A246437 -id:A246437 - cp's OEIS frontend

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

1, 1, 3, 4, 11, 16, 42, 64, 163, 256, 638, 1024, 2510, 4096, 9908, 16384, 39203, 65536, 155382, 262144, 616666, 1048576, 2449868, 4194304, 9740686, 16777216, 38754732, 67108864, 154276028, 268435456, 614429672, 1073741824, 2448023843

Offset: 0

Clark Kimberling

Inverse binomial transform of A027914. Hankel transform (see A001906 for definition) is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 21 2005
Number of walks of length n on a line that starts at the origin and ends at or above 0. - Benjamin Phillabaum, Mar 05 2011
Number of binary integers (i.e., with a leading 1 bit) of length n+1 which have a majority of 1-bits. E.g., for n+1=4: (1011, 1101, 1110, 1111) a(3)=4. - Toby Gottfried, Dec 11 2011
Number of distinct symmetric staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum > 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A345917. For example, the a(0) = 1 through a(4) = 11 compositions are:
  (1)  (2)  (3)    (4)    (5)
            (21)   (31)   (32)
            (111)  (112)  (41)
                   (211)  (113)
                          (122)
                          (212)
                          (221)
                          (311)
                          (1121)
                          (2111)
                          (11111)
The following relate to these compositions:
- The unordered version is A027193.
- The complement is counted by A058622.
- The reverse unordered version is A086543.
- The version for alternating sum >= 0 is A116406.
- The version for alternating sum < 0 is A294175.
- Ranked by A345917. (End)
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 07 2022

			From _Gus Wiseman_, Aug 20 2021: (Start)
The a(0) = 1 through a(4) = 11 binary numbers with a majority of 1-bits (Gottfried's comment) are:
  1   11   101   1011   10011
           110   1101   10101
           111   1110   10110
                 1111   10111
                        11001
                        11010
                        11011
                        11100
                        11101
                        11110
                        11111
The version allowing an initial zero is A058622.
(End)

A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.1.6)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
F. Disanto, A. Frosini, and S. Rinaldi, Square involutions, J. Int. Seq. 14 (2011) # 11.3.5.
Zachary Hamaker and Eric Marberg, Atoms for signed permutations, arXiv:1802.09805 [math.CO], 2018.
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.

a(n) = Sum{(k+1)T(n, m-k)}, 0<=k<=[ (n+1)/2 ], T given by A008315.
Column k=2 of A226873. - Alois P. Heinz, Jun 21 2013
Cf. A008949, A248574, A001405, A246437.
The even bisection is A000302.
The odd bisection appears to be A032443.
Cf. A000984, A001700, A001791, A008549, A011782, A088218, A097805, A163493, A182616, A345197.

List([0..35],n->Sum([0..Int(n/2)],k->Binomial(n,k))); # Muniru A Asiru, Nov 27 2018

a027306 n = a008949 n (n `div` 2)  -- Reinhard Zumkeller, Nov 14 2014

[2^(n-1)+(1+(-1)^n)/4*Binomial(n, n div 2): n in [0..40]]; // Vincenzo Librandi, Jun 19 2016

a:= proc(n) add(binomial(n, j), j=0..n/2) end:
seq(a(n), n=0..32); # Zerinvary Lajos, Mar 29 2009

Table[Sum[Binomial[n, k], {k, 0, Floor[n/2]}], {n, 1, 35}]
(* Second program: *)
a[0] = a[1] = 1; a[2] = 3; a[n_] := a[n] = (2(n-1)(2a[n-2] + a[n-1]) - 8(n-2) a[n-3])/n; Array[a, 33, 0] (* Jean-François Alcover, Sep 04 2016 *)

a(n)=if(n<0,0,(2^n+if(n%2,0,binomial(n, n/2)))/2)

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k).
Odd terms are 2^(n-1). Also a(2n) - 2^(2n-1) is given by A001700. a(n) = 2^n + (n mod 2)*binomial(n, (n-1)/2).
E.g.f.: (exp(2x) + I_0(2x))/2.
O.g.f.: 2*x/(1-2*x)/(1+2*x-((1+2*x)*(1-2*x))^(1/2)). - Vladeta Jovovic, Apr 27 2003
a(n) = A008949(n, floor(n/2)); a(n) + a(n-1) = A248574(n), n > 0. - Reinhard Zumkeller, Nov 14 2014
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 2*x - 1/(1 - x) )^n.
O.g.f.: (1/2)*( 1/sqrt(1 - 4*x^2) + 1/(1 - 2*x) ).
Inverse binomial transform is (-1)^n*A246437(n).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + ... is the o.g.f. for A001405. (End)
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n-1,(2n+1-(-1)^n)/4 -k). - Anthony Browne, Jun 18 2016
D-finite with recurrence: n*a(n) + 2*(-n+1)*a(n-1) + 4*(-n+1)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Aug 09 2017

Better description from Robert G. Wilson v, Aug 30 2000 and from Yong Kong (ykong(AT)curagen.com), Dec 28 2000

A113682 Expansion of 2/(sqrt(1-2x-3x^2)(1+x+sqrt(1-2x-3*x^2))).

Original entry on oeis.org

1, 1, 4, 9, 26, 70, 197, 553, 1570, 4476, 12827, 36894, 106471, 308113, 893804, 2598313, 7567466, 22076404, 64498427, 188689684, 552675365, 1620567763, 4756614062, 13974168190, 41088418151, 120906613075, 356035078102

Offset: 0

Paul Barry, Nov 04 2005

Convolution of A002426 and A005043. Diagonal sums of A094531.
Hankel transform is A164611. - Paul Barry, Aug 17 2009
David Scambler observed that [1,0,a(n-2)] for n>=2 count the Dyck paths of semilength n such that the number of peaks equals the number of hills plus the number of returns. - Peter Luschny, Oct 22 2012
Conjectural congruences (working with an offset of 1): a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. - Peter Bala, Mar 15 2020

G. C. Greubel, Table of n, a(n) for n = 0..1000
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.

Cf. A002426, A005043, A094531, A217539, A217540, A246437, A005717.

[(Evaluate(GegenbauerPolynomial(n+1, -n-1), -1/2) + (-1)^n)/2: n in [0..40]];  // G. C. Greubel, Apr 04 2024

ex[x_]:=Module[{sx=Sqrt[1-2x-3x^2]},2/(sx (1+x+sx))]; CoefficientList[ Series[ ex[x],{x,0,40}],x] (* Harvey P. Dale, May 28 2012 *)
Flatten[{1, Table[Coefficient[Sum[(1 + x + x^2)^k, {k, 0, n}], x^n], {n, 1, 30}]}] (* Vaclav Kotesovec, Jan 08 2016 *)

makelist((ultraspherical(n+1,-n-1,-1/2)+(-1)^n)/2,n,0,12); /* Emanuele Munarini, Dec 20 2016 */

x='x+O('x^50); Vec(2/(sqrt(1-2*x-3*x^2)*(1+x+sqrt(1-2*x-3*x^2)))) \\ G. C. Greubel, Feb 28 2017

[(gegenbauer(n+1,-n-1,-1/2) +(-1)^n)/2 for n in range(41)] # G. C. Greubel, Apr 04 2024

a(n) = Sum_{k=0..floor(n/2)} ( Sum_{i=0..n-k} C(n-2k-i, i)*C(n-k, k+i) ).
a(n) = Sum_{k=0..n} A002426(k)*A005043(n-k).
a(n) = Sum_{k=0..n} C(n+1,k+1)*C(k,n-k). - Paul Barry, Aug 21 2007
a(n) = (A002426(n+1) + (-1)^n)/2. - Paul Barry, Aug 17 2009
G.f.: d/dx log(1/(1-x*A005043(x))). - Vladimir Kruchinin, Apr 18 2011
D-finite with recurrence: (n+1)*a(n) +(-n-1)*a(n-1) +(-5*n+1)*a(n-2) +3*(-n+1)*a(n-3)=0. - R. J. Mathar, Nov 26 2012
Recurrence: (n+4)*a(n+3)-(n+4)*a(n+2)-(5*n+14)*a(n+1)-3*(n+2)*a(n)=0. Remark: this recurrence can be obtained using the identity a(n) = (t(n+1)+(-1)^n)/2 and the recurrence of the central trinomial coefficients t(n) = A002426(n). So, the above P-finite recurrences are true. - Emanuele Munarini, Dec 20 2016
a(n) = (-1)^(n+1) * (hypergeom([1/2, -n-1], [1], 4) - 1)/2. - Vladimir Reshetnikov, Apr 25 2016
a(n) = (-1)^n + A246437(n+1). - Vladimir Reshetnikov, Apr 25 2016

A104507 Row sums of triangle A104505, which is equal to the right-hand side of the triangle A084610 of coefficients in (1+x-x^2)^n.

Original entry on oeis.org

1, 0, -2, -3, 2, 15, 19, -28, -134, -129, 353, 1254, 791, -4238, -11818, -3123, 49162, 110007, -17783, -554458, -996323, 690932, 6096792, 8624747, -12287153, -65419110, -69285296, 178655307, 684550946, 483569751, -2354830741, -6970706252, -2324044054, 29195280375, 68793790705

Offset: 0

Paul D. Hanna, Mar 11 2005

Cf. A104505, A1086223, A246437, A370616, A386548.

CoefficientList[Series[(x/((1 - x)) + 1/((-Sqrt[5 x^2 - 2 x + 1] + x + 1)) x (1 - (5 x - 1)/(Sqrt[5 x^2 - 2 x + 1]))), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 05 2015 *)

a(n):=sum((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j),j,0,n/2); /* Vladimir Kruchinin, Oct 04 2015 */

a(n)=sum(k=0,n,polcoeff((1+x-x^2)^n,n+k))

a(n) = sum(k=0, n/2, (-1)^k*binomial(n,k)*binomial(n-k-1,n-2*k));
vector(40, n, a(n-1)) \\ Altug Alkan, Oct 04 2015

G.f.: (x/((1-x))+1/((-sqrt(5*x^2-2*x+1)+x+1))*x*(1-(5*x-1)/(sqrt(5*x^2-2*x+1)))). - Vladimir Kruchinin, Oct 04 2015
a(n) = Sum_{j=0..n/2}((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j)). - Vladimir Kruchinin, Oct 04 2015
From Peter Bala, Jul 24 2025: (Start)
a(n) = [x^n] (1 - x^2/(1 - x))^n. Cf. A246437.
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all primes p and all positive integers n and k.
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 - x^2 - x^3 + x^4 + 4*x^5 + ... is the g.f. of A108623.(End)

A110166 Row sums of Riordan array A110165.

Original entry on oeis.org

1, 4, 18, 85, 410, 1999, 9807, 48304, 238570, 1180615, 5851253, 29033074, 144190943, 716652070, 3564079250, 17734184365, 88280673770, 439625873215, 2189988826125, 10912480440850, 54389237971285, 271142650382080

Offset: 0

Paul Barry, Jul 14 2005

Number of 5-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
From Peter Bala, Jan 09 2022: (Start)
Conjectures: for k >= 2, the number of k-ary words of length n such that the number of 1's <= the number of 0's is equal to the coefficient of x^n in the expansion of ( k*x + 1/(1 + x) )^n, and satisfies the recurrence u(0) = 1, u(1) = k-1 and n*u(n) = (k-2)*(2*n-1)*u(n-1) - k*(k-4)*(n-1)* u(n-2) + k^(n-1) for n >= 2.
For cases see A027306 (k = 2), A027914 (k = 3) and A032443 (k = 4). (End)

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

Row sums A110165. Cf. A026375, A032443, A246437, A242586.

seq( (1/2)*(5^n + add(binomial(n,k)*binomial(2*k,k), k = 0..n)), n = 0..30); # Peter Bala, Jan 08 2022

Table[Sum[Sum[Binomial[n,j]Binomial[2j,j+k],{j,0,n}],{k,0,n}],{n,0,25}] (* Harvey P. Dale, Dec 16 2011 *)

G.f.: (1/sqrt(1-6*x+5*x^2))/(1-(1-3*x-sqrt(1-6*x+5*x^2))/(2*x)).
a(n) = Sum_{k = 0..n} Sum_{j = 0..n} C(n, j)*C(2*j, j+k).
Recurrence: n*a(n) = (11*n-8)*a(n-1) - 5*(7*n-10)*a(n-2) + 25*(n-2)*a(n-3). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ 5^n/2*(1+sqrt(5)/(2*sqrt(Pi*n))). - Vaclav Kotesovec, Oct 18 2012
From Peter Bala, Jan 08 2022: (Start)
a(n) = (1/2)*(5^n + A026375(n)) = (1/2)*(5^n + Sum_{k = 0..n} binomial(n,k) *binomial(2*k,k)).
a(n) = (1/2)*(5^n)*(1 + Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(-1/5)^k).
a(n) = [x^n] ( 5*x + 1/(1 + x) )^n.
a(0) = 1, a(1) = 4 and n*a(n) = 3*(2*n-1)*a(n-1) - 5*(n-1)*a(n-2) + 5^(n-1) for n >= 2.
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.
Binomial transform of A032443. (End)

A262440 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+k-1,n-k).

Original entry on oeis.org

1, 1, 5, 22, 101, 476, 2282, 11075, 54245, 267592, 1327580, 6617128, 33110090, 166215895, 836761343, 4222640822, 21354409445, 108193910000, 549084400088, 2790744368660, 14203023709276, 72371208424880, 369170645788840, 1885051297844624

Offset: 0

Vladimir Kruchinin, Sep 23 2015

nonn

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

Cf. A109081, A246437, A262441, A262442.

[&+[Binomial(n, k)*Binomial(n+k-1, n-k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Sep 13 2015

Join[{1}, Table[Sum[ Binomial[n,k] Binomial[n+k-1, n-k], {k, n}], {n, 25}]] (* Vincenzo Librandi, Sep 23 2015 *)

a(n):=sum(binomial(n,k)*binomial(n+k-2,n-k-1),k,0,n-1)/n;
A(x):=sum(a(n)*x^n,n,1,30);
taylor(diff(A(x),x)/A(x)*x,x,0,10);

a(n)=sum(k=0,n,(binomial(n,k)*binomial(n+k-1,n-k))) \\ Anders Hellström, Sep 23 2015

G.f.: x*A'(x)/A(x), where A(x) is g.f. of A109081.
Recurrence: 2*n*(2*n-1)*(38*n^3 - 210*n^2 + 377*n - 219)*a(n) = 2*(380*n^5 - 2480*n^4 + 5998*n^3 - 6598*n^2 + 3219*n - 540)*a(n-1) + 2*(n-2)*(76*n^4 - 382*n^3 + 572*n^2 - 300*n + 45)*a(n-2) + 3*(n-3)*(n-2)*(38*n^3 - 96*n^2 + 71*n - 14)*a(n-3). - Vaclav Kotesovec, Sep 23 2015
a(n) = n^2*hypergeom([1-n, 1-n, n+1], [3/2, 2], 1/4) for n >= 1. - Peter Luschny, Mar 06 2022
a(n) = [x^n] ( (1 - x + x^2) / (1 - x)^2 )^n. - Seiichi Manyama, Apr 29 2024
a(n) ~ sqrt((513 - 67*sqrt(57))^(1/3) + (513 + 67*sqrt(57))^(1/3)) * (10 + (1261 - 57*sqrt(57))^(1/3) + (1261 + 57*sqrt(57))^(1/3))^n / (19^(1/3) * sqrt(Pi*n) * 2^(n + 5/6) * 3^(n + 1/3)). - Vaclav Kotesovec, Apr 30 2024

A290380 Analog of Motzkin sums for Coxeter type D.

Original entry on oeis.org

1, 4, 12, 36, 105, 306, 889, 2584, 7515, 21880, 63778, 186132, 543855, 1590876, 4658580, 13655472, 40065243, 117654876, 345786396, 1017040380, 2993498739, 8816790906, 25984489545, 76625467128, 226085062525, 667415280376, 1971209865654, 5824651789852

Offset: 3

F. Chapoton, Jul 28 2017

nonn

See proposition 3.3 of the Athanasiadis-Savvidou reference.

Christos A. Athanasiadis and Christina Savvidou, The Local h-Vector of the Cluster Subdivision of a Simplex, Séminaire Lotharingien de Combinatoire 66 (2012), Article B66c.

Cf. A001006, A005043 (type A), A246437 (type B).

a:= proc(n) option remember; `if`(n<5, [0$2, 1, 4][n],
      ((n-2)*(2*n-3)*(n-4)*a(n-1)+3*(n-2)*(n-3)^2*
      a(n-2))/((n-3)*(n-4)*n))
    end:
seq(a(n), n=3..35);  # Alois P. Heinz, Jul 28 2017

Table[Sum[(n - 2)/i*Binomial[2i - 2, i - 1] Binomial[n - 2, 2i - 2], {i, n/2}], {n, 3, 50}] (* Indranil Ghosh, Jul 29 2017 *)
a[n_] := (n - 2) Hypergeometric2F1[1 - n/2, 3/2 - n/2, 2, 4];
Table[a[n], {n, 3, 30}] (* Peter Luschny, Jan 23 2018 *)

def A290380(n):
    return sum(ZZ(n - 2) / i * binomial(2 * i - 2, i - 1) *
               binomial(n - 2, 2 * i - 2)
               for i in range(1, n // 2 + 1))

a(n) = Sum_{i=1..n/2} (n-2)/i*binomial(2*i-2, i-1)*binomial(n-2, 2*i-2).
From Peter Luschny, Jan 23 2018: (Start)
a(n) = (n - 2)*hypergeom([1 - n/2, 3/2 - n/2], [2], 4).
a(n) = (-1)^n (n - 2)*hypergeom([2 - n, 3/2], [3], 4).
a(n) = (n - 2)*A001006(n-2). (End)
G.f.: ((x-2)*sqrt(-3*x^2-2*x+1)-3*x^2-3*x+2)/(2*sqrt(-3*x^2-2*x+1)). - Vladimir Kruchinin, Jun 21 2024

A370616 Coefficient of x^n in the expansion of ( (1-x) / (1-x-x^2) )^n.

Original entry on oeis.org

1, 0, 2, 3, 14, 35, 125, 371, 1238, 3909, 12847, 41580, 136577, 447187, 1473341, 4855703, 16053830, 53138243, 176233967, 585202261, 1945964079, 6478043120, 21588979876, 72016891508, 240452892569, 803489258285, 2686964354375, 8991840800136, 30110638705889

Offset: 0

Seiichi Manyama, Apr 30 2024

nonn

Cf. A046736, A213684, A104507, A246437, A386548.

Table[Sum[Binomial[-1 - k + n, -2*k + n] Binomial[-1 + k + n, k], {k, 0, n/2}], {n, 0, 30}] (* Vaclav Kotesovec, Jul 30 2025 *)

a(n, s=2, t=1, u=1) = sum(k=0, n\s, binomial(t*n+k-1, k)*binomial((t-u+1)*n-(s-1)*k-1, n-s*k));

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k-1,k) * binomial(n-k-1,n-2*k).
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-x-x^2) / (1-x) ).
From Peter Bala, 26 Jul 2025: (Start)
a(n) = n * hypergeom([1 + n, 1 - n/2, 3/2 - n/2], [2, 2 - n], -4) for n >= 3.
P-recursive: 5*n*(74*n^3-493*n^2+1075*n-766)*(n-1)*a(n) = 2*(n-1)*(296*n^4-2120*n^3+5393*n^2-5716*n+2100)*a(n-1) + 2*(1184*n^5-10256*n^4+34088*n^3-53995*n^2+40397*n-11250)*a(n-2) - 2*(n-3)*(2*n-5)*(74*n^3-271*n^2+311*n-110)*a(n-3) with a(0) = 1, a(1) = 0 and a(2) = 2.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and all positive integers n and k. (End)
a(n) ~ sqrt(1/12 + sqrt(10/37)*(sin(arcsin((13*sqrt(37/10))/40)/3)/3)) * (8*((1 + sqrt(34)*cos(arccos(2461/(1088*sqrt(34)))/3))/15))^n / sqrt(Pi*n). - Vaclav Kotesovec, Jul 30 2025

A386548 a(n) = [x^n] ((1 - x)/(1 - x + x^2))^n.

Original entry on oeis.org

1, 0, -2, -3, 6, 25, 1, -147, -218, 591, 2223, -484, -14871, -18759, 68353, 222697, -116058, -1629671, -1656989, 8275203, 23266031, -20154144, -184550412, -141418628, 1019061001, 2468408775, -3122976521, -21213927840, -10837119735, 126256071125, 262294667301, -456407675223

Offset: 0

Peter Bala, Jul 25 2025

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and all positive integers n and k.

Conjecture: the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all primes p >= 5 and all positive integers n and k.

Cf. A104507, A246437, A364374, A370616.

a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 0 elif n = 2 then -2 else
( 2*(n-1)*(2*n-3)*(19*n^2-60*n+36)*a(n-1) - 2*(190*n^4-1170*n^3+2519*n^2-2229*n+666)*a(n-2) - 2*(n-3)*(2*n-3)*(19*n^2-41*n+18)*a(n-3) )/(3*n*(n-1)*(19*n^2-79*n+78)) fi; end:
seq(a(n), n = 0..30);

a[n_]:=SeriesCoefficient[((1 - x)/(1 - x + x^2))^n,{x,0,n}]; Array[a,32,0] (* Stefano Spezia, Jul 29 2025 *)

a(n) = my(x='x+O('x^(n+1))); polcoef(((1 - x)/(1 - x + x^2))^n, n); \\ Michel Marcus, Aug 03 2025

a(n) = Sum_{k = 0..floor(n/2)} binomial(-n, k)*binomial(n-k-1, n-2*k) = Sum_{k = 0
..floor(n/2)} (-1)^k*binomial(n+k-1, k)*binomial(n-k-1, n-2*k). Cf. A246437.
a(n) = -n*hypergeom([n+1, 1 - (1/2)*n, 3/2 - (1/2)*n], [2, 2 - n], 4) for n >= 3.
P-recursive: 3*n*(n - 1)*(19*n^2 - 79*n + 78)*a(n) = 2*(n - 1)*(2*n - 3)*(19*n^2 - 60*n + 36)*a(n-1) - 2*(190*n^4 - 1170*n^3 + 2519*n^2 - 2229*n + 666)*a(n-2) - 2*(n - 3)*(2*n - 3)*(19*n^2 - 41*n + 18)*a(n-3) with a(0) = 1, a(1) = 0 and a(2) = -2.
exp( Sum_{n >= 1} a(n)*(-x)^n/n ) = 1 - x^2 + x^3 + 2*x^4 - 6*x^5 - x^6 +  ... is the g.f. of A364374.

A242586 Expansion of 1/(2sqrt(1-x))(1/sqrt(1-x)+1/(sqrt(1-5*x))).

Original entry on oeis.org

1, 2, 6, 23, 98, 437, 1995, 9242, 43258, 204053, 968441, 4619012, 22120631, 106300508, 512321438, 2475395303, 11986728458, 58156146653, 282640193313, 1375737276788, 6705522150973, 32724071280518, 159878425878848

Offset: 0

Vladimir Kruchinin, May 18 2014

Binomial transform of A088218.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A007317, A088218, A105477, A110166, A246437.

a := n -> hypergeom([1/2,-n],[1], -4)/2 + 1/2;
seq(round(evalf(a(n),32)), n=0..22); # Peter Luschny, May 18 2014

CoefficientList[Series[1/(2Sqrt[1-x]) (1/Sqrt[1-x]+1/Sqrt[1-5x]),{x,0,30}],x] (* Harvey P. Dale, Mar 19 2020 *)

a(n):=sum(binomial(2*j-1,j)*binomial(n,j),j,0,n);

a(n) = Sum_{j = 0..n} binomial(2*j-1,j)*binomial(n,j).
G.f. A(x) = x*F'(x)/F(x), where F(x) is g.f. of A007317.
a(n) = T(2*n,n) for n>0, where T(n,k) is triangle of A105477.
a(n) = hypergeom([1/2,-n],[1],-4)/2 + 1/2. - Peter Luschny, May 18 2014
D-finite with recurrence: n*a(n) + (-7*n+4)*a(n-1) + (11*n-14)*a(n-2) + 5*(-n+2)*a(n-3) = 0. - R. J. Mathar, May 23 2014
2*a(n) = 1 + A026375(n). - R. J. Mathar, Jan 26 2020
From Peter Bala, Jan 09 2022: (Start)
a(n) = [x^n] ( x + 1/(1 - x) )^n.
a(0) = 1, a(1) = 2 and n*a(n) = 3*(2*n-1)*a(n-1) - 5*(n-1)*a(n-2) - 1 for n >= 2.
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. (End)

A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

Original entry on oeis.org

1, 2, 1, 1, 3, 3, 4, 2, 1, 4, 6, 10, 9, 7, 3, 1, 5, 10, 20, 25, 26, 19, 11, 4, 1, 6, 15, 35, 55, 71, 70, 56, 34, 16, 5, 1, 7, 21, 56, 105, 161, 196, 197, 160, 106, 55, 22, 6, 1, 8, 28, 84, 182, 322, 462, 554, 553, 463, 321, 183, 83, 29, 7, 1, 9, 36, 120, 294, 588, 966, 1338, 1569, 1570, 1337, 967, 587, 295, 119, 37, 8, 1, 10, 45, 165, 450, 1002, 1848, 2892, 3873, 4477, 4476, 3874

Offset: 0

Werner Schulte, Nov 08 2015

The triangle is related to the triangle of trinomial coefficients.

			The irregular triangle T(n,k) begins:
n\k:  0   1   2    3    4    5    6    7    8    9   10  11  12  13  14  ...
0     1;
1     2   1   1;
2     3   3   4    2    1;
3     4   6  10    9    7    3    1;
4     5  10  20   25   26   19   11    4    1;
5     6  15  35   55   71   70   56   34   16    5    1;
6     7  21  56  105  161  196  197  160  106   55   22   6   1;
7     8  28  84  182  322  462  554  553  463  321  183  83  29   7   1;
etc.
The polynomial corresponding to row 2 is p(2,x) = 3+3*x+4*x^2+2*x^3+x^4.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A000027 (col 0), A000217 (col 1), A000292 (col 2), A001590, A002426, A004524, A005582 (col 3), A008937, A027907, A095662 (col 5), A113682, A246437.

A027907[n_, k_] := Sum[Binomial[n, j]*Binomial[j, k - j], {j, 0, n}]; Table[ Sum[A027907[j, k], {j, 0, n}], {n,0,10}, {k, 0, 2*n} ] // Flatten (* G. C. Greubel, Mar 07 2017 *)

T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k) + T(n,k-1) + T(n,k-2) for k > 0.
T(n,k) = Sum_{j=0..n} A027907(j,k) for 0 <= k <= 2*n.
T(n,k) = Sum_{j=0..k} (-1)^(k-j)*A027907(n+1,j+1) for 0 <= k <= 2*n.
T(n,k) = T(n,2*n-1-k) + (-1)^k for 0 <= k < 2*n.
p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x+x^2)^k for n >= 0.
p(n,x) = ((1+x+x^2)^(n+1)-1)/(x+x^2), p(n,0) = p(n,-1) = n+1 for n >= 0.
p(n+1,x) = (1+x+x^2)*p(n,x)+1 for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t)*(1-t*(1+x+x^2))).
T(n,2*n) = 1, and T(n,n) = A113682(n) for n >= 0.
T(n,n-1) = A246437(n+1), and T(n,n-1)+T(n,n) = A002426(n+1) for n > 0.
If d(n) is n-th antidiagonal sum of the triangle then: d(n) = A008937(n+1), and d(n+2)-d(n) = A001590(n+5) for n >= 0.
Conjecture: If a(n) is n-th antidiagonal alternating sum of the triangle then: a(n) = A004524(n+3).
Sum_{k=0..2*n} (-1)^k*T(n,k)^2 = (3^(n+1)-1)/2 for n >= 0.
Sum_{k=0..2*n} (-1)^k*(y*k+1)*T(n,k) = Sum{k=0..n} y*k+1 = (n+1)*(y*n+2)/2 for real y and n >= 0.
Conjecture of linear recurrence for column k: Sum_{m=0..k+2} (-1)^m*T(n+m,k)* binomial(k+2,m) = 0 for k >= 0 and n >= 0.

cp's OEIS Frontend

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

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A113682 Expansion of 2/(sqrt(1-2*x-3*x^2)*(1+x+sqrt(1-2*x-3*x^2))).

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A104507 Row sums of triangle A104505, which is equal to the right-hand side of the triangle A084610 of coefficients in (1+x-x^2)^n.

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A110166 Row sums of Riordan array A110165.

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A262440 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+k-1,n-k).

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A290380 Analog of Motzkin sums for Coxeter type D.

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A370616 Coefficient of x^n in the expansion of ( (1-x) / (1-x-x^2) )^n.

A113682 Expansion of 2/(sqrt(1-2x-3x^2)(1+x+sqrt(1-2x-3*x^2))).

A242586 Expansion of 1/(2sqrt(1-x))(1/sqrt(1-x)+1/(sqrt(1-5*x))).