A109955 -id:A109955 - cp's OEIS frontend

A109956 Inverse of Riordan array (1/(1-x), x/(1-x)^3), A109955.

1, -1, 1, 3, -4, 1, -12, 18, -7, 1, 55, -88, 42, -10, 1, -273, 455, -245, 75, -13, 1, 1428, -2448, 1428, -510, 117, -16, 1, -7752, 13566, -8379, 3325, -910, 168, -19, 1, 43263, -76912, 49588, -21252, 6578, -1472, 228, -22, 1, -246675, 444015, -296010, 134550, -45630, 11700, -2223, 297, -25, 1

Offset: 0

Paul Barry, Jul 06 2005

Riordan array (g,f) where f/(1-f)^3=x and g=1-f.
First column is (-1)^n*binomial(3n,n)/(2n+1), a signed version of A001764.
Second column is a signed version of A006629.
Diagonal sums are A109957.

			Triangle begins:
      1;
     -1,   1;
      3,  -4,    1;
    -12,  18,   -7,   1;
     55, -88,   42, -10,   1;
   -273, 455, -245,  75, -13, 1;
   ...

Paul Barry, d-orthogonal polynomials, Fuss-Catalan matrices and lattice paths, arXiv:2505.16718 [math.CO], 2025. See p. 13.
Paul Drube, Generalized Path Pairs and Fuss-Catalan Triangles, arXiv:2007.01892 [math.CO], 2020. See Figure 4 p. 8 (up to signs).

Cf. A001764, A109955, A109957, A321620.

# Function RiordanSquare defined in A321620.
tt := sin(arcsin(3*sqrt(x*3/4))/3)/sqrt(x*3/4): R := RiordanSquare(tt, 11):
seq(seq(LinearAlgebra:-Row(R,n)[k]*(-1)^(n+k), k=1..n), n=1..11); # Peter Luschny, Nov 27 2018

T[n_, k_] := (-1)^(n - k)((3k + 1)/(2n + k + 1)) Binomial[3n, n - k];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019 *)

tabl(nn) = {my(m = matrix(nn, nn, n, k, if (nMichel Marcus, Nov 20 2015

Number triangle T(n, k) = (-1)^(n-k)*((3k+1)/(2n+k+1))*binomial(3n, n-k).
From Werner Schulte, Oct 27 2015: (Start)
If u(m,n) = (-1)^n*(Sum_{k=0..n} T(n,k)*((m+1)*k+1)) and v(m,n) = (-1)^n*(Sum_{k=0..n} (-1)^k*T(n,k)*m^k) and D(x) is the g.f. of A001764 then P(m,x) = Sum_{n>=0} u(m,n)*x^n = 1-(m+1)*x*D(x)^2 and Q(m,x) = Sum_{n>=0} v(m,n)*x^n = 1/P(m,x).
If G(k,x) is the g.f. of column k (k>=0) then G(k,x) = G(0,x)^(3*k+1). (End)

A049086 Number of tilings of 4 X 3n rectangle by 1 X 3 rectangles. Rotations and reflections are considered distinct tilings.

Original entry on oeis.org

1, 3, 13, 57, 249, 1087, 4745, 20713, 90417, 394691, 1722917, 7520929, 32830585, 143313055, 625594449, 2730863665, 11920848033, 52037243619, 227154537661, 991581805481, 4328482658041, 18894822411423, 82480245888473, 360045244866137, 1571680309076689, 6860746056673507

Offset: 0

Allan C. Wechsler

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mudit Aggarwal and Samrith Ram, Generating functions for straight polyomino tilings of narrow rectangles, arXiv:2206.04437 [math.CO], 2022.
R. J. Mathar, Paving Rectangular Regions with Rectangular Tiles: Tatami and Non-Tatami Tilings, arXiv:1311.6135 [math.CO], 2013, Table 19.
R. J. Mathar, Tilings of rectangular regions by rectangular tiles: Counts derived from transfer matrices, arXiv:1406.7788 (2014), eq. (10)
Index entries for linear recurrences with constant coefficients, signature (5,-3,1).

Cf. A000930, A005178.

a[0]:=1:a[1]:=3:a[2]:=13: for n from 3 to 25 do a[n]:=5*a[n-1]-3*a[n-2]+a[n-3] od: seq(a[n],n=0..25); # Emeric Deutsch, Feb 15 2005
a := n -> hypergeom([(n+1)/2, n/2+1, -n], [1/3, 2/3], -8/27):
seq(simplify(a(n)), n=0..25); # Peter Luschny, Dec 09 2020

LinearRecurrence[{5,-3,1},{1,3,13},50] (* Vincenzo Librandi, Feb 18 2012 *)
CoefficientList[Series[(1-x)^2/(1-5x+3x^2-x^3), {x, 0, 40}], x] (* M. Poyraz Torcuk, Nov 06 2021 *)

a(n) = 5*a(n-1) - 3*a(n-2) + a(n-3).
a(n)/a(n-1) tends to 4.3652300134..., an eigenvalue of the matrix M and an inverse root of the polynomial x^3 - 3x^2 + 5x - 1. [a(n-2), a(n-1), a(n)] = M^n * [1 1 1], where M = the 3 X 3 matrix [ 5 -3 1 / 1 0 0 / 0 1 0]. E.g., a(3), a(4), a(5) = 57, 249, 1087. M^5 * [1 1 1] = [57, 249, 1087] - Gary W. Adamson, Apr 25 2004
G.f.: (1-x)^2/(1-5*x+3*x^2-x^3). - Colin Barker, Feb 03 2012
a(n) = Sum_{k=0..n} A109955(n,k)*2^k. - Philippe Deléham, Feb 18 2012
a(n) = hypergeom([(n+1)/2, n/2+1, -n], [1/3, 2/3], -8/27). - Peter Luschny, Dec 09 2020

More terms from Emeric Deutsch, Feb 15 2005

A185963 Row sums of number triangle A185962.

Original entry on oeis.org

1, 0, -2, -3, 0, 7, 11, 1, -24, -40, -7, 82, 145, 37, -279, -524, -174, 945, 1888, 767, -3185, -6783, -3244, 10676, 24301, 13330, -35567, -86823, -53615, 117672, 309366, 212101, -386224, -1099385, -827997, 1255937, 3896480, 3197152, -4039199, -13773374

Offset: 0

Paul Barry, Feb 07 2011

			G.f. = 1 - 2*x^2 - 3*x^3 + 7*x^5 + 11*x^6 + x^7 - 24*x^8 - 40*x^9 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Taras Goy and Mark Shattuck, Toeplitz-Hessenberg determinant formulas for the sequence F_n-1, Online J. Anal. Comb. (2025) Vol. 19, Paper 1, 1-26. See p. 12.
Index entries for linear recurrences with constant coefficients, signature (2,-3,1).

Cf. A000931.

a := n -> hypergeom([(n+1)/2, n/2+1, -n], [1/3, 2/3], 4/27):
seq(simplify(a(n)), n=0..39); # Peter Luschny, Nov 03 2017

LinearRecurrence[{2,-3,1},{1,0,-2},50] (* Vincenzo Librandi, Feb 18 2012 *)

x='x+O('x^50); Vec((1-x)^2/(1-2*x+3*x^2-x^3)) \\ G. C. Greubel, Jul 23 2017

G.f.: (1-x)^2/(1-2x+3x^2-x^3).
a(n) = Sum_{k=0..n} Sum_{i=0..(2k+2)} C(2k+2,i)*Sum_{j=0..(n-k-i)} C(k+j,j)*C(j,n-k-i-j)*(-1)^(n-k-j).
a(n) = Sum_{k=0..n} binomial(n+2k,3k)*(-1)^k = Sum_{k=0..n} A109955(n,k)*(-1)^k. - Philippe Deléham, Feb 18 2012
a(n) = A000931(-3*n). - Michael Somos, Sep 18 2012
a(n) = hypergeom([(n+1)/2, n/2+1, -n], [1/3, 2/3], 4/27). - Peter Luschny, Nov 03 2017

More terms from Philippe Deléham, Feb 07 2012

A243116 a(n) = Sum_{k=0..n} C(n + 2k, 3k) * C(3k, 2k).

Original entry on oeis.org

1, 4, 28, 220, 1816, 15424, 133456, 1169872, 10354528, 92331904, 828204928, 7464652672, 67547774464, 613295870464, 5584367987968, 50974595472640, 466307503244800, 4273832891668480, 39237007284226048, 360768875975526400, 3321625537178669056, 30619908430235828224, 282578914501599305728

Offset: 0

Paul D. Hanna, Aug 20 2014

nonn

Compare to: Sum_{k=0..n} (-1)^k * C(n+2*k,3*k) * C(3*k,2*k) = (-2)^n for n>=0.

			G.f.: A(x) = 1 + 4*x + 28*x^2 + 220*x^3 + 1816*x^4 + 15424*x^5 +...
where
A(x) = 1/(1-x) + 3*x/(1-x)^4 + 15*x^2/(1-x)^7 + 84*x^3/(1-x)^10 + 495*x^4/(1-x)^13 + 3003*x^5/(1-x)^16 + 18564*x^6/(1-x)^19 + 116280*x^7/(1-x)^22 + 735471*x^8/(1-x)^25 +...+ C(3*n, n)*x^n/(1-x)^(3*n+1) +...
ILLUSTRATION OF TERMS.
The sequence A005809(k) = C(3*k, 2*k) begins:
  [1, 3, 15, 84, 495, 3003, 18564, 116280, 735471, 4686825, ...];
the triangle A109955(n,k) = C(n + 2*k, 3*k) begins:
  1;
  1, 1;
  1, 4, 1;
  1, 10, 7, 1;
  1, 20, 28, 10, 1;
  1, 35, 84, 55, 13, 1;
  1, 56, 210, 220, 91, 16, 1;
  1, 84, 462, 715, 455, 136, 19, 1; ...
where a(n) = Sum_{k=0..n} A109955(n,k) * A005809(k):
  a(1) = 1*1 + 1*3 = 4;
  a(2) = 1*1 + 4*3 + 1*15 = 28;
  a(3) = 1*1 + 10*3 + 7*15 + 1*84 = 220;
  a(4) = 1*1 + 20*3 + 28*15 + 10*84 + 1*495 = 1816; ...
compare to: Sum_{k=0..n} (-1)^k * A109955(n,k) * A005809(k) = (-2)^n.

Michael De Vlieger, Table of n, a(n) for n = 0..1027
Hacène Belbachir and Abdelghani Mehdaoui, Diagonal sums in Pascal pyramid (1, 2, r), Les Annales RECITS (2019) Vol. 6, 45-52.

Cf. A109955, A005809.

Table[Sum[Binomial[n + 2*k, 3*k] * Binomial[3*k, 2*k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Aug 21 2014 *)

{a(n)=sum(k=0,n, binomial(n+2*k,3*k) * binomial(3*k,2*k))}
for(n=0,30,print1(a(n),", "))

{a(n)=-(-2)^n + 2*sum(k=0, n\2, binomial(n+4*k, 6*k) * binomial(6*k, 4*k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=local(A=1); A=sum(m=0, n, binomial(3*m, m) * x^m/(1-x +x*O(x^n))^(3*m+1)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

G.f.: Sum_{n>=0} C(3*n, n) * x^n / (1-x)^(3*n+1).  - Paul D. Hanna, Aug 30 2014
G.f.: 1/(1-x) / ( 3 / G(x/(1-x)^3) - 2 ), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764. - Paul D. Hanna, Aug 30 2014
G.f. satisfies: A(x) = 1 + (4-3*x)*A(x) - (4 - 39*x + 12*x^2 - 4*x^3)*A(x)^3. - Paul D. Hanna, Sep 05 2014
a(n) = Sum_{k=0..n} A109955(n,k) * A005809(k).
a(n) = -(-2)^n + 2*Sum_{k=0..[n/2]} C(n+4*k, 6*k) * C(6*k, 4*k).
Recurrence: 2*n*(2*n-1)*(3*n-4)*a(n) = (3*n-2)*(39*n^2 - 65*n + 18)*a(n-1) - 2*(n-1)*(18*n^2 - 33*n + 10)*a(n-2) + 4*(n-2)*(n-1)*(3*n-1)*a(n-3). - Vaclav Kotesovec, Aug 21 2014
From Peter Bala, Mar 11 2022: (Start)
a(n) = Sum_{k = 0..floor(n/4)} (-1)^k*binomial(n,k)*binomial(4*n-4*k,3*n).
a(n) = [x^n] ( (1 + x)^4 - x^4 )^n. Cf. A122868(n) = [x^n] ( (1 + x)^3 - x^3 )^n.
It follows that the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)
a(n) = [x^n] (1 + x + (1 + x)^3)^n. - Seiichi Manyama, Nov 25 2024

cp's OEIS Frontend

A109956 Inverse of Riordan array (1/(1-x), x/(1-x)^3), A109955.

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A049086 Number of tilings of 4 X 3n rectangle by 1 X 3 rectangles. Rotations and reflections are considered distinct tilings.

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A185963 Row sums of number triangle A185962.

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A243116 a(n) = Sum_{k=0..n} C(n + 2*k, 3*k) * C(3*k, 2*k).

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A243116 a(n) = Sum_{k=0..n} C(n + 2k, 3k) * C(3k, 2k).