A001873 -id:A001873 - cp's OEIS frontend

A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

2, 7, 22, 70, 222, 705, 2238, 7105, 22556, 71608, 227332, 721705, 2291178, 7273743, 23091762, 73308814, 232731578, 738846865, 2345597854, 7446508273, 23640235416, 75050038224, 238259397096, 756395887969, 2401310279090, 7623377054503, 24201736119310

Offset: 0

Clark Kimberling, Sep 04 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
In the following guide to p-INVERT sequences using s = (1,1,0,0,0,...) = A019590, in some cases t(1,1,0,0,0,...) is a shifted version of the cited sequence:
p(S)                     t(1,1,0,0,0,...)
1 - S                       A000045 (Fibonacci numbers)
1 - S^2                     A094686
1 - S^3                     A115055
1 - S^4                     A291379
1 - S^5                     A281380
1 - S^6                     A281381
1 - 2 S                     A002605
1 - 3 S                     A125145
(1 - S)^2                   A001629
(1 - S)^3                   A001628
(1 - S)^4                   A001629
(1 - S)^5                   A001873
(1 - S)^6                   A001874
1 - S - S^2                 A123392
1 - 2 S - S^2               A291382
1 - S - 2 S^2               A124861
1 - 2 S - S^2               A291383
(1 - 2 S)^2                 A073388
(1 - 3 S)^2                 A291387
(1 - 5 S)^2                 A291389
(1 - 6 S)^2                 A291391
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 3 S)            A291394
(1 - 2 S)(1 - 3 S)          A291395
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 2 S)(1 - 3 S)   A291396
1 - S - S^3                 A291397
1 - S^2 - S^3               A291398
1 - S - S^2 - S^3           A186812
1 - S - S^2 - S^3 - S^4     A291399
1 - S^2 - S^4               A291400
1 - S - S^4                 A291401
1 - S^3 - S^4               A291402
1 - 2 S^2 - S^4             A291403
1 - S^2 - 2 S^4             A291404
1 - 2 S^2 - 2 S^4           A291405
1 - S^3 - S^6               A291407
(1 - S)(1 - S^2)            A291408
(1 - S^2)(1 - S)^2          A291409
1 - S - S^2 - 2 S^3         A291410
1 - 2 S - S^2 + S^3         A291411
1 - S - 2 S^2 + S^3         A291412
1 - 3 S + S^2 + S^3         A291413
1 - 2 S + S^3               A291414
1 - 3 S + S^2               A291415
1 - 4 S + S^2               A291416
1 - 4 S + 2 S^2             A291417

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, 3, 2, 1)

Cf. A019590, A290890, A291000, A291219, A291728, A292479, A292480.

z = 60; s = x + x^2; p = 1 - 2 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291382 *)

G.f.: (-2 - 3 x - 2 x^2 - x^3)/(-1 + 2 x + 3 x^2 + 2 x^3 + x^4).

a(n) = 2*a(n-1) + 3*a(n-2) + 2*a(n-3) + a(n-4) for n >= 5.

A377152 a(n) = Sum_{k=0..n} binomial(k+4,4) * binomial(k,n-k)^2.

Original entry on oeis.org

1, 5, 20, 95, 400, 1561, 5915, 21610, 76585, 265075, 898622, 2992235, 9810290, 31727815, 101379175, 320464280, 1003259080, 3113576320, 9586763720, 29305985800, 88997753446, 268642069750, 806394498200, 2408144329250, 7157177344225, 21177323087891

Offset: 0

Seiichi Manyama, Oct 18 2024

nonn

Robert Israel, Table of n, a(n) for n = 0..2357

Cf. A051286, A182884, A377145, A377148, A377153, A377158, A377159.

Cf. A001873, A089627.

f:= proc(n) local k; add(binomial(k+4,4)*binomial(k,n-k)^2,k=0..n) end proc:
map(f, [$0..50]); # Robert Israel, Dec 05 2024

a(n) = sum(k=0, n, binomial(k+4, 4)*binomial(k, n-k)^2);

a089627(n, k) = n!/((n-2*k)!*k!^2);
my(N=4, M=30, x='x+O('x^M), X=1-x-x^2, Y=3); Vec(sum(k=0, N\2, a089627(N, k)*X^(N-2*k)*x^(Y*k))/(X^2-4*x^Y)^(N+1/2))

G.f.: (Sum_{k=0..2} A089627(4,k) * (1-x-x^2)^(4-2*k) * x^(3*k)) / ((1-x-x^2)^2 - 4*x^3)^(9/2).

A178821 Triangle read by rows: T(n,k) = binomial(n+4,4) * binomial(n,k), 0 <= k <= n.

Original entry on oeis.org

1, 5, 5, 15, 30, 15, 35, 105, 105, 35, 70, 280, 420, 280, 70, 126, 630, 1260, 1260, 630, 126, 210, 1260, 3150, 4200, 3150, 1260, 210, 330, 2310, 6930, 11550, 11550, 6930, 2310, 330, 495, 3960, 13860, 27720, 34650, 27720, 13860, 3960, 495, 715, 6435, 25740, 60060, 90090, 90090, 60060, 25740, 6435, 715

Offset: 0

Harlan J. Brothers, Jun 19 2010

The product of the pentatope numbers (A000332, beginning with fifth term) and Pascal's triangle (A007318). Also level 5 of Pascal's prism (A178819) read by rows: (i+4; 4, i-j, j), i >= 0, 0 <= j <= i.

			Triangle begins:
   1;
   5,   5;
  15,  30,  15;
  35, 105, 105,  35;
  70, 280, 420, 280,  70;

G. C. Greubel, Rows n=0..100 of triangle, flattened
H. J. Brothers, Pascal's prism, The Mathematical Gazette, 96 (July 2012), 213-220.
H. J. Brothers, Pascal's Prism: Supplementary Material

Cf. A000332, A007318, A178819, A178820, A178822.

Rows sum to A003472, shallow diagonals sum to A001873.

T:=Flat(List([0..10], n-> List([0..n], k-> Binomial(n+4, 4)* Binomial(n, k) ))); # G. C. Greubel, Jan 22 2019

/* As triangle */ [[Binomial(n+4,4)*Binomial(n,k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Oct 23 2017

T:=(n,k)->binomial(n+4,4)*binomial(n,k): seq(seq(T(n,k),k=0..n),n=0..9); # Muniru A Asiru, Jan 22 2019

Table[Multinomial[4, i-j, j], {i, 0, 9}, {j, 0, i}]//Column

{T(n,k) = binomial(n+4, 4)*binomial(n, k)}; \\ G. C. Greubel, Jan 22 2019

[[binomial(n+4, 4)*binomial(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jan 22 2019

T(n,k) = C(n+4,4) * C(n,k), 0 <= k <= n.
For element a in A178819: a_(5, i, j) = (i+3; 4, i-j, j-1), i >= 1, 1 <= j <= i.
G.f.: 1/(1 - x - x*y)^5. - Ilya Gutkovskiy, Mar 20 2020

A124137 A signed aerated and skewed version of A038137.

Original entry on oeis.org

1, 0, 1, -1, 0, 2, 0, -2, 0, 3, 1, 0, -5, 0, 5, 0, 3, 0, -10, 0, 8, -1, 0, 9, 0, -20, 0, 13, 0, -4, 0, 22, 0, -38, 0, 21, 1, 0, -14, 0, 51, 0, -71, 0, 34, 0, 5, 0, -40, 0, 111, 0, -130, 0, 55, -1, 0, 20, 0, -105, 0, 233, 0, -235, 0, 89

Offset: 0

Philippe Deléham, Nov 30 2006

			Triangle begins:
1;
0, 1;
-1, 0, 2;
0, -2, 0, 3;
1, 0, -5, 0, 5;
0, 3, 0, -10, 0, 8;
-1, 0, 9, 0, -20, 0, 13;
0, -4, 0, 22, 0, -38, 0, 21;
1, 0, -14, 0, 51, 0, -71, 0, 34;
0, 5, 0, -40, 0, 111, 0, -130, 0, 55;

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

Cf. A038137, A208336, A000045, A001629, A001628, A001872, A001873

T[0, 0]:= 1; T[n_, n_]:= Fibonacci[n + 1]; T[n_, k_]:= T[n, k] = If[k < 0 || n < k, 0, T[n - 1, k - 1] + T[n - 2, k - 2] - T[n - 2, k]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten  (* G. C. Greubel, May 27 2018 *)

{T(n,k) = if(n==0 && k==0, 1, if(k==n, fibonacci(n+1), if(k<0 || nG. C. Greubel, May 27 2018

T(n,k) = T(n-1,k-1) + T(n-2,k-2) - T(n-2,k), T(0,0)=1, T(n,k)=0 if k<0 or if nA000045(n+1).

Sum_{0<=k<=n} x^k*T(n,k)= A014983(n+1), A033999(n), A056594(n), A000012(n), A015518(n+1), A015525(n+1) for x=-2, -1, 0, 1, 2, 3 respectively.

Corrected and extended by Philippe Deléham, Apr 05 2012

A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 5, 1, 8, 10, 3, 13, 20, 9, 1, 21, 38, 22, 4, 34, 71, 51, 14, 1, 55, 130, 111, 40, 5, 89, 235, 233, 105, 20, 1, 144, 420, 474, 256, 65, 6, 233, 744, 942, 594, 190, 27, 1, 377, 1308, 1836, 1324, 511, 98, 7, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 987, 3970

Offset: 0

Emeric Deutsch, Feb 18 2007

Row sums are the Jacobsthal numbers (A001045). Column 0 yields the Fibonacci numbers (A000045); the other columns yield convolved Fibonacci numbers (A001629, A001628, A001872, A001873, etc.). Sum_{k=0..floor(n/2)} k*T(n,k) = A073371(n-2).
Triangle T(n,k), with zeros omitted, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 24 2012
Riordan array (1/(1-x-x^2), x^2/(1-x-x^2)), with zeros omitted. - Philippe Deléham, Feb 06 2012
Diagonal sums are A000073(n+2) (tribonacci numbers). - Philippe Deléham, Feb 16 2014
Number of induced subgraphs of the Fibonacci cube Gamma(n-1) that are isomorphic to the hypercube Q_k. Example: row n=4 is 5, 5, 1; indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; it has 5 vertices (i.e., Q_0's), 5 edges (i.e., Q_1's) and 1 square (i.e., Q_2). - Emeric Deutsch, Aug 12 2014
Row n gives the coefficients of the polynomial p(n,x) defined as the numerator of the rational function given by f(n,x) = 1 + (x + 1)/f(n-1,x), where f(x,0) = 1. Conjecture: for n > 2, p(n,x) is irreducible if and only if n is a (prime - 2). - Clark Kimberling, Oct 22 2014

			Triangle starts:
   1;
   1;
   2,  1;
   3,  2;
   5,  5,  1;
   8, 10,  3;
  13, 20,  9,  1;
  21, 38, 22,  4;
From _Philippe Deléham_, Jan 24 2012: (Start)
Triangle (1, 1, -1, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, ...) begins:
   1;
   1,  0;
   2,  1,  0;
   3,  2,  0,  0;
   5,  5,  1,  0,  0;
   8, 10,  3,  0,  0,  0;
  13, 20,  9,  1,  0,  0,  0;
  21, 38, 22,  4,  0,  0,  0,  0; (End)
From _Clark Kimberling_, Oct 22 2014: (Start)
Here are the first 4 polynomials p(n,x) as in Comment and generated by Mathematica program:
  1
  2 +  x
  3 + 2x
  5 + 5x + x^2. (End)

C.-P. Chou and H. A. Witek, An Algorithm and FORTRAN Program for Automatic Computation of the Zhang-Zhang Polynomial of Benzenoids, MATCH: Commun. Math. Comput. Chem, 68 (2012) 3-30. See Eq. (9). - From _N. J. A. Sloane_, Dec 23 2012
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105. - _Emeric Deutsch_, Aug 12 2014

Cf. A001045, A000045, A001629, A001628, A001872, A001873, A073371.

Cf. A109466, A119473.

G:=1/(1-z-(1+t)*z^2): Gser:=simplify(series(G,z=0,19)): for n from 0 to 16 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 16 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form

p[x_, n_] := 1 + (x + 1)/p[x, n - 1]; p[x_, 1] = 1;
Numerator[Table[Factor[p[x, n]], {n, 1, 20}]]  (* Clark Kimberling, Oct 22 2014 *)

G.f.: 1/(1-z-(1+t)z^2).
Sum_{k=0..n} T(n,k)*x^k = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, and -13, respectively. - Philippe Deléham, Jan 24 2012
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Jan 24 2012
G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
T(n,k) = Sum_{i=k..floor(n/2)} binomial(n-i,i)*binomial(i,k). See Corollary 3.3 in the Klavzar et al. link. - Emeric Deutsch, Aug 12 2014

A152440 Riordan matrix (1/(1-x-x^2),x/(1-x-x^2)^2).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 9, 5, 1, 5, 22, 20, 7, 1, 8, 51, 65, 35, 9, 1, 13, 111, 190, 140, 54, 11, 1, 21, 233, 511, 490, 255, 77, 13, 1, 34, 474, 1295, 1554, 1035, 418, 104, 15, 1, 55, 942, 3130, 4578, 3762, 1925, 637, 135, 17, 1, 89, 1836, 7285, 12720, 12573, 7865, 3276

Offset: 0

Emanuele Munarini, Dec 04 2008, Dec 05 2008

From Philippe Deléham, Feb 20 2014: (Start)
T(n,0)   = A000045(n+1);
T(n+1,1) = A001628(n);
T(n+2,2) = A001873(n);
T(n+3,3) = A001875(n).
Row sums are A238236(n). (End)

			Triangle begins:
1;
1, 1;
2, 3, 1;
3, 9, 5, 1;
5, 22, 20, 7, 1;
8, 51, 65, 35, 9, 1;
13, 111, 190, 140, 54, 11, 1;
21, 233, 511, 490, 255, 77, 13, 1, etc.
- _Philippe Deléham_, Feb 20 2014

The first row is given by A000045.

Cf. A037027, A238241.

a(n,k) = sum( binomial(n-j-k,2k) binomial(n-j-k,j), j=0...(n-k)/2 )
a(n,k) = sum( binomial(i+2k,2k) binomial(n-i+k,i+2k), i=0...(n - k)/2 )
Recurrence: a(n+4,k+1) - 2 a(n+3,k+1) - a(n+3,k) - a(n+2,k+1) + 2 a(n+1,k+1) + a(n,k+1) = 0
GF for columns: 1/(1-x-x^2)(x/(1-x-x^2)^2)^k
GF: (1-x-x^2)/((1-x-x^2)^2-xy)
T(n,k) = A037027(n+k, 2*k). - Philippe Deléham, Feb 20 2014

A036683 T(n+4,4) with T as in A036355.

Original entry on oeis.org

5, 20, 71, 207, 556, 1390, 3310, 7576, 16807, 36331, 76850, 159575, 326092, 657124, 1307992, 2575180, 5020570, 9702043, 18599391, 35397328, 66918850, 125738650, 234930380, 436660010, 807690455, 1487269940, 2727149885, 4981046893

Offset: 0

Floor van Lamoen

Index entries for linear recurrences with constant coefficients, signature (5,-5,-10,15,11,-15,-10,5,5,1).

LinearRecurrence[{5,-5,-10,15,11,-15,-10,5,5,1},{5,20,71,207,556,1390,3310,7576,16807,36331},30] (* Harvey P. Dale, May 26 2025 *)

G.f.: (x^4+2x^3-4x^2-5x+5)/(1-x-x^2)^5.

a(n) = A001628(n)+3*A001872(n)+A001873(n). - R. J. Mathar, Jul 03 2022

A036684 T(n+5,5) with T as in A036355.

Original entry on oeis.org

8, 38, 149, 478, 1390, 3736, 9496, 23080, 54127, 123230, 273653, 594878, 1269532, 2665912, 5518900, 11280856, 22797331, 45599656, 90362560, 177550600, 346157050, 670060100, 1288497590, 2462607020, 4679908400, 8846662634

Offset: 0

Floor van Lamoen

Index entries for linear recurrences with constant coefficients, signature (6,-9,-10,30,6,-41,-6,30,10,-9,-6,-1).

G.f.: (3x^4+6x^3-7x^2-10x+8)/(1-x-x^2)^6.

a(n)=3*A001872(n)+4*A001873(n)+A001874(n). - R. J. Mathar, Jul 03 2022

A132883 Triangle read by rows: T(n,k) is the number of paths in the first quadrant from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k U steps (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 5, 9, 2, 8, 22, 10, 13, 51, 40, 5, 21, 111, 130, 35, 34, 233, 380, 175, 14, 55, 474, 1022, 700, 126, 89, 942, 2590, 2450, 756, 42, 144, 1836, 6260, 7770, 3570, 462, 233, 3522, 14570, 22890, 14490, 3234, 132, 377, 6666, 32870, 63600, 52668

Offset: 0

Emeric Deutsch, Sep 03 2007

Row n has 1+floor(n/2) terms. T(n,0) = A000045(n+1) (the Fibonacci numbers). T(2n,n) = binomial(2n,n)/(n+1) = A000108(n) (the Catalan numbers). Row sums yield A118720. Column k has g.f. = c(k)z^(2k)/(1-z-z^2)^(2k+1), where c(k) = binomial(2k,k)/(k+1) are the Catalan numbers; accordingly, T(n,1) = A001628(n-2), T(n,2) = 2*A001873(n-4), T(n,3) = 5*A001875(n-6). Sum_{k>=0} k*T(n,k) = A106050(n+1).

			Triangle starts:
   1;
   1;
   2,  1;
   3,  3;
   5,  9,  2;
   8, 22, 10;
  13, 51, 40,  5;
T(3,1)=3 because we have hUD, UhD and UDh.

Cf. A000045, A118720, A106050, A000108.

G:=((1-z-z^2-sqrt(1-2*z-z^2+2*z^3+z^4-4*t*z^2))*1/2)/(t*z^2): Gser:=simplify(series(G, z = 0, 17)): for n from 0 to 13 do P[n]:=sort(coeff(Gser,z,n)) end do: for n from 0 to 13 do seq(coeff(P[n],t,j),j=0..floor((1/2)*n)) end do; # yields sequence in triangular form

G.f.: G = G(t,z) satisfies G = 1 + zG + z^2*G + tz^2*G^2 (see explicit expression at the Maple program).

A182880 Triangle read by rows: T(n,k) is the number of weighted lattice paths in L_n having k (1,1)-steps. L_n is the set of lattice paths of weight n that start at (0,0) and end on the horizontal axis and whose steps are of the following four kinds: a (1,0)-step with weight 1; a (1,0)-step with weight 2; a (1,1)-step with weight 2; a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.

Original entry on oeis.org

1, 1, 2, 3, 2, 5, 6, 8, 18, 13, 44, 6, 21, 102, 30, 34, 222, 120, 55, 466, 390, 20, 89, 948, 1140, 140, 144, 1884, 3066, 700, 233, 3672, 7770, 2800, 70, 377, 7044, 18780, 9800, 630, 610, 13332, 43710, 31080, 3780, 987, 24946, 98610, 91560, 17850, 252, 1597, 46218, 216732, 254400, 72450, 2772

Offset: 0

Emeric Deutsch, Dec 11 2010

Sum of entries in row n is A051286(n).

			T(3,1)=2. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and u=(1,1), d=(1,-1), the five paths of weight 3 are ud, du, hH, Hh, and hhh; two of them, ud and du, have exactly one u step.
Triangle starts:
   1;
   1;
   2;
   3,  2;
   5,  6;
   8, 18;
  13, 44, 6;

M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177.

Cf. A051286, A000045, A182881.

G:=1/sqrt(1-2*z-z^2+2*z^3+z^4-4*t*z^3): Gser:=simplify(series(G,z=0,18)): for n from 0 to 16 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 16 do seq(coeff(P[n],t,k),k=0..floor(n/3)) od; # yields sequence in triangular form

T(n,0) = A000045(n+1) (the Fibonacci numbers).
Sum_{k=0..n} k*T(n,k) = A182881(n).
G.f.: G(t,z) = 1/sqrt(1 - 2*z - z^2 + 2*z^3 + z^4 - 4*t*z^3).
The g.f. of column k is binomial(2n,n)*z^(3n)/(1-z-z^2)^(2n+1).
Apparently, T(n,1) = 2*A001628(n-3), T(n,2) = 6*A001873(n-6), T(n,3) = 20*A001875(n-9). - R. J. Mathar, Dec 11 2010

cp's OEIS Frontend

A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

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A377152 a(n) = Sum_{k=0..n} binomial(k+4,4) * binomial(k,n-k)^2.

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Maple

PARI

PARI

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A178821 Triangle read by rows: T(n,k) = binomial(n+4,4) * binomial(n,k), 0 <= k <= n.

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GAP

Magma

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Sage

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A124137 A signed aerated and skewed version of A038137.

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Extensions

A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

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A152440 Riordan matrix (1/(1-x-x^2),x/(1-x-x^2)^2).

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A036683 T(n+4,4) with T as in A036355.

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A036684 T(n+5,5) with T as in A036355.

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