A079908 -id:A079908 - cp's OEIS frontend

A061989 Number of ways to place 3 nonattacking queens on a 3 X n board.

0, 0, 0, 0, 4, 14, 36, 76, 140, 234, 364, 536, 756, 1030, 1364, 1764, 2236, 2786, 3420, 4144, 4964, 5886, 6916, 8060, 9324, 10714, 12236, 13896, 15700, 17654, 19764, 22036, 24476, 27090, 29884, 32864, 36036, 39406, 42980, 46764, 50764

Offset: 0

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 29 2001

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Vaclav Kotesovec, Ways of placing non-attacking queens and kings..., part of "Between chessboard and computer", 1996, pp. 204 - 206.
E. Lucas, Recreations mathematiques I, Albert Blanchard, Paris, 1992, p. 231.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A061990, A117937, A229183.

Essentially the same as A079908.

[0,0,0] cat [(n-3)*(n^2-6*n+12): n in [3..50]]; // G. C. Greubel, Apr 29 2022

A061989 := proc(n)
    if n >= 3 then
        (n-3)*(n^2-6*n+12) ;
    else
        0;
    end if;
end proc:
seq(A061989(n),n=0..30) ; # R. J. Mathar, Aug 16 2019

CoefficientList[Series[2*x^4*(2-x+2*x^2)/(1-x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, May 02 2013 *)

[0,0,0]+[(n-3)*((n-3)^2 +3) for n in (3..50)] # G. C. Greubel, Apr 29 2022

G.f.: 2*x^4*(2-x+2*x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n >= 7.
Explicit formula (H. Tarry, 1890): a(n) = (n-3)*(n^2-6*n+12), n >= 3.
(4, 14, 36, ...) is the binomial transform of row 4 of A117937: (4, 10, 12, 6). - Gary W. Adamson, Apr 09 2006
a(n) = 2*A229183(n-3). - R. J. Mathar, Aug 16 2019
E.g.f.: 36 + 14*x + 2*x^2 + (-36 + 22*x - 6*x^2 + x^3)*exp(x). - G. C. Greubel, Apr 29 2022

A079928 Solution to the Dancing School Problem with n girls and n+9 boys: f(n,9).

Original entry on oeis.org

10, 91, 756, 5766, 40688, 268108, 1666000, 9856552, 56010096, 308055528, 1650607040, 8641355080, 44321202192, 223245029176, 1106667645872, 5409040818588, 26108349881112, 124618084809174, 588875653169028

Offset: 1

Jaap Spies, Jan 28 2003

nonn

f(g,h) = per(B), the permanent of the (0,1)-matrix B of size g X g+h with b(i,j)=1 if and only if i <= j <= i+h. See A079908 for more information.

Jaap Spies, Dancing School Problems, Nieuw Archief voor Wiskunde 5/7 nr. 4, Dec 2006, pp. 283-285.
Jaap Spies, Dancing School Problems, Permanent solutions of Problem 29.

Cf. A079908-A079927.

Corrected by Jaap Spies, Feb 01 2004

More terms Dec 15 2006

A119467 A masked Pascal triangle.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 0, 6, 0, 1, 0, 5, 0, 10, 0, 1, 1, 0, 15, 0, 15, 0, 1, 0, 7, 0, 35, 0, 21, 0, 1, 1, 0, 28, 0, 70, 0, 28, 0, 1, 0, 9, 0, 84, 0, 126, 0, 36, 0, 1, 1, 0, 45, 0, 210, 0, 210, 0, 45, 0, 1, 0, 11, 0, 165, 0, 462, 0, 330, 0, 55, 0, 1, 1, 0, 66, 0, 495, 0, 924

Offset: 0

Paul Barry, May 21 2006

Row sums are A011782. Diagonal sums are F(n+1)*(1+(-1)^n)/2 (aerated version of A001519). Product by Pascal's triangle A007318 is A119468. Schur product of (1/(1-x),x/(1-x)) and (1/(1-x^2),x).
Exponential Riordan array (cosh(x),x). Inverse is (sech(x),x) or A119879. - Paul Barry, May 26 2006
Rows give coefficients of polynomials p_n(x) = Sum_{k=0..n} (k+1 mod 2)*binomial(n,k)*x^(n-k) having e.g.f. exp(x*t)*cosh(t)= 1*(t^0/0!) + x*(t^1/1!) + (1+x^2)*(t^2/2!) + ... - Peter Luschny, Jul 14 2009
Inverse of the coefficient matrix of the Swiss-Knife polynomials in ascending order of x^i (reversed and aerated rows of A153641). - Peter Luschny, Jul 16 2012
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*... is equal to A136630 but with the first row and column omitted. - Peter Bala, Jul 28 2014
The row polynomials SKv(n,x) = [(x+1)^n + (x-1)^n]/2 , with e.g.f. cosh(t)*exp(xt), are the umbral compositional inverses of the row polynomials of A119879 (basically the Swiss Knife polynomials SK(n,x) of A153641); i.e., umbrally SKv(n,SK(.,x)) = x^n = SK(n,SKv(.,x)). Therefore, this entry's matrix and A119879 are an inverse pair. Both sequences of polynomials are Appell sequences, i.e., d/dx P(n,x) = n * P(n-1,x) and (P(.,x)+y)^n = P(n,x+y). In particular, (SKv(.,0)+x)^n = SKv(n,x), reflecting that the first column has the e.g.f. cosh(t). The raising operator is R = x + tanh(d/dx); i.e., R SKv(n,x) = SKv(n+1,x). The coefficients of this operator are basically the signed and aerated zag numbers A000182, which can be expressed as normalized Bernoulli numbers. The triangle is formed by multiplying the n-th diagonal of the lower triangular Pascal matrix by the Taylor series coefficient a(n) of cosh(x). More relations for this type of triangle and its inverse are given by the formalism of A133314. - Tom Copeland, Sep 05 2015
The signed version of this matrix has the e.g.f. cos(t) e^{xt}, generating Appell polynomials that have only real, simple zeros and whose extrema are maxima above the x-axis and minima below and situated above and below the zeros of the next lower degree polynomial. The bivariate versions appear on p. 27 of Dimitrov and Rusev in conditions for entire functions that are cosine transforms of a class of functions to have only real zeros. - Tom Copeland, May 21 2020
The n-th row of the triangle is obtained by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probabilities of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms of the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k-1) as the (2k-1)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k-1). - Luca Onnis, Oct 29 2023

			Triangle begins
  1,
  0, 1,
  1, 0,  1,
  0, 3,  0,  1,
  1, 0,  6,  0,   1,
  0, 5,  0, 10,   0,   1,
  1, 0, 15,  0,  15,   0,   1,
  0, 7,  0, 35,   0,  21,   0,  1,
  1, 0, 28,  0,  70,   0,  28,  0,  1,
  0, 9,  0, 84,   0, 126,   0, 36,  0, 1,
  1, 0, 45,  0, 210,   0, 210,  0, 45, 0, 1
p[0](x) = 1
p[1](x) = x
p[2](x) = 1 + x^2
p[3](x) = 3*x + x^3
p[4](x) = 1 + 6*x^2 + x^4
p[5](x) = 5*x + 10*x^3 + x^5
Connection with A136630: With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1        \/1        \      /1         \
|0 1      ||0 1      ||0 1      |      |0 1       |
|1 0 1    ||0 0 1    ||0 0 1    |... = |1 0  1    |
|0 3 0 1  ||0 1 0 1  ||0 0 0 1  |      |0 4  0 1  |
|1 0 6 0 1||0 0 3 0 1||0 0 1 0 1|      |1 0 10 0 1|
|...      ||...      ||...      |      |...       |
- _Peter Bala_, Jul 28 2014

Paul and Tatjana Ehrenfest, Über zwei bekannte Einwände gegen das Boltzmannsche H-Theorem, Physikalische Zeitschrift, vol. 8 (1907), pp. 311-314.

Reinhard Zumkeller, Rows n = 0..125 of table, flattened
Paul Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 28.
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.
Tom Copeland, Skipping over Dimensions, Juggling Zeros in the Matrix, 2020.
D. Dimitrov and P. Rusev, Zeros of entire Fourier transforms, East Journal on Approximations, Vol. 17, No. 1, p. 1-108, 2011.
Miguel Méndez and Rafael Sánchez, On the combinatorics of Riordan arrays and Sheffer polynomials: monoids, operads and monops, arXiv:1707.00336 [math.CO], 2017, Section 4.3, Example 4.
Miguel A. Méndez and Rafael Sánchez Lamoneda, Monops, Monoids and Operads: The Combinatorics of Sheffer Polynomials, The Electronic Journal of Combinatorics 25(3) (2018), #P3.25.
Luca Onnis, Animation of the Ehrenfest model.
Wikipedia, Ehrenfest model.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A131047, A153641, A162590.
From Peter Luschny, Jul 14 2009: (Start)
Cf. A034839, A162590.
p[n](k), n=0,1,...
k= 0: 1,  0,   1,    0,    1,   0, ... A128174
k= 1: 1,  1,   2,    4,    8,  16, ... A011782
k= 2: 1,  2,   5,   14,   41, 122, ... A007051
k= 3: 1,  3,  10,   36,  136,      ... A007582
k= 4: 1,  4,  17,   76,  353,      ... A081186
k= 5: 1,  5,  26,  140,  776,      ... A081187
k= 6: 1,  6,  37,  234, 1513,      ... A081188
k= 7: 1,  7,  50,  364, 2696,      ... A081189
k= 8: 1,  8,  65,  536, 4481,      ... A081190
k= 9: 1,  9,  82,  756, 7048,      ... A060531
k=10: 1, 10, 101, 1030,            ... A081192
p[n](k), k=0,1,...
p[0]: 1,1,1,1,1,1, ....... A000012
p[1]: 0,1,2,3,4,5, ....... A001477
p[2]: 1,2,5,10,17,26, .... A002522
p[3]: 0,4,14,36,76,140, .. A079908 (End)
Cf. A000182, A133314, A153641.

a119467 n k = a119467_tabl !! n !! k
a119467_row n = a119467_tabl !! n
a119467_tabl = map (map (flip div 2)) $
               zipWith (zipWith (+)) a007318_tabl a130595_tabl
-- Reinhard Zumkeller, Mar 23 2014

/* As triangle */ [[Binomial(n, k)*(1 + (-1)^(n - k))/2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 26 2015

# Polynomials: p_n(x)
p := proc(n,x) local k, pow; pow := (n,k) -> `if`(n=0 and k=0,1,n^k);
add((k+1 mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;
# Coefficients: a(n)
seq(print(seq(coeff(i!*coeff(series(exp(x*t)*cosh(t),t,16),t,i),x,n),n=0..i)),i=0..8); # Peter Luschny, Jul 14 2009

Table[Binomial[n, k] (1 + (-1)^(n - k))/2, {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 06 2015 *)
n = 15; "n-th row"
mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];
mat[[1, 2]] = 1;
mat[[n + 1, n]] = 1;
For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];
For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];
mat // MatrixForm;
P2 = Dot[mat, mat];
R1 = Simplify[
  Eigenvectors[Transpose[P2]][[1]]/
   Total[Eigenvectors[Transpose[P2]][[1]]]]
R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]
odd = R2*2^(n - 1) (* _Luca Onnis *)

@CachedFunction
def A119467_poly(n):
    R = PolynomialRing(ZZ, 'x')
    x = R.gen()
    return R.one() if n==0 else R.sum(binomial(n,k)*x^(n-k) for k in range(0,n+1,2))
def A119467_row(n):
    return list(A119467_poly(n))
for n in (0..10) : print(A119467_row(n)) # Peter Luschny, Jul 16 2012

G.f.: (1-x*y)/(1-2*x*y-x^2+x^2*y^2);
T(n,k) = C(n,k)*(1+(-1)^(n-k))/2;
Column k has g.f. (1/(1-x^2))*(x/(1-x^2))^k*Sum_{j=0..k+1} binomial(k+1,j)*sin((j+1)*Pi/2)^2*x^j.
Column k has e.g.f. cosh(x)*x^k/k!. - Paul Barry, May 26 2006
Let Pascal's triangle, A007318 = P; then this triangle = (1/2) * (P + 1/P). Also A131047 = (1/2) * (P - 1/P). - Gary W. Adamson, Jun 12 2007
Equals A007318 - A131047 since the zeros of the triangle are masks for the terms of A131047. Thus A119467 + A131047 = Pascal's triangle. - Gary W. Adamson, Jun 12 2007
T(n,k) = (A007318(n,k) + A130595(n,k))/2, 0<=k<=n. - Reinhard Zumkeller, Mar 23 2014

Edited by N. J. A. Sloane, Jul 14 2009

A197602 Floor((n+1/n)^3).

Original entry on oeis.org

8, 15, 37, 76, 140, 234, 364, 536, 756, 1030, 1364, 1764, 2236, 2786, 3420, 4144, 4964, 5886, 6916, 8060, 9324, 10714, 12236, 13896, 15700, 17654, 19764, 22036, 24476, 27090, 29884, 32864, 36036, 39406, 42980, 46764, 50764, 54986, 59436, 64120, 69044, 74214, 79636, 85316, 91260, 97474, 103964, 110736, 117796, 125150

Offset: 1

Vincenzo Librandi, Oct 17 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A079908, A014052.

Magma
```
[Floor((n+1/n)^3): n in [1..60]]
```

a(n)=n^3+3*n+(3*n^2+1)\n^3 \\ Charles R Greathouse IV, Oct 07 2015

From Bruno Berselli, Oct 17 2011: (Start)
G.f.: x*(8-17*x+25*x^2-14*x^3+6*x^4-3*x^5+x^6)/(1-x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>7.
a(n) = A079908(n) for n>3. (End)

A319284 The profiles of the backtrack tree for the n queens problem, triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 0, 1, 3, 2, 0, 1, 4, 6, 4, 2, 1, 5, 12, 14, 12, 10, 1, 6, 20, 36, 46, 40, 4, 1, 7, 30, 76, 140, 164, 94, 40, 1, 8, 42, 140, 344, 568, 550, 312, 92, 1, 9, 56, 234, 732, 1614, 2292, 2038, 1066, 352, 1, 10, 72, 364, 1400, 3916, 7552, 9632, 7828, 4040, 724, 1, 11, 90, 536, 2468, 8492, 21362, 37248, 44148, 34774, 15116, 2680

Offset: 0

Peter Luschny, Sep 16 2018

The profile (p_0, p_1, ..., p_n) is the number of nodes at each level of the tree.

			[1]
[1,  1]
[1,  2,  0]
[1,  3,  2,    0]
[1,  4,  6,    4,    2]
[1,  5,  12,  14,   12,    10]
[1,  6,  20,  36,   46,    40,     4]
[1,  7,  30,  76,  140,   164,    94,     40]
[1,  8,  42, 140,  344,   568,   550,    312,     92]
[1,  9,  56, 234,  732,  1614,  2292,   2038,   1066,    352]
[1, 10,  72, 364, 1400,  3916,  7552,   9632,   7828,   4040,    724]
[1, 11,  90, 536, 2468,  8492, 21362,  37248,  44148,  34774,  15116,  2680]
[1, 12, 110, 756, 4080, 16852, 52856, 120104, 195270, 222720, 160964, 68264, 14200]

D. E. Knuth, The Art of Computer Programming, Volume 4, Pre-fascicle 5B, Introduction to Backtracking, 7.2.2. Backtrack programming. 2018.

Peter Luschny, Rows n = 0..19, flattened
Candida Bowtell and Peter Keevash, The n-queens problem, arXiv:2109.08083 [math.CO] 2021.
V. Kotesovec, Ways of placing non-attacking queens and kings..., part of "Between chessboard and computer", 1996, pp. 204 - 206.
Peter Luschny, Julia implementation of the n queens problem with profiles
Michael Simkin, The number of n-queens configurations, arXiv:2107.13460 [math.CO] 2021.
Wikipedia, Backtracking
Wikipedia, Eight queens puzzle

Cf. A000170 (T(n,n)), A319283 (row sums), A319288 (indices of the row maxima).

Cf. A000012 (col. 0), A000027 (col. 1), A002378 (col. 2), A061989 and A079908 (col. 3), A061990 (col. 4), A061991 (col. 5), A061992 (col. 6), A061993 (col. 7), A172449 (col. 8).

Julia
```
# See the link section.
```

A180354 a(n) = n^4 + 4*n.

Original entry on oeis.org

0, 5, 24, 93, 272, 645, 1320, 2429, 4128, 6597, 10040, 14685, 20784, 28613, 38472, 50685, 65600, 83589, 105048, 130397, 160080, 194565, 234344, 279933, 331872, 390725, 457080, 531549, 614768, 707397, 810120, 923645, 1048704, 1186053

Offset: 0

Odimar Fabeny, Aug 30 2010

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

Cf. A079908.

Table[n^4+4n,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,5,24,93,272},40] (* Harvey P. Dale, Jun 12 2017 *)

a(n) = n^4 + 4*n; \\ Michel Marcus, Jan 11 2014

From Chai Wah Wu, Oct 15 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4.
G.f.: x*(3*x^3 - 23*x^2 + x - 5)/(x - 1)^5. (End)

a(0) corrected by R. J. Mathar, Sep 19 2010

A247495 Generalized Motzkin numbers: Square array read by descending antidiagonals, T(n, k) = k![x^k](exp(nx)* BesselI_{1}(2*x)/x), n>=0, k>=0.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 2, 4, 5, 3, 1, 0, 9, 14, 10, 4, 1, 5, 21, 42, 36, 17, 5, 1, 0, 51, 132, 137, 76, 26, 6, 1, 14, 127, 429, 543, 354, 140, 37, 7, 1, 0, 323, 1430, 2219, 1704, 777, 234, 50, 8, 1, 42, 835, 4862, 9285, 8421, 4425, 1514, 364, 65, 9, 1

Offset: 0

Peter Luschny, Dec 11 2014

This two-dimensional array of numbers can be seen as a generalization of the Motzkin numbers A001006 for two reasons: The case n=1 reduces to the Motzkin numbers and the columns are the values of the Motzkin polynomials M_{k}(x) = sum_{j=0..k} A097610(k,j)*x^j evaluated at the nonnegative integers.

			Square array starts:
[n\k][0][1] [2]  [3]   [4]   [5]    [6]     [7]      [8]
[0]   1, 0,  1,   0,    2,    0,     5,      0,      14, ...  A126120
[1]   1, 1,  2,   4,    9,   21,    51,    127,     323, ...  A001006
[2]   1, 2,  5,  14,   42,  132,   429,   1430,    4862, ...  A000108
[3]   1, 3, 10,  36,  137,  543,  2219,   9285,   39587, ...  A002212
[4]   1, 4, 17,  76,  354, 1704,  8421,  42508,  218318, ...  A005572
[5]   1, 5, 26, 140,  777, 4425, 25755, 152675,  919139, ...  A182401
[6]   1, 6, 37, 234, 1514, 9996, 67181, 458562, 3172478, ...  A025230
A000012,A001477,A002522,A079908, ...
.
Triangular array starts:
              1,
             0, 1,
           1, 1, 1,
          0, 2, 2, 1,
        2, 4, 5, 3, 1,
      0, 9, 14, 10, 4, 1,
   5, 21, 42, 36, 17, 5, 1,
0, 51, 132, 137, 76, 26, 6, 1.

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Main diagonal gives A247496.

Cf. A126120, A001006, A000108, A002212, A005572, A182401, A025230, A002522, A079908, A055151, A097610, A247497, A306684.

# RECURRENCE
T := proc(n,k) option remember; if k=0 then 1 elif k=1 then n else
(n*(2*k+1)*T(n,k-1)-(n-2)*(n+2)*(k-1)*T(n,k-2))/(k+2) fi end:
seq(print(seq(T(n,k),k=0..9)),n=0..6);
# OGF (row)
ogf := n -> (1-n*x-sqrt(((n-2)*x-1)*((n+2)*x-1)))/(2*x^2):
seq(print(seq(coeff(series(ogf(n),x,12),x,k),k=0..9)),n=0..6);
# EGF (row)
egf := n -> exp(n*x)*hypergeom([],[2],x^2):
seq(print(seq(k!*coeff(series(egf(n),x,k+2),x,k),k=0..9)),n=0..6);
# MOTZKIN polynomial (column)
A097610 := proc(n,k) if type(n-k,odd) then 0 else n!/(k!*((n-k)/2)!^2* ((n-k)/2+1)) fi end: M := (k,x) -> add(A097610(k,j)*x^j,j=0..k):
seq(print(seq(M(k,n),n=0..9)),k=0..6);
# OGF (column)
col := proc(n, len) local G; G := A247497_row(n); (-1)^(n+1)* add(G[k+1]/(x-1)^(k+1), k=0..n); seq(coeff(series(%, x, len+1),x,j), j=0..len) end: seq(print(col(n,8)), n=0..6); # Peter Luschny, Dec 14 2014

T[0, k_] := If[EvenQ[k], CatalanNumber[k/2], 0];
T[n_, k_] := n^k*Hypergeometric2F1[(1 - k)/2, -k/2, 2, 4/n^2];
Table[T[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 03 2017 *)

def A247495(n,k):
    if n==0: return(k//2+1)*factorial(k)/factorial(k//2+1)^2 if is_even(k) else 0
    return n^k*hypergeometric([(1-k)/2,-k/2],[2],4/n^2).simplify()
for n in (0..7): print([A247495(n,k) for k in range(11)])

T(n,k) = (n*(2*k+1)*T(n,k-1)-(n-2)*(n+2)*(k-1)*T(n,k-2))/(k+2) for k>=2.
T(n,k) = Sum_{j=0..floor(k/2)} n^(k-2*j)*binomial(k,2*j)*binomial(2*j,j)/(j+1).
T(n,k) = n^k*hypergeom([(1-k)/2,-k/2], [2], 4/n^2) for n>0.
T(n,n) = A247496(n).
O.g.f. for row n: (1-n*x-sqrt(((n-2)*x-1)*((n+2)*x-1)))/(2*x^2).
O.g.f. for row n: R(x)/x where R(x) is series reversion of x/(1+n*x+x^2).
E.g.f. for row n: exp(n*x)*hypergeom([],[2],x^2).
O.g.f. for column k: the k-th column consists of the values of the k-th Motzkin polynomial M_{k}(x) evaluated at x = 0,1,2,...; M_{k}(x) = sum_{j=0..k} A097610(k,j)*x^j = sum_{j=0..k} (-1)^j*binomial(k,j)*A001006(j)*(x+1)^(k-j).
O.g.f. for column k: sum_{j=0..k} (-1)^(k+1)*A247497(k,j)/(x-1)^(j+1). - Peter Luschny, Dec 14 2014
O.g.f. for row n: 1/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
T(n,k) is the coefficient of x^k in the expansion of 1/(k+1) * (1 + n*x + x^2)^(k+1). - Seiichi Manyama, May 07 2019

A180355 a(n) = n^5 + 5n.

Original entry on oeis.org

0, 6, 42, 258, 1044, 3150, 7806, 16842, 32808, 59094, 100050, 161106, 248892, 371358, 537894, 759450, 1048656, 1419942, 1889658, 2476194, 3200100, 4084206, 5153742, 6436458, 7962744, 9765750, 11881506, 14349042, 17210508, 20511294

Offset: 0

Odimar Fabeny, Aug 30 2010

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A132411, A079908, A180354.

a(n)=n^5+5*n \\ Charles R Greathouse IV, Oct 16 2015

From R. J. Mathar, Sep 24 2010: (Start)
G.f.: 6*x*(1+x+16*x^2+x^3+x^4)/(x-1)^6.
a(n) = +6*a(n-1) -15*a(n-2) +20*a(n-3) -15*a(n-4) +6*a(n-5) -a(n-6). (End)

First term corrected by Odimar Fabeny, Sep 23 2010

a(0) corrected by R. J. Mathar, Sep 24 2010

A352362 Array read by ascending antidiagonals. T(n, k) = L(k, n) where L are the Lucas polynomials.

Original entry on oeis.org

2, 2, 0, 2, 1, 2, 2, 2, 3, 0, 2, 3, 6, 4, 2, 2, 4, 11, 14, 7, 0, 2, 5, 18, 36, 34, 11, 2, 2, 6, 27, 76, 119, 82, 18, 0, 2, 7, 38, 140, 322, 393, 198, 29, 2, 2, 8, 51, 234, 727, 1364, 1298, 478, 47, 0, 2, 9, 66, 364, 1442, 3775, 5778, 4287, 1154, 76, 2

Offset: 0

Peter Luschny, Mar 18 2022

			Array starts:
n\k 0, 1,  2,   3,    4,     5,      6,       7,        8, ...
--------------------------------------------------------------
[0] 2, 0,  2,   0,    2,     0,      2,       0,        2, ... A010673
[1] 2, 1,  3,   4,    7,    11,     18,      29,       47, ... A000032
[2] 2, 2,  6,  14,   34,    82,    198,     478,     1154, ... A002203
[3] 2, 3, 11,  36,  119,   393,   1298,    4287,    14159, ... A006497
[4] 2, 4, 18,  76,  322,  1364,   5778,   24476,   103682, ... A014448
[5] 2, 5, 27, 140,  727,  3775,  19602,  101785,   528527, ... A087130
[6] 2, 6, 38, 234, 1442,  8886,  54758,  337434,  2079362, ... A085447
[7] 2, 7, 51, 364, 2599, 18557, 132498,  946043,  6754799, ... A086902
[8] 2, 8, 66, 536, 4354, 35368, 287298, 2333752, 18957314, ... A086594
[9] 2, 9, 83, 756, 6887, 62739, 571538, 5206581, 47430767, ... A087798
A007395|A059100|
    A001477 A079908

Eric Weisstein's World of Mathematics, Lucas Polynomial

Rows: A010673, A000032, A002203, A006497, A014448, A087130, A085447, A086902, A086594, A087798.
Columns: A007395, A001477, A059100, A079908.
Cf. A320570 (main diagonal), A114525, A309220 (variant), A117938 (variant), A352361 (Fibonacci polynomials), A350470 (Jacobsthal polynomials).

T := (n, k) -> (n/2 + sqrt((n/2)^2 + 1))^k + (n/2 - sqrt((n/2)^2 + 1))^k:
seq(seq(simplify(T(n - k, k)), k = 0..n), n = 0..10);

Table[LucasL[k, n], {n, 0, 9}, {k, 0, 9}] // TableForm
(* or *)
T[ 0, k_] := 2 Mod[k+1, 2]; T[n_, 0] := 2;
T[n_, k_] := n^k Hypergeometric2F1[1/2 - k/2, -k/2, 1 - k, -4/n^2];
Table[T[n, k], {n, 0, 9}, {k, 0, 8}] // TableForm

T(n, k) = ([0, 1; 1, k]^n*[2; k])[1, 1] ;
export(T)
for(k = 0, 9, print(parvector(10, n, T(n - 1, k))))

T(n, k) = Sum_{j=0..floor(k/2)} binomial(k-j, j)*(k/(k-j))*n^(k-2*j) for k >= 1.
T(n, k) = (n/2 + sqrt((n/2)^2 + 1))^k + (n/2 - sqrt((n/2)^2 + 1))^k.
T(n, k) = [x^k] ((2 - n*x)/(1 - n*x - x^2)).
T(n, k) = n^k*hypergeom([1/2 - k/2, -k/2], [1 - k], -4/n^2) for n,k >= 1.

cp's OEIS Frontend

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A119467 A masked Pascal triangle.

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A247495 Generalized Motzkin numbers: Square array read by descending antidiagonals, T(n, k) = k![x^k](exp(nx)* BesselI_{1}(2*x)/x), n>=0, k>=0.