A046854 -id:A046854 - cp's OEIS frontend

A123148 Triangle read by rows: T(n,k) is the coefficient of x^k in the polynomial p(n,x) defined by p(0,x) = -1, p(1,x) = x-2, p(n,x) = -xp(n-1,x) + 2p(n-2,x) for n >= 3 and 0 <= k <= n.

-1, -2, 1, -2, 2, -1, -4, 4, -2, 1, -4, 8, -6, 2, -1, -8, 12, -12, 8, -2, 1, -8, 24, -24, 16, -10, 2, -1, -16, 32, -48, 40, -20, 12, -2, 1, -16, 64, -80, 80, -60, 24, -14, 2, -1, -32, 80, -160, 160, -120, 84, -28, 16, -2, 1, -32, 160, -240, 320, -280, 168, -112, 32, -18, 2, -1

Offset: 0

Roger L. Bagula, Oct 01 2006

			The first few polynomials, p(n,x), are:
  p(0,x) = -1;
  p(1,x) = -2 +   x;
  p(2,x) = -2 + 2*x -   x^2;
  p(3,x) = -4 + 4*x - 2*x^2 +   x^3;
  p(4,x) = -4 + 8*x - 6*x^2 + 2*x^3 - x^4;
The triangle, T(n, k) = [x^k] p(n, x), begins as:
  -1;
  -2,  1;
  -2,  2,  -1;
  -4,  4,  -2,  1;
  -4,  8,  -6,  2,  -1;
  -8, 12, -12,  8,  -2,  1;
  -8, 24, -24, 16, -10,  2, -1;

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

Cf. A000034, A001045, A016116, A038754, A046854, A171647.

A123148:=func< n,k | (-1)^(k+1)*2^Floor((n-k+1)/2)*Binomial( Floor((n+k)/2), k) >;
[A123148(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 17 2023

p[0]:=-1: p[1]:=x-2: for n from 2 to 10 do p[n]:=sort(expand(-x*p[n-1]+2*p[n-2])) od: for n from 0 to 10 do seq(coeff(p[n],x,k),k=0..n) od; # yields sequence in triangular form

p[0,x]= -1; p[1,x]= x-2; p[k_, x_]:= p[k,x]= -x*p[k-1,x] + 2*p[k-2,x];
T[n_, k_]:= Coefficient[p[n, x], x, k];
Table[T[n,k], {n,0,12},{k,0,n}]//Flatten

def A123148(n,k): return (-1)^(k+1)*2^((n-k+1)//2)*binomial((n+k)//2, k)
flatten([[A123148(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 17 2023

T(n, k) = [x^k]( p(n,x) ), where p(0,x) = -1, p(1,x) = x-2, p(n,x) = -x*p(n-1,x) + 2*p(n-2,x).
Sum_{k=0..n} T(n, k) = -1.
Sum_{k=0..n} (-1)^k * T(n,k) = -A001045(n+2).
From G. C. Greubel, Jul 17 2023: (Start)
T(n,k) = (-1)^(k+1)*2^Floor((n-k+1)/2)*Binomial( Floor((n+k)/2), k).
T(n,k) = (-1)^(k+1)*2^Floor((n-k+1)/2)*A046854(n,k).
T(n,0) = -A016116(n+1).
T(n,1) = A171647(n).
Sum_{k=0..n} (-1)^k * abs(T(n,k)) = 1.
Sum_{k=0..floor(n/2)} T(n-k,k) = - A000034(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k,k) = -A038754(n). (End)

Edited by N. J. A. Sloane, Oct 29 2006

A126127 Inverse square of A061554.

Original entry on oeis.org

1, -2, 1, -1, -2, 1, 5, -3, -2, 1, 2, 9, -5, -2, 1, -13, 9, 13, -7, -2, 1, -5, -33, 20, 17, -9, -2, 1, 34, -27, -61, 35, 21, -11, -2, 1, 13, 111, -73, -97, 54, 25, -13, -2, 1, -89, 80, 248, -151, -141, 77, 29, -15, -2, 1, -34, -355, 252, 461, -269, -193, 104, 33, -17, -2, 1

Offset: 0

Gary W. Adamson, Dec 17 2006

Inverse of A061554 = A046854; therefore A126127 = (A046854)^2.

			First few rows of the triangle are:
1;
-2, 1;
-1, -2, 1;
5, -3, -2, 1;
2, 9, -5, -2, 1;
-13, 9, 13, -7, -2, 1;
...

Robert Israel, Table of n, a(n) for n = 0..10009 (rows 0 to 140, flattened)

Cf. A061554, A046854.

T:= Matrix(20,20,(n,k) -> binomial(n-1, floor((n)/2 - (-1)^(n-k)*(k)/2)), shape=triangular[lower]):
A:= T^(-2):
seq(seq(A[i,k],k=1..i),i=1..20); # Robert Israel, Jul 07 2019

Given M = Pascal's triangle with descending row terms, (A061554); A126127 = M^(-2).

G.f. as triangle (conjectured): (1-x)*(1-x+x^2)/(1-x*y+3*x^2-x^3*y+x^4). - Robert Israel, Jul 07 2019

A153764 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,-1,0,0,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 3, 1, 1, 0, 1, 3, 3, 4, 1, 1, 0, 1, 3, 6, 4, 5, 1, 1, 0, 1, 4, 6, 10, 5, 6, 1, 1, 0, 1, 4, 10, 10, 15, 6, 7, 1, 1, 0, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 0, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 0, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 0

Offset: 0

Philippe Deléham, Jan 01 2009

A130595*A153342 as infinite lower triangular matrices. Reflected version of A103631. Another version of A046854. Row sums are Fibonacci numbers (A000045).

A055830*A130595 as infinite lower triangular matrices.

			Triangle begins:
  1;
  1, 0;
  1, 1, 0;
  1, 1, 1, 0;
  1, 2, 1, 1, 0;
  1, 2, 3, 1, 1, 0;
  1, 3, 3, 4, 1, 1, 0;
  ...

G. C. Greubel, Table of n, a(n) for the first 45 rows

Cf. A046854, A103631.

/* As triangle */ [[Binomial(Floor((n+k-1)/2),k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 28 2016

Table[Binomial[Floor[(n + k - 1)/2], k], {n, 0, 45}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 27 2016 *)

T(n,k) = binomial(floor((n+k-1)/2),k).
Sum_{k=0..n} T(n,k)*x^k = A122335(n-1), A039834(n-2), A000012(n), A000045(n+1), A001333(n), A003688(n), A015448(n), A015449(n), A015451(n), A015453(n), A015454(n), A015455(n), A015456(n), A015457(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively. - Philippe Deléham, Dec 17 2011
Sum_{k=0..n} T(n,k)*x^(n-k) = A152163(n), A000007(n), A000045(n+1), A026597(n), A122994(n+1), A158608(n), A122995(n+1), A158797(n), A122996(n+1), A158798(n), A158609(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Dec 17 2011
G.f.: (1+(1-y)*x)/(1-y*x-x^2). - Philippe Deléham, Dec 17 2011
T(n,k) = T(n-1,k-1) + T(n-2,k), T(0,0) = T(1,0) = T(2,0) = T(2,1) = 1, T(1,1) = T(2,2) = 0, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 09 2013

A213211 Triangular array read by rows: T(n,k) is the number of size k subsets of {1,2,...,n} such that (when the elements are arranged in increasing order) the smallest element is congruent to 1 mod 3 and the difference of every pair of successive elements is also congruent to 1 mod 3.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 3, 4, 1, 1, 1, 1, 3, 3, 4, 5, 1, 1, 1, 1, 3, 6, 4, 5, 6, 1, 1, 1, 1, 3, 6, 10, 5, 6, 7, 1, 1, 1, 1, 4, 6, 10, 15, 6, 7, 8, 1, 1, 1, 1, 4, 10, 10, 15, 21, 7, 8, 9, 1, 1, 1, 1, 4, 10, 20, 15, 21, 28, 8, 9, 10, 1, 1, 1

Offset: 0

Geoffrey Critzer, Mar 02 2013

Row sums are A000930.

			T(6,3) = 4 because we have: {1,2,3}, {1,2,6}, {1,5,6}, {4,5,6}.
1;
1, 1;
1, 1, 1;
1, 1, 1,  1;
1, 2, 1,  1,  1;
1, 2, 3,  1,  1, 1;
1, 2, 3,  4,  1, 1, 1;
1, 3, 3,  4,  5, 1, 1, 1;
1, 3, 6,  4,  5, 6, 1, 1, 1;
1, 3, 6, 10,  5, 6, 7, 1, 1, 1;
1, 4, 6, 10, 15, 6, 7, 8, 1, 1, 1;

Combinatorial Enumeration, I. Goulden and D. Jackson, John Wiley and Sons, 1983, page 56.

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A046854.

T:= (n, k)-> binomial(k+floor((n-k)/3), k):
seq(seq(T(n,k), k=0..n), n=0..14);  # Alois P. Heinz, Mar 02 2013

nn=10;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[ (1+x+x^2)/(1-x^3-y x),{x,0,nn}],{x,y}]]//Grid

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

T(n,k) = C(k+floor((n-k)/3),k). - Alois P. Heinz, Mar 02 2013

A274742 Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's that begin with 1 and have exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 4, 3, 1, 3, 6, 6, 4, 1, 3, 9, 12, 8, 5, 1, 4, 12, 18, 20, 10, 6, 1, 4, 16, 30, 30, 30, 12, 7, 1, 5, 20, 40, 60, 45, 42, 14, 8, 1, 5, 25, 60, 80, 105, 63, 56, 16, 9, 1, 6, 30, 75, 140, 140, 168, 84, 72, 18, 10, 1, 6, 36, 105, 175, 280, 224, 252, 108, 90, 20, 11, 1, 7, 42, 126, 280, 350, 504, 336, 360, 135, 110, 22, 12, 1

Offset: 3

Jeremy Dover, Jul 04 2016

It appears that the row sums give the positive integers of A001629. - Omar E. Pol, Jul 09 2016

			n=3 => 100 -> T(3,0) = 1.
n=4 => 1001 -> T(4,0) = 1; 1100 -> T(4,1) = 1.
n=5 => 10010, 10100 -> T(5,0) = 1; 10011, 11001 -> T(5,1) = 2;
       11100 -> T(5,2) = 1.
Triangle starts:
1
1, 1
2, 2, 1
2, 4, 3, 1
3, 6, 6, 4, 1
3, 9, 12, 8, 5, 1
4, 12, 18, 20, 10, 6, 1
4, 16, 30, 30, 30, 12, 7, 1
5, 20, 40, 60, 45, 42, 14, 8, 1
5, 25, 60, 80, 105, 63, 56, 16, 9, 1
6, 30, 75, 140, 140, 168, 84, 72, 18, 10, 1
6, 36, 105, 175, 280, 224, 252, 108, 90, 20, 11, 1
7, 42, 126, 280, 350, 504, 336, 360, 135, 110, 22, 12, 1

Cf. A046854, A274228.

Columns: A008619, A087811.

Table[Binomial[Floor[(n + k - 2)/2], k] Floor[(n - k - 1)/2], {n, 3, 15}, {k, 0, n - 3}] // Flatten (* Michael De Vlieger, Jul 05 2016 *)

t(n, k) = binomial(floor((n+k-2)/2), k) * floor((n-k-1)/2)
trianglerows(n) = for(x=3, n+2, for(y=0, x-3, print1(t(x, y), ", ")); print(""))
trianglerows(13) \\ Felix Fröhlich, Jul 05 2016

T(n,k) = binomial(floor((n+k-2)/2),k)*floor((n-k-1)/2).

A322596 Square array read by descending antidiagonals (n >= 0, k >= 0): let b(n,k) = (n+k)!/((n+1)!*k!); then T(n,k) = b(n,k) if b(n,k) is an integer, and T(n,k) = floor(b(n,k)) + 1 otherwise.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 4, 3, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 4, 7, 9, 7, 4, 1, 1, 1, 4, 10, 14, 14, 10, 4, 1, 1, 1, 5, 12, 21, 26, 21, 12, 5, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 6, 19, 42, 66, 77, 66, 42, 19, 6, 1, 1, 1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1, 1

Offset: 0

Franck Maminirina Ramaharo, Jan 22 2019

For n >= 1, T(n,k) is the number of nodes in n-dimensional space for Mysovskikh's cubature formula which is exact for any polynomial of degree k of n variables.

			Array begins:
  1, 1, 1,  1,  1,   1,   1,    1,    1,    1, ...
  1, 1, 2,  2,  3,   3,   4,    4,    5,    5, ...
  1, 1, 2,  4,  5,   7,  10,   12,   15,   19, ...
  1, 1, 3,  5,  9,  14,  21,   30,   42,   55, ...
  1, 1, 3,  7, 14,  26,  42,   66,   99,  143, ...
  1, 1, 4, 10, 21,  42,  77,  132,  215,  334, ...
  1, 1, 4, 12, 30,  66, 132,  246,  429,  715, ...
  1, 1, 5, 15, 42,  99, 215,  429,  805, 1430, ...
  1, 1, 5, 19, 55, 143, 334,  715, 1430, 2702, ...
  1, 1, 6, 22, 72, 201, 501, 1144, 2431, 4862, ...
  ...
As triangular array, this begins:
  1;
  1, 1;
  1, 1,  1;
  1, 2,  1,  1;
  1, 2,  2,  1,  1;
  1, 3,  4,  3,  1,  1;
  1, 3,  5,  5,  3,  1,  1;
  1, 4,  7,  9,  7,  4,  1,  1;
  1, 4, 10, 14, 14, 10,  4,  1, 1;
  1, 5, 12, 21, 26, 21, 12,  5, 1, 1;
  1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1;
  ...

Ronald Cools, Encyclopaedia of Cubature Formulas
Ivan P. Mysovskikh, On the construction of cubature formulae for very simple domains, USSR Computational Mathematics and Mathematical Physics, Volume 4, Issue 1, 1964, 1-17.

Main diagonal: A000108.

Cf. A007318, A037306, A046854, A047996, A065941, A241926, A267632.

b(n, k) := (n + k)!/((n + 1)!*k!)$
T(n, k) := if integerp(b(n, k)) then b(n, k) else floor(b(n, k)) + 1$
create_list(T(k, n - k), n, 0, 15, k, 0, n);

A091650 Let M = the 4 X 4 matrix [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 -1 3 2]. Set seed vector = [1 1 1 1] = first row, then take M[1 1 1 1] = [1 1 1 3] then M [1 1 1 3], etc. Sequence gives terms in rightmost column.

Original entry on oeis.org

1, 3, 7, 21, 59, 171, 491, 1415, 4073, 11729, 33771, 97241, 279993, 806209, 2321385, 6684163, 19246279, 55417453, 159568195, 459458307, 1322957467, 3809304207, 10968454313, 31582405473, 90937912211, 261845282321, 753953441489, 2170922412257, 6250921954449

Offset: 1

Gary W. Adamson, Jan 25 2004

a(n)/a(n-1) tends to a 9-Gon diagonal.
The other 3 columns are offsets of 1, 3, 7, 21, 59, ... starting with 1's.
The characteristic equation of the 4 X 4 matrix is x^4 - 2x^3 - 3x^4 + x + 1 (coefficients may be found in A066170) with roots 2.879385241..., -1, -.5320888862... and .65270364466... An alternative matrix giving the same eigenvalues (refer to A046854) relates to the 9-Gon: [1 1 1 1 / 1 1 1 0 / 1 1 0 0 / 1 0 0 0] since the eigenvalue 2.8793852...is the longest diagonal of the 9-Gon given edge = 1. Or, 2.879385... = 1/(2*cos(k*Pi/9)), k = 4.

			a(5) = 59 since M*[1 1 1 1] then 4 iterates = [3 7 21 59]. a(5) = rightmost term.
a(10)/a(9) = 11729/4073 = 2.8796955...

Index entries for linear recurrences with constant coefficients, signature (2,3,-1,-1).

Cf. A066170, A046854.

Rest[CoefficientList[Series[x (1+x-2x^2-x^3)/(1-2x-3x^2+x^3+x^4),{x,0,40}],x]] (* or *) LinearRecurrence[{2,3,-1,-1},{1,3,7,21},40] (* Harvey P. Dale, Feb 17 2012 *)

Vec((1+x-2*x^2-x^3)/(1-2*x-3*x^2+x^3+x^4)+O(x^99)) \\ Charles R Greathouse IV, Jan 31 2012

G.f.: x*(1+x-2*x^2-x^3)/(1-2*x-3*x^2+x^3+x^4). - Colin Barker, Jan 31 2012

a(1)=1, a(2)=3, a(3)=7, a(4)=21, a(n)=2*a(n-1)+3*a(n-2)-a(n-3)-a(n-4). - Harvey P. Dale, Feb 17 2012

More terms from Harvey P. Dale, Feb 17 2012

A306210 T(n,k) = binomial(n + k, n) - binomial(n + floor(k/2), n) + 1, square array read by descending antidiagonals (n >= 0, k >= 0).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 3, 4, 4, 1, 1, 3, 8, 7, 5, 1, 1, 4, 10, 17, 11, 6, 1, 1, 4, 16, 26, 31, 16, 7, 1, 1, 5, 19, 47, 56, 51, 22, 8, 1, 1, 5, 27, 65, 112, 106, 78, 29, 9, 1, 1, 6, 31, 101, 176, 232, 183, 113, 37, 10, 1, 1, 6, 41, 131, 296, 407, 435, 295, 157, 46, 11, 1

Offset: 0

Franck Maminirina Ramaharo, Jan 29 2019

There are at most T(n,k) possible values for the number of knots in an interpolatory cubature formula of degree k for an integral over an n-dimensional region.

			Square array begins:
  1, 1,  1,   1,   1,    1,    1,    1,     1,  ...
  1, 2,  2,   3,   3,    4,    4,    5,     5,  ...
  1, 3,  4,   8,  10,   16,   19,   27,    31,  ...
  1, 4,  7,  17,  26,   47,   65,  101,   131,  ...
  1, 5, 11,  31,  56,  112,  176,  296,   426,  ...
  1, 6, 16,  51, 106,  232,  407,  737,  1162,  ...
  1, 7, 22,  78, 183,  435,  841, 1633,  2794,  ...
  1, 8, 29, 113, 295,  757, 1597, 3313,  6106,  ...
  1, 9, 37, 157, 451, 1243, 2839, 6271, 12376,  ...
  ...
As triangular array, this begins:
  1;
  1, 1;
  1, 2,  1;
  1, 2,  3,  1;
  1, 3,  4,  4,  1;
  1, 3,  8,  7,  5,  1;
  1, 4, 10, 17, 11,  6,  1;
  1, 4, 16, 26, 31, 16,  7, 1;
  1, 5, 19, 47, 56, 51, 22, 8, 1;
  ...

Ronald Cools, Encyclopaedia of Cubature Formulas
Ronald Cools, A Survey of Methods for Constructing Cubature Formulae, In: Espelid T.O., Genz A. (eds), Numerical Integration, NATO ASI Series (Series C: Mathematical and Physical Sciences), Vol. 357, 1991, Springer, Dordrecht, pp. 1-24.
T. N. L. Patterson, On the Construction of a Practical Ermakov-Zolotukhin Multiple Integrator, In: Keast P., Fairweather G. (eds), Numerical Integration, NATO ASI Series (Series C: Mathematical and Physical Sciences), Vol. 203, 1987, Springer, Dordrecht, pp. 269-290.

Cf. A007318, A046854, A322596.

T[n_, k_] = Binomial[n + k, n] - Binomial[n + Floor[k/2], n] + 1;
Table[T[k, n - k], {k, 0, n}, {n, 0, 20}] // Flatten

T(n, k) := binomial(n + k, n) - binomial(n + floor(k/2), n) + 1$
create_list(T(k, n - k), n, 0, 20, k, 0, n);

T(n,k) = A007318(n+k,n) - A046854(n+k,n) + 1.

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

cp's OEIS Frontend

A123148 Triangle read by rows: T(n,k) is the coefficient of x^k in the polynomial p(n,x) defined by p(0,x) = -1, p(1,x) = x-2, p(n,x) = -x*p(n-1,x) + 2*p(n-2,x) for n >= 3 and 0 <= k <= n.

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A126127 Inverse square of A061554.

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A153764 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,-1,0,0,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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A213211 Triangular array read by rows: T(n,k) is the number of size k subsets of {1,2,...,n} such that (when the elements are arranged in increasing order) the smallest element is congruent to 1 mod 3 and the difference of every pair of successive elements is also congruent to 1 mod 3.

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A274742 Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's that begin with 1 and have exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.

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A322596 Square array read by descending antidiagonals (n >= 0, k >= 0): let b(n,k) = (n+k)!/((n+1)!*k!); then T(n,k) = b(n,k) if b(n,k) is an integer, and T(n,k) = floor(b(n,k)) + 1 otherwise.

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A091650 Let M = the 4 X 4 matrix [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 -1 3 2]. Set seed vector = [1 1 1 1] = first row, then take M*[1 1 1 1] = [1 1 1 3] then M * [1 1 1 3], etc. Sequence gives terms in rightmost column.

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A123148 Triangle read by rows: T(n,k) is the coefficient of x^k in the polynomial p(n,x) defined by p(0,x) = -1, p(1,x) = x-2, p(n,x) = -xp(n-1,x) + 2p(n-2,x) for n >= 3 and 0 <= k <= n.

A091650 Let M = the 4 X 4 matrix [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 -1 3 2]. Set seed vector = [1 1 1 1] = first row, then take M[1 1 1 1] = [1 1 1 3] then M [1 1 1 3], etc. Sequence gives terms in rightmost column.