x-1)/(sqrt(4*x+1)*(x^2+3*x+1)-4*x^2-5*x-1))) \\ G. C. Greubel, Mar 24 2017

G.f.: (1/2)*(sqrt(4*x+1)*(1+x)-3*x-1)/(sqrt(4*x+1)*(x^2+3*x+1)-4*x^2-5*x-1). - Vladimir Kruchinin, Mar 28 2016
a(n) ~ (-1)^(n+1) *2^(2*n+1) / (5*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 28 2016
Conjecture: +2*n*a(n) +8*n*a(n-1) +(-n+20)*a(n-2) +5*(-n+4)*a(n-3) +2*(-2*n+5)*a(n-4)=0. - R. J. Mathar, Jun 14 2016

A237619 Riordan array (1/(1+xc(x)), xc(x)) where c(x) is the g.f. of Catalan numbers (A000108).

Original entry on oeis.org

1, -1, 1, 0, 0, 1, -1, 1, 1, 1, -2, 2, 3, 2, 1, -6, 6, 8, 6, 3, 1, -18, 18, 24, 18, 10, 4, 1, -57, 57, 75, 57, 33, 15, 5, 1, -186, 186, 243, 186, 111, 54, 21, 6, 1, -622, 622, 808, 622, 379, 193, 82, 28, 7, 1, -2120, 2120, 2742, 2120, 1312, 690, 311, 118, 36, 8, 1

Offset: 0

Philippe Deléham, Feb 10 2014

			Triangle begins:
    1;
   -1,  1;
    0,  0,  1;
   -1,  1,  1,  1;
   -2,  2,  3,  2,  1;
   -6,  6,  8,  6,  3,  1;
  -18, 18, 24, 18, 10,  4, 1;
  -57, 57, 75, 57, 33, 15, 5, 1;
Production matrix begins:
  -1, 1;
  -1, 1, 1;
  -1, 1, 1, 1;
  -1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1, 1, 1;
  -1, 1, 1, 1, 1, 1, 1, 1, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Diagonals: A000012, A000217, A023443, A166830.
Cf. A000957, A000958, A001558, A001559, A104629, A126983.
Cf. A065602, A167772.

A065602[n_, k_]:= A065602[n, k]= Sum[(k-1+2*j)*Binomial[2*(n-j)-k-1, n-1]/(2*(n - j) -k-1), {j,0,(n-k)/2}];
T[n_, k_]:= If[k==0, A065602[n, 0], If[n==1 && k==1, 1, A065602[n, k]]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 27 2022 *)

def A065602(n, k): return sum( (k+2*j-1)*binomial(2*n-2*j-k-1, n-1)/(2*n-2*j-k-1) for j in (0..(n-k)//2) )
def A237619(n, k):
    if (n<2): return (-1)^(n-k)
    elif (k==0): return A065602(n, 0)
    else: return A065602(n, k)
flatten([[A237619(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 27 2022

Sum_{k=0..n} T(n,k)*x^k = A126983(n), A000957(n+1), A026641(n) for x = 0, 1, 2 respectively.
T(n, k) = A167772(n-1, k-1) for k > 0, with T(n, 0) = A167772(n, 0).
T(n, 0) = A126983(n).
T(n+1, 1) = A000957(n+1).
T(n+2, 2) = A000958(n+1).
T(n+3, 3) = A104629(n) = A000957(n+3).
T(n+4, 4) = A001558(n).
T(n+5, 5) = A001559(n).
T(n, k) = A065602(n, k) for k > 0, with T(n, k) = (-1)^(n-k), for n < 2, and T(n, 0) = A065602(n, 0). - G. C. Greubel, May 27 2022

A360211 a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(2n-3k,n-2*k).

Original entry on oeis.org

1, 2, 5, 17, 61, 221, 812, 3021, 11344, 42899, 163146, 623320, 2390653, 9198879, 35494701, 137290466, 532149805, 2066501909, 8038146035, 31312535610, 122140123201, 477002869614, 1864912495716, 7298427590543, 28588888586743, 112080607196843, 439744801379594

Offset: 0

Seiichi Manyama, Jan 30 2023

nonn

Cf. A005317, A024718, A176332, A360185.
Cf. A000108, A176287.
Cf. A026641, A360212.

A360211 := proc(n)
    add((-1)^k*binomial(2*n-3*k,n-2*k),k=0..floor(n/2)) ;
end proc:
seq(A360211(n),n=0..40) ; # R. J. Mathar, Mar 02 2023

a(n) = sum(k=0, n\2, (-1)^k*binomial(2*n-3*k, n-2*k));

my(N=30, x='x+O('x^N)); Vec(1/(sqrt(1-4*x)*(1+2*x^2/(1+sqrt(1-4*x)))))

G.f.: 1 / ( sqrt(1-4*x) * (1 + x^2 * c(x)) ), where c(x) is the g.f. of A000108.
a(n) ~ 2^(2*n+3) / (9*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 18 2023
D-finite with recurrence 2*n*a(n) +(-5*n+2)*a(n-1) +(-11*n+12)*a(n-2) +2*(-n+5)*a(n-3) +(-7*n+2)*a(n-4) +2*(-2*n+5)*a(n-5)=0. - R. J. Mathar, Mar 02 2023

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A181933 a(n) = Sum_{k=0..n} binomial(n+k,k)sin(Pi(n+k)/2).

A237619 Riordan array (1/(1+xc(x)), xc(x)) where c(x) is the g.f. of Catalan numbers (A000108).

A360211 a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(2n-3k,n-2*k).