A200139 -id:A200139 - cp's OEIS frontend

A097805 Number of compositions of n with k parts, T(n, k) = binomial(n-1, k-1) for n, k >= 1 and T(n, 0) = 0^n, triangle read by rows for n >= 0 and 0 <= k <= n.

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 5, 1, 0, 1, 6, 15, 20, 15, 6, 1, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Offset: 0

Paul Barry, Aug 25 2004

Previous name was: Riordan array (1, 1/(1-x)) read by rows.
Note this Riordan array would be denoted (1, x/(1-x)) by some authors.
Columns have g.f. (x/(1-x))^k. Reverse of A071919. Row sums are A011782. Antidiagonal sums are Fibonacci(n-1). Inverse as Riordan array is (1, 1/(1+x)). A097805=B*A059260*B^(-1), where B is the binomial matrix.
(0,1)-Pascal triangle. - Philippe Deléham, Nov 21 2006
(n+1) * each term of row n generates triangle A127952: (1; 0, 2; 0, 3, 3; 0, 4, 8, 4; ...). - Gary W. Adamson, Feb 09 2007
Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2008
From Paul Weisenhorn, Feb 09 2011: (Start)
Triangle read by rows: T(r,c) is the number of unordered partitions of n=r*(r+1)/2+c into (r+1) parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2.
Triangle read by rows: T(r,c) is the number of unordered partitions of the number n=r*(r+1)/2+(c-1) into r parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2. (End)
Triangle read by rows: T(r,c) is the number of ordered partitions (compositions) of r into c parts. - Juergen Will, Jan 04 2016
From Tom Copeland, Oct 25 2012: (Start)
Given a basis composed of a sequence of polynomials p_n(x) characterized by ladder (creation / annihilation, or raising / lowering) operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x) with p_0(x)=1, giving the number operator # p_n(x) = RL p_n(x) = n p_n(x), the lower triangular padded Pascal matrix Pd (A097805) serves as a matrix representation of the operator exp(R^2*L) = exp(R#) =
  1) exp(x^2D) for the set x^n  and
  2) D^(-1) exp(t*x)D for the set x^n/n! (see A218234).
  (End)
From James East, Apr 11 2014: (Start)
Square array a(m,n) with m,n=0,1,2,... read by off-diagonals.
a(m,n) gives the number of order-preserving functions f:{1,...,m}->{1,...,n}. Order-preserving means that xa(n,n)=A088218(n) is the size of the semigroup O_n of all order-preserving transformations of {1,...,n}.
Read as a triangle, this sequence may be obtained by augmenting Pascal's triangle by appending the column 1,0,0,0,... on the left.
(End)
A formula based on the partitions of n with largest part k is given as a Sage program below. The 'conjugate' formula leads to A048004. - Peter Luschny, Jul 13 2015
From Wolfdieter Lang, Feb 17 2017: (Start)
The transposed of this lower triangular Riordan matrix of the associated type T provides the transition matrix between the monomial basis {x^n}, n >= 0, and the basis {y^n}, n >= 0, with y = x/(1-x): x^0 = 1 = y^0, x^n = Sum_{m >= n} Ttrans(n,m) y^m, for n >= 1, with Ttrans(n,m) = binomial(m-1,n-1).
Therefore, if a transformation with this Riordan matrix from a sequence {a} to the sequence {b} is given by b(n) = Sum_{m=0..n} T(n, m)*a(m), with T(n, m) = binomial(n-1, m-1), for n >= 1, then Sum_{n >= 0} a(n)*x^n = Sum_{n >= 0} b(n)*y^n, with y = x/(1-x) and vice versa. This is a modified binomial transformation; the usual one belongs to the Pascal Riordan matrix A007318. (End)
From Gus Wiseman, Jan 23 2022: (Start)
Also the number of compositions of n with alternating sum k, with k ranging from -n to n in steps of 2. For example, row n = 6 counts the following compositions (empty column indicated by dot):
  .  (15)  (24)    (33)      (42)     (51)   (6)
           (141)   (132)     (123)    (114)
           (1113)  (231)     (222)    (213)
           (1212)  (1122)    (321)    (312)
           (1311)  (1221)    (1131)   (411)
                   (2112)    (2121)
                   (2211)    (3111)
                   (11121)   (11112)
                   (12111)   (11211)
                   (111111)  (21111)
The reverse-alternating version is the same. Counting compositions by all three parameters (sum, length, alternating sum) gives A345197. Compositions of 2n with alternating sum 2k with k ranging from -n + 1 to n are A034871. (End)
Also the convolution triangle of A000012. - Peter Luschny, Oct 07 2022
From Sergey Kitaev, Nov 18 2023: (Start)
Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with k right-to-left maxima. A right-to-left maximum in a permutation a(1)a(2)...a(n) is position i such that a(j) < a(i) for all i < j.
Number of permutations of length n avoiding simultaneously the patterns 231 and 312 with k right-to-left minima (resp., left-to-right maxima). A right-to-left minimum (resp., left-to-right maximum) in a permutation a(1)a(2)...a(n) is position i such that a(j) > a(i) for all j > i (resp., a(j) < a(i) for all j < i).
Number of permutations of length n avoiding simultaneously the patterns 213 and 312 with k right-to-left maxima (resp., left-to-right maxima).
Number of permutations of length n avoiding simultaneously the patterns 213 and 231 with k right-to-left maxima (resp., right-to-left minima). (End)

			G.f. = 1 + x * (x + x^3 * (1 + x) + x^6 * (1 + x)^2 + x^10 * (1 + x)^3 + ...). - _Michael Somos_, Aug 20 2006
The triangle T(n, k) begins:
n\k  0 1 2  3  4   5   6  7  8 9 10 ...
0:   1
1:   0 1
2:   0 1 1
3:   0 1 2  1
4:   0 1 3  3  1
5:   0 1 4  6  4   1
6:   0 1 5 10 10   5   1
7:   0 1 6 15 20  15   6  1
8:   0 1 7 21 35  35  21  7  1
9:   0 1 8 28 56  70  56 28  8 1
10:  0 1 9 36 84 126 126 84 36 9  1
... reformatted _Wolfdieter Lang_, Jul 31 2017
From _Paul Weisenhorn_, Feb 09 2011: (Start)
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+c = 18 with (r+1)=6 summands: (5+5+4+2+1+1), (5+5+3+3+1+1), (5+4+4+3+1+1), (5+5+3+2+2+1), (5+4+4+2+2+1), (5+4+3+3+2+1).
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+(c-1) = 17 with r=5 summands: (5+5+4+2+1), (5+5+3+3+1), (5+5+3+2+2), (5+4+4+3+1), (5+4+4+2+2), (5+4+3+3+2).  (End)
From _James East_, Apr 11 2014: (Start)
a(0,0)=1 since there is a unique (order-preserving) function {}->{}.
a(m,0)=0 for m>0 since there is no function from a nonempty set to the empty set.
a(3,2)=4 because there are four order-preserving functions {1,2,3}->{1,2}: these are [1,1,1], [2,2,2], [1,1,2], [1,2,2]. Here f=[a,b,c] denotes the function defined by f(1)=a, f(2)=b, f(3)=c.
a(2,3)=6 because there are six order-preserving functions {1,2}->{1,2,3}: these are [1,1], [1,2], [1,3], [2,2], [2,3], [3,3].
(End)

D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Part 1, Section 7.2.1.3, 2011.

Alois P. Heinz, Rows n = 0..140, flattened
Tom Copeland, Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras
James East and Peter J. McNamara, On the work performed by a transformation semigroup, Australas. J. Combin. 49 (2011), 95-109.
Tian Han and Sergey Kitaev, Joint distributions of statistics over permutations avoiding two patterns of length 3, arXiv:2311.02974 [math.CO], 2023.
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 8.
P. A. MacMahon, Memoir on the Theory of the Compositions of Numbers, Phil. Trans. Royal Soc. London A, 184 (1893), 835-901. - _Juergen Will_, Jan 02 2016
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.

Case m=0 of the polynomials defined in A278073.
Cf. A000012 (diagonal), A011782 (row sums), A088218 (central terms).
Cf. A007318, A048004, A127952, A118800, A200139, A130595, A167374.
The terms just left of center in odd-indexed rows are A001791, even A002054.
The odd-indexed rows are A034871.
Row sums without the center are A058622.
The unordered version is A072233, without zeros A008284.
Right half without center has row sums A027306(n-1).
Right half with center has row sums A116406(n).
Left half without center has row sums A294175(n-1).
Left half with center has row sums A058622(n-1).
A025047 counts alternating compositions.
A098124 counts balanced compositions, unordered A047993.
A106356 counts compositions by number of maximal anti-runs.
A344651 counts partitions by sum and alternating sum.
A345197 counts compositions by sum, length, and alternating sum.
Cf. A000346, A000984, A001700, A008549, A008965, A114121, A124754, A238279, A345907, A345908, A346632.

b:= proc(n, i, p) option remember; `if`(n=0, p!, `if`(i<1, 0,
      expand(add(b(n-i*j, i-1, p+j)/j!*x^j, j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, 0)):
seq(T(n), n=0..20);  # Alois P. Heinz, May 25 2014
# Alternatively:
T := proc(k,n) option remember;
if k=n then 1 elif k=0 then 0 else
add(T(k-1,n-i), i=1..n-k+1) fi end:
A097805 := (n,k) -> T(k,n):
for n from 0 to 12 do seq(A097805(n,k), k=0..n) od; # Peter Luschny, Mar 12 2016
# Uses function PMatrix from A357368.
PMatrix(10, n -> 1);  # Peter Luschny, Oct 07 2022

T[0, 0] = 1; T[n_, k_] := Binomial[n-1, k-1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 03 2014, after Paul Weisenhorn *)
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&]],{n,0,10},{k,0,n}] (* Gus Wiseman, Jan 23 2022 *)

{a(n) = my(m); if( n<2, n==0, n--; m = (sqrtint(8*n + 1) - 1)\2; binomial(m-1, n - m*(m + 1)/2))}; /* Michael Somos, Aug 20 2006 */

T(n,k) = if (k==0, 0^n, binomial(n-1, k-1)); \\ Michel Marcus, May 06 2022

row(n) = vector(n+1, k, k--; if (k==0, 0^n, binomial(n-1, k-1))); \\ Michel Marcus, May 06 2022

from math import comb
def T(n, k): return comb(n-1, k-1) if k != 0 else k**n  # Peter Luschny, May 06 2022

# Illustrates a basic partition formula, is not efficient as a program for large n.
def A097805_row(n):
    r = []
    for k in (0..n):
        s = 0
        for q in Partitions(n, max_part=k, inner=[k]):
            s += mul(binomial(q[j],q[j+1]) for j in range(len(q)-1))
        r.append(s)
    return r
[A097805_row(n) for n in (0..9)] # Peter Luschny, Jul 13 2015

Number triangle T(n, k) defined by T(n,k) = Sum_{j=0..n} binomial(n, j)*if(k<=j, (-1)^(j-k), 0).
T(r,c) = binomial(r-1,c-1), 0 <= c <= r. - Paul Weisenhorn, Feb 09 2011
G.f.: (-1+x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015
a(0,0) = 1, a(n,k) = binomial(n-1,n-k) = binomial(n-1,k-1) Juergen Will, Jan 04 2016
G.f.: (x^1 + x^2 + x^3 + ...)^k = (x/(1-x))^k. - Juergen Will, Jan 04 2016
From Tom Copeland, Nov 15 2016: (Start)
E.g.f.: 1 + x*[e^((x+1)t)-1]/(x+1).
This padded Pascal matrix with the odd columns negated is NpdP = M*S = S^(-1)*M^(-1) = S^(-1)*M, where M(n,k) = (-1)^n A130595(n,k), the inverse Pascal matrix with the odd rows negated, S is the summation matrix A000012, the lower triangular matrix with all elements unity, and S^(-1) = A167374, a finite difference matrix. NpdP is self-inverse, i.e., (M*S)^2 = the identity matrix, and has the e.g.f. 1 - x*[e^((1-x)t)-1]/(1-x).
M = NpdP*S^(-1) follows from the well-known recursion property of the Pascal matrix, implying NpdP = M*S.
The self-inverse property of -NpdP is implied by the self-inverse relation of its embedded signed Pascal submatrix M (cf. A130595). Also see A118800 for another proof.
Let P^(-1) be A130595, the inverse Pascal matrix. Then T = A200139*P^(-1) and T^(-1) = padded P^(-1) = P*A097808*P^(-1). (End)
The center (n>0) is T(2n+1,n+1) = A000984(n) = 2*A001700(n-1) = 2*A088218(n) = A126869(2n) = 2*A138364(2n-1). - Gus Wiseman, Jan 25 2022

Corrected by Philippe Deléham, Oct 05 2005

New name using classical terminology by Peter Luschny, Feb 05 2019

A028297 Coefficients of Chebyshev polynomials of the first kind: triangle of coefficients in expansion of cos(n*x) in descending powers of cos(x).

Original entry on oeis.org

1, 1, 2, -1, 4, -3, 8, -8, 1, 16, -20, 5, 32, -48, 18, -1, 64, -112, 56, -7, 128, -256, 160, -32, 1, 256, -576, 432, -120, 9, 512, -1280, 1120, -400, 50, -1, 1024, -2816, 2816, -1232, 220, -11, 2048, -6144, 6912, -3584, 840, -72, 1, 4096, -13312, 16640, -9984

Offset: 0

N. J. A. Sloane

Rows are of lengths 1, 1, 2, 2, 3, 3, ... (A008619).
This triangle is generated from A118800 by shifting down columns to allow for (1, 1, 2, 2, 3, 3, ...) terms in each row. - Gary W. Adamson, Dec 16 2007
Unsigned triangle = A034839 * A007318. - Gary W. Adamson, Nov 28 2008
Triangle, with zeros omitted, given by (1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, -1, 1, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 16 2011
From Wolfdieter Lang, Aug 02 2014: (Start)
This irregular triangle is the row reversed version of the Chebyshev T-triangle A053120 given by A039991 with vanishing odd-indexed columns removed.
If zeros are appended in each row n >= 1, in order to obtain a regular triangle (see the Philippe Deléham comment, g.f. and example) this becomes the Riordan triangle (1-x)/(1-2*x), -x^2/(1-2*x). See also the unsigned version A201701 of this regular triangle.
(End)
Apparently, unsigned diagonals of this array are rows of A200139. - Tom Copeland, Oct 11 2014
It appears that the coefficients are generated by the following: Let SM_k = Sum( d_(t_1, t_2)* eM_1^t_1 * eM_2^t_2) summed over all length 2 integer partitions of k, i.e., 1*t_1 + 2*t_2 = k, where SM_k are the averaged k-th power sum symmetric polynomials in 2 data (i.e., SM_k = S_k/2 where S_k are the k-th power sum symmetric polynomials, and where eM_k are the averaged k-th elementary symmetric polynomials, eM_k = e_k/binomial(2,k) with e_k being the k-th elementary symmetric polynomials. The data d_(t_1, t_2) form an irregular triangle, with one row for each k value starting with k=1. Thus this procedure and associated OEIS sequences A287768, A288199, A288207, A288211, A288245, A288188 are generalizations of Chebyshev polynomials of the first kind. - Gregory Gerard Wojnar, Jul 01 2017

			Letting c = cos x, we have: cos 0x = 1, cos 1x = 1c; cos 2x = 2c^2-1; cos 3x = 4c^3-3c, cos 4x = 8c^4-8c^2+1, etc.
T4 = 8x^4 - 8x^2 + 1 = 8, -8, +1 = 2^(3) - (4)(2) + [2^(-1)](4)/2.
From _Wolfdieter Lang_, Aug 02 2014: (Start)
The irregular triangle T(n,k) begins:
n\k     1      2     3      4     5     6   7   8 ....
0:      1
1:      1
2:      2     -1
3:      4     -3
4:      8     -8     1
5:     16    -20     5
6:     32    -48    18     -1
7:     64   -112    56     -7
8:    128   -256   160    -32     1
9:    256   -576   432   -120     9
10:   512  -1280  1120   -400    50    -1
11:  1024  -2816  2816  -1232   220   -11
12:  2048  -6144  6912  -3584   840   -72   1
13:  4096 -13312 16640  -9984  2912  -364  13
14:  8192 -28672 39424 -26880  9408 -1568  98  -1
15: 16384 -61440 92160 -70400 28800 -6048 560 -15
...
T(4,x) = 8*x^4 -8*x^2 + 1*x^0, T(5,x) = 16*x^5 - 20*x^3 + 5*x^1, with Chebyshev's T-polynomials (A053120). (End)
From _Philippe Deléham_, Dec 16 2011: (Start)
The triangle (1,1,0,0,0,0,...) DELTA (0,-1,1,0,0,0,0,...) includes zeros and begins:
   1;
   1,   0;
   2,  -1,  0;
   4,  -3,  0,  0;
   8,  -8,  1,  0, 0;
  16, -20,  5,  0, 0, 0;
  32, -48, 18, -1, 0, 0, 0; (End)

I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, 5th ed., Section 1.335, p. 35.
S. Selby, editor, CRC Basic Mathematical Tables, CRC Press, 1970, p. 106. [From Rick L. Shepherd, Jul 06 2010]

Alois P. Heinz, Rows n = 0..200, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy] p. 795.
Pantelis A. Damianou, A Beautiful Sine Formula, Amer. Math. Monthly 121 (2014), no. 2, 120--135. MR3149030.
Daniel J. Greenhoe, Frames and Bases: Structure and Design, Version 0.20, Signal Processing ABCs series (2019) Vol. 4, see page 172.
Daniel J. Greenhoe, A Book Concerning Transforms, Version 0.10, Signal Processing ABCs series (2019) Vol. 5, see page 94.
Tian-Xiao He, Peter J.-S. Shiue, Zihan Nie, and Minghao Chen, Recursive sequences and Girard-Waring identities with applications in sequence transformation, Electronic Research Archive (2020) Vol. 28, No. 2, 1049-1062.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
G. G. Wojnar, D. Sz. Wojnar, and L. Q. Brin, Universal Peculiar Linear Mean Relationships in All Polynomials, Table GW.n=2, p. 22, arXiv:1706.08381 [math.GM], 2017.

Cf. A028298.
Reflection of A008310, the main entry. With zeros: A039991.
Cf. A053120 (row reversed table including zeros).
Cf. A118800, A034839, A081277, A124182.
Cf. A001333 (row sums 1), A001333 (alternating row sums). - Wolfdieter Lang, Aug 02 2014

b:= proc(n) b(n):= `if`(n<2, 1, expand(2*b(n-1)-x*b(n-2))) end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
seq(T(n), n=0..15);  # Alois P. Heinz, Sep 04 2019

t[n_] := (Cos[n x] // TrigExpand) /. Sin[x]^m_ /; EvenQ[m] -> (1 - Cos[x]^2)^(m/2) // Expand; Flatten[Table[ r = Reverse @ CoefficientList[t[n], Cos[x]]; If[OddQ[Length[r]], AppendTo[r,0]]; Partition[r,2][[All, 1]],{n, 0, 13}] ][[1 ;; 53]] (* Jean-François Alcover, May 06 2011 *)
Tpoly[n_] := HypergeometricPFQ[{(1 - n)/2, -n/2}, {1/2}, 1 - x];
Table[CoefficientList[Tpoly[n], x], {n, 0, 12}] // Flatten (* Peter Luschny, Feb 03 2021 *)

cos(n*x) = 2 * cos((n-1)*x) * cos(x) - cos((n-2)*x) (from CRC's Multiple-angle relations). - Rick L. Shepherd, Jul 06 2010
G.f.: (1-x) / (1-2x+y*x^2). - Philippe Deléham, Dec 16 2011
Sum_{k=0..n} T(n,k)*x^k = A011782(n), A000012(n), A146559(n), A087455(n), A138230(n), A006495(n), A138229(n) for x = 0, 1, 2, 3, 4, 5, 6, respectively. - Philippe Deléham, Dec 16 2011
T(n,k) = [x^k] hypergeom([1/2 - n/2, -n/2], [1/2], 1 - x). - Peter Luschny, Feb 03 2021
T(n,k) = (-1)^k * 2^(n-1-2*k) * A034807(n,k). - Hoang Xuan Thanh, Jun 21 2025

More terms from David W. Wilson

Row length sequence and link to Abramowitz-Stegun added by Wolfdieter Lang, Aug 02 2014

A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, 4, 3, 1, 0, 3, 6, 7, 4, 1, 1, 3, 9, 13, 11, 5, 1, 0, 4, 12, 22, 24, 16, 6, 1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1

Offset: 0

N. J. A. Sloane, Jan 23 2001

Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Diagonal sums give A008346. - Paul Barry, Sep 23 2004
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
From Robert Coquereaux, Oct 01 2014: (Start)
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
Cf. A193844. (End)
From Peter Bala, Feb 07 2024: (Start)
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
From Miquel A. Fiol, Oct 04 2024: (Start)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0). (End)

			Triangle begins
  1;
  0,  1;
  1,  1,  1;
  0,  2,  2,  1;
  1,  2,  4,  3,  1;
  0,  3,  6,  7,  4,  1;
  1,  3,  9, 13, 11,  5,  1;
  0,  4, 12, 22, 24, 16,  6,  1;
  1,  4, 16, 34, 46, 40, 22,  7,  1;
  0,  5, 20, 50, 80, 86, 62, 29,  8,  1;
Sequences obtained with _Miquel A. Fiol_'s Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6):
A000007  A000012  A000027  A025747  A000292* A000332* A000389*
A059841  A008619  A087811* A002623  A001752  A001753  A001769
A193356  A008794* A005993  A005994  -------  -------  -------
-------  -------  -------  A005995  A018210  -------  A052267
-------  -------  -------  -------  A018211  A018212  -------
-------  -------  -------  -------  -------  A018213  A018214
-------  -------  -------  -------  -------  -------  A062136
*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
     22--12---24---14---44
          | \    / |
         23   11   34
            \  |  /
              13
               |
              33

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
Joseph Briggs, Alex Parker, Coy Schwieder, and Chris Wells, Frogs, hats and common subsequences, arXiv:2404.07285 [math.CO], 2024. See p. 28.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, J. of Geometry and Physics, 1995, Vol 15, pp 333-352.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, arXiv:hep-th/9310147, 1993.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 8.
R. H. Hammack and G. D. Smith, Cycle bases of reduced powers of graphs, Ars Math. Contemp. 12 (2017) 183-203.
Christian Kassel, A Künneth formula for the cyclic cohomology of Z2-graded algebras, Math. Ann. 275 (1986) 683.
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, Numerical Algorithms, June 2012, Volume 60, Issue 2, pp 297-314. - From _N. J. A. Sloane_, Oct 12 2012
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, preprint, Centro de Matemática da Universidade do Porto.
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mark Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.

Cf. A059259. Row sums give A001045.
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001
Cf. A026641, A014300.
Cf. A007318, A097805, A097808, A130595, A200139, A062135.

read transforms; 1/(1-y-x*y-x^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);

t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)

T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
for(n=0, 12, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 11 2017

from sympy import binomial
def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017

def A059260_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
for n in (1..9): print(A059260_row(n)) # Peter Luschny, Mar 16 2016

G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson, Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Feb 17 2007
Equals A130595*A097805*A007318 = (inverse Pascal matrix)*(padded Pascal matrix)*(Pascal matrix) = A130595*A200139. Inverse is A097808 = A130595*(padded A130595)*A007318. - Tom Copeland, Nov 14 2016
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
From Miquel A. Fiol, Sep 30 2024: (Start)
The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below. (End)

Formula corrected by Philippe Deléham, Jan 11 2014

A081038 3rd binomial transform of (1,2,0,0,0,0,0,0,...).

Original entry on oeis.org

1, 5, 21, 81, 297, 1053, 3645, 12393, 41553, 137781, 452709, 1476225, 4782969, 15411789, 49424013, 157837977, 502211745, 1592728677, 5036466357, 15884240049, 49977243081, 156905298045, 491636600541, 1537671920841

Offset: 0

Paul Barry, Mar 03 2003

a(n) is the number of distinguished parts in all compositions of n+1 in which some (possibly all or none) of the parts have been distinguished. a(1) = 2 because we have: 2', 1'+1, 1+1', 1'+1' where we see 5's marking the distinguished parts. With offset=1, a(n) = Sum_{k=1..n} A200139(n,k)*k. - Geoffrey Critzer, Jan 12 2013
For n>=1, a(n-1) the number of ternary strings of length 2n containing the block 11..12 with n ones where no runs of length larger than n are permitted. - Marko Riedel, Mar 08 2016
Binomial transform of {A001787(n + 1)}{n >= 0}. - _Wolfdieter Lang, Oct 01 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See p. 13.
Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021. See p. 67.
Index entries for linear recurrences with constant coefficients, signature (6,-9).

Cf. A001787, A001792, A081039, A081040, A086972, A200139.
First differences of A027471.
Cf. A269914, A269915, A269916, A269917.

[(2*n+3)*3^(n-1): n in [0..30]]; // Vincenzo Librandi, Jun 09 2011

A081038:=n->(2*n+3)*3^(n-1): seq(A081038(n), n=0..30); # Wesley Ivan Hurt, Mar 07 2016

LinearRecurrence[{6,-9},{1,5},40] (* Harvey P. Dale, Jun 22 2012 *)

G.f.: (1-x)/(1-3*x)^2.
a(n) = 6*a(n-1) - 9*a(n-2), with a(0)=1, a(1)=5.
a(n) = (2*n+3)*3^(n-1).
a(n) = Sum_{k=0..n} (k+1)*2^k*binomial(n, k).
a(n) = 2*A086972(n) - 1. - Lambert Herrgesell (zero815(AT)googlemail.com), Feb 10 2008
From Amiram Eldar, May 17 2022: (Start)
Sum_{n>=0} 1/a(n) = 9*(sqrt(3)*arctanh(1/sqrt(3)) - 1).
Sum_{n>=0} (-1)^n/a(n) = 9 - 3*sqrt(3)*Pi/2. (End)
E.g.f.: exp(3*x)*(1 + 2*x). - Stefano Spezia, Jan 31 2025

A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 2, -3, 1, 4, -8, 5, -1, 8, -20, 18, -7, 1, 16, -48, 56, -32, 9, -1, 32, -112, 160, -120, 50, -11, 1, 64, -256, 432, -400, 220, -72, 13, -1, 128, -576, 1120, -1232, 840, -364, 98, -15, 1, 256, -1280, 2816, -3584, 2912, -1568, 560, -128, 17, -1, 512, -2816, 6912, -9984, 9408, -6048, 2688, -816, 162, -19, 1

Offset: 0

Paul D. Hanna, May 02 2006

The matrix square, T^2, consists of columns that are all the same.
Matrix inverse is triangle A118801. Row sums form {0^n, n>=0}.
Unsigned row sums equal A025192(n) = 2*3^(n-1), n>=1.
Row squared sums equal A051708.
Antidiagonal sums equals all 1's.
Unsigned antidiagonal sums form A078057 (with offset).
Antidiagonal squared sums form A002002(n) = Sum_{k=0..n-1} C(n,k+1)*C(n+k,k), n>=1.
From Paul Barry, Nov 10 2008: (Start)
T is [1,1,0,0,0,...] DELTA [ -1,0,0,0,0,...] or C(1,n) DELTA -C(0,n). (DELTA defined in A084938).
The positive matrix T_p is [1,1,0,0,0,...] DELTA [1,0,0,0,0,...]. T_p*C^-1 is
[0,1,0,0,0,....] DELTA [1,0,0,0,0,...] which is C(n-1,k-1) for n,k>=1. (End)
The triangle formed by deleting the minus signs is the mirror of the self-fusion of Pascal's triangle; see Comments at A081277 and A193722. - Clark Kimberling, Aug 04 2011
Riordan array ( (1 - x)/(1 - 2*x), -x/(1 - 2*x) ). Cf. A209149. The matrix square is the Riordan array ( (1 - x)^2/(1 - 2*x), x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013
From Peter Bala, Feb 23 2019: (Start)
There is a 1-parameter family of solutions to the simultaneous equations C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle. Let T(k) denote the Riordan array ( (1 - k*x)/(1 - (k + 1)*x), -x/(1 - (k + 1)*x) ) so that T(1) = T. Then C*T(k)*C = T(k)^-1 and T(k)*C*T(k) = C^-1, for arbitrary k. For arbitrary m, the Riordan arrays (T(k)*C^m)^2 and (C^m*T(k))^2 both belong to the Appell subgroup of the Riordan group.
More generally, given a fixed m, we can ask for a lower triangular array X solving the simultaneous equations (C^m)*X*(C^m) = X^-1 and X*(C^m)*X = C^(-m). A 1-parameter family of solutions is given by the Riordan arrays X = ( (1 - m*k*x)/(1 - m*(k + 1)*x), -x/(1 - m*(k + 1)*x) ). The Riordan arrays X^2 , (X*C^n)^2 and (C^n*X)^2, for arbitrary n, all belong to the Appell subgroup of the Riordan group. (End)

			Triangle begins:
     1;
     1,    -1;
     2,    -3,     1;
     4,    -8,     5,     -1;
     8,   -20,    18,     -7,     1;
    16,   -48,    56,    -32,     9,     -1;
    32,  -112,   160,   -120,    50,    -11,     1;
    64,  -256,   432,   -400,   220,    -72,    13,    -1;
   128,  -576,  1120,  -1232,   840,   -364,    98,   -15,    1;
   256, -1280,  2816,  -3584,  2912,  -1568,   560,  -128,   17,   -1;
   512, -2816,  6912,  -9984,  9408,  -6048,  2688,  -816,  162,  -19,  1;
  1024, -6144, 16640, -26880, 28800, -21504, 11424, -4320, 1140, -200, 21, -1;
  ...
The matrix square, T^2, equals:
   1;
   0,  1;
   1,  0,  1;
   2,  1,  0,  1;
   4,  2,  1,  0,  1;
   8,  4,  2,  1,  0,  1;
  16,  8,  4,  2,  1,  0,  1;
  32, 16,  8,  4,  2,  1,  0,  1;
  64, 32, 16,  8,  4,  2,  1,  0,  1; ...
where all columns are the same.

Paul D. Hanna, Rows 0..45 of triangle, flattened.
Florian Luca, Attila Pethő, and László Szalay, Duplications in the k-generalized Fibonacci sequences, New York J. Math. (2021) Vol. 27, 1115-1133.

Cf. A118801 (inverse), A025192 (unsigned row sums), A051708 (row squared sums), A078057 (unsigned antidiagonal sums), A002002 (antidiagonal squared sums).

Cf. A135387, A209149, A000012, A097805, A130595, A167374, A200139, A239473, A321620.

(* This program generates A118800 as the mirror of the self-fusion of Pascal's triangle. *)
z = 8; a = 1; b = 1; c = 1; d = 1;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n;
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, 0];
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, _] = 1;
g[n_] := CoefficientList[w[n, -x], x];
TableForm[Table[Reverse[Abs@g[n]], {n, -1, z}]]
Flatten[Table[Reverse[Abs@g[n]], {n, -1, z}]] (* A081277 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A118800 *)
(* Clark Kimberling, Aug 04 2011 *)
T[ n_, k_] := If[ n<0 || k<0, 0, (-1)^k 2^(n-k) (Binomial[ n, k] + Binomial[ n-1, n-k]) / 2]; (* Michael Somos, Nov 25 2016 *)

{T(n,k)=if(n==0&k==0,1,(-1)^k*2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

/* Chebyshev Polynomials as Antidiagonals: */
{T(n,k)=local(Ox=x*O(x^(2*k))); polcoeff(((1+sqrt(1-x^2+Ox))^(n+k)+(1-sqrt(1-x^2+Ox))^(n+k))/2,2*k,x)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

# uses[riordan_square from A321620]
# Computes the unsigned triangle.
riordan_square((1-x)/(1-2*x), 8) # Peter Luschny, Jan 03 2019

T(n,k) = (-1)^k * 2^(n-k) * ( C(n,k) + C(n-1,k-1) )/2 for n>=k>=0 with T(0,0) = 1. Antidiagonals form the coefficients of Chebyshev polynomials: T(n,k) = [x^(2*n)] [(1+sqrt(1-x^2))^(n+k) + (1-sqrt(1-x^2))^(n+k)]/2.
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
O.g.f.: (1 - t)/(1 + t*(x - 2)) = 1 + (1 - x)*t + (2 - 3*x + x^2)^t^2 + (4 - 8*x + 5*x^2 - x^3)*t^3 + ....  Row polynomial R(n,x) = (1 - x)*(2 - x)^(n-1) for n >= 1. - Peter Bala, Jul 17 2013
T(n,k)=2*T(n-1,k)-T(n-1,k-1) with T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 25 2013
G.f. for row n (n>=1): Sum_{k=0..n} T(n,k)*x^k = (1-x)*(2-x)^(n-1). - Philippe Deléham, Nov 25 2013
From Tom Copeland, Nov 15 2016: (Start)
E.g.f. is [1 + (1-x)e^((2-x)t)]/(2-x), so the row polynomials are p_n(x) = (1-q,(x))^n, umbrally, where (q.(x))^k = q_k(x) are the row polynomials of A239473, or, equivalently, T = M*A239473, where M is the inverse Pascal matrix C^(-1) = A130595 with the odd rows negated, i.e., M(n,k) = (-1)^n C^(-1)(n,k) with e.g.f. exp[(1-x)t]. Cf. A200139: A200139(n,k) = (-1)^k* A118800(n,k).
TCT = C^(-1) = A130595 and A239473 = A000012*C^(-1) = S*C^(-1) imply (M*S)^2 = Identity matrix, i.e., M*S = (M*S)^(-1) = S^(-1)*M^(-1) = A167374*M^(-1). Note that M = M^(-1). Cf. A097805. (End)

A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -7, 5, -1, 1, -15, 17, -7, 1, 1, -31, 49, -31, 9, -1, 1, -63, 129, -111, 49, -11, 1, 1, -127, 321, -351, 209, -71, 13, -1, 1, -255, 769, -1023, 769, -351, 97, -15, 1, 1, -511, 1793, -2815, 2561, -1471, 545, -127, 17, -1, 1, -1023, 4097, -7423, 7937, -5503, 2561, -799, 161, -19, 1

Offset: 0

Paul D. Hanna, May 02 2006

Matrix inverse is triangle A118800. Row sums are: (1-n). Unsigned row sums equal A007051(n) = (3^n + 1)/2. Row squared sums equal A118802. Antidiagonal sums equal A080956(n) = (n+1)(2-n)/2. Unsigned antidiagonal sums form A024537 (with offset).
T = C^2*D^-1 where matrix product D = C^-1*T*C = T^-1*C^2 has only 2 nonzero diagonals: D(n,n)=-D(n+1,n)=(-1)^n, with zeros elsewhere. Also, [B^-1]*T*[B^-1] = B*[T^-1]*B forms a self-inverse matrix, where B^2 = C and B(n,k) = C(n,k)/2^(n-k). - Paul D. Hanna, May 04 2006
Riordan array ( 1/(1 - x), -x/(1 - 2*x) ) The matrix square is the Riordan array ( (1 - 2*x)/(1 - x)^2, x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013

			Formulas for initial columns are, for n>=0:
T(n+1,1) = 1 - 2^(n+1);
T(n+2,2) = 1 + 2^(n+1)*n;
T(n+3,3) = 1 - 2^(n+1)*(n*(n+1)/2 + 1);
T(n+4,4) = 1 + 2^(n+1)*(n*(n+1)*(n+2)/6 + n);
T(n+5,5) = 1 - 2^(n+1)*(n*(n+1)*(n+2)*(n+3)/24 + n*(n+1)/2 + 1).
Triangle begins:
1;
1,-1;
1,-3,1;
1,-7,5,-1;
1,-15,17,-7,1;
1,-31,49,-31,9,-1;
1,-63,129,-111,49,-11,1;
1,-127,321,-351,209,-71,13,-1;
1,-255,769,-1023,769,-351,97,-15,1;
1,-511,1793,-2815,2561,-1471,545,-127,17,-1;
1,-1023,4097,-7423,7937,-5503,2561,-799,161,-19,1; ...
The matrix square, T^2, starts:
1;
0,1;
-1,0,1;
-2,-1,0,1;
-3,-2,-1,0,1;
-4,-3,-2,-1,0,1; ...
where all columns are the same.
The matrix product C^-1*T*C = T^-1*C^2 is:
1;
-1,-1;
0, 1, 1;
0, 0,-1,-1;
0, 0, 0, 1, 1; ...
where C(n,k) = n!/(n-k)!/k!.

M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Q., 48 (2010), 290-297.

Cf. A118800 (inverse), A007051 (unsigned row sums), A118802 (Row squared sums), A080956 (antidiagonal sums), A024537 (unsigned antidiagonal sums).
A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011
Cf. A007318, A032807, A097805, A112857, A130595, A156644, A167374, A200139, A239473.

Table[(1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]) - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

{T(n,k)=if(n==0&k==0,1,1+(-1)^k*2^(n-k+1)*sum(j=0,k\2,binomial(n-2*j-2,k-2*j-1)))}

T(n,k) = 1 + (-1)^k*2^(n-k+1)*Sum_{j=0..[k/2]} C(n-2j-2,k-2j-1) for n>=k>=0 with T(0,0) = 1.
For k>0, T(n,k) = -T(n-1,k-1) + 2*T(n-1,k). - Gerald McGarvey, Aug 05 2006
O.g.f.: (1 - 2*t)/(1 - t) * 1/(1 + t*(x - 2)) = 1 + (1 - x)*t + (1 - 3*x + x^2)*t^2 + (1 - 7*x + 5*x^2 - x^3)*t^3 + .... - Peter Bala, Jul 17 2013
From Tom Copeland, Nov 17 2016: (Start)
Let M = A200139^(-1) = (unsigned A118800)^(-1) and NpdP be the signed padded Pascal matrix defined in A097805. Then T(n,k) = (-1)^n* M(n,k) and T = P*NpdP = (A239473)^(-1)*P^(-1) = P*A167374*P^(-1) = A156644*P^(-1), where P is the Pascal matrix A007318 with inverse A130595. Cf. A112857.
Signed P^2 = signed A032807 = T*A167374. (End)

A201701 Riordan triangle ((1-x)/(1-2x), x^2/(1-2x)).

Original entry on oeis.org

1, 1, 0, 2, 1, 0, 4, 3, 0, 0, 8, 8, 1, 0, 0, 16, 20, 5, 0, 0, 0, 32, 48, 18, 1, 0, 0, 0, 64, 112, 56, 7, 0, 0, 0, 0, 128, 256, 160, 32, 1, 0, 0, 0, 0, 256, 576, 432, 120, 9, 0, 0, 0, 0, 0, 512, 1280, 1120, 400, 50, 1, 0, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Dec 03 2011

Triangle T(n,k), read by rows, given by (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.
Skewed version of triangle in A200139.
Triangle without zeros: A207537.
For the version with negative odd numbered columns, which is Riordan ((1-x)/(1-2*x), -x^2/(1-2*x)) see comments on A028297 and A039991. - Wolfdieter Lang, Aug 06 2014
This is an example of a stretched Riordan array in the terminology of Section 2 of Corsani et al. - Peter Bala, Jul 14 2015

			The triangle T(n,k) begins:
  n\k      0     1     2     3     4    5   6  7 8 9 10 11 ...
  0:       1
  1:       1     0
  2:       2     1     0
  3:       4     3     0     0
  4:       8     8     1     0     0
  5:      16    20     5     0     0    0
  6:      32    48    18     1     0    0   0
  7:      64   112    56     7     0    0   0  0
  8:     128   256   160    32     1    0   0  0 0
  9:     256   576   432   120     9    0   0  0 0 0
  10:    512  1280  1120   400    50    1   0  0 0 0  0
  11:   1024  2816  2816  1232   220   11   0  0 0 0  0  0
  ...  reformatted and extended. - _Wolfdieter Lang_, Aug 06 2014

C. Corsani, D. Merlini, and R. Sprugnoli, Left-inversion of combinatorial sums, Discrete Mathematics, 180 (1998) 107-122.

Columns include A011782, A001792, A001793, A001794, A006974, A006975, A006976.
Diagonals sums are in A052980.
Cf. A028297, A081265, A124182, A131577, A039991 (zero-columns deleted, unsigned and zeros appended).
Cf. A098158, A200139, A207537.
Cf. A028297 (signed version, zeros deleted). Cf. A034839.

(* The function RiordanArray is defined in A256893. *)
RiordanArray[(1 - #)/(1 - 2 #)&, #^2/(1 - 2 #)&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

T(n,k) = 2*T(n-1,k) + T(n-2,k-1) with T(0,0) = T(1,0) = 1, T(1,1) = 0 and T(n,k) = 0 for k<0 or for n

Sum_{k=0..n} T(n,k)^2 = A002002(n) for n>0.

Sum_{k=0..n} T(n,k)*x^k = A138229(n), A006495(n), A138230(n), A087455(n), A146559(n), A000012(n), A011782(n), A001333(n), A026150(n), A046717(n), A084057(n), A002533(n), A083098(n), A084058(n), A003665(n), A002535(n), A133294(n), A090042(n), A125816(n), A133343(n), A133345(n), A120612(n), A133356(n), A125818(n) for x = -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17 respectively.

G.f.: (1-x)/(1-2*x-y*x^2). - Philippe Deléham, Mar 03 2012

From Peter Bala, Jul 14 2015: (Start)

Factorizes as A034839 * A007318 = (1/(1 - x), x^2/(1 - x)^2) * (1/(1 - x), x/(1 - x)) as a product of Riordan arrays.

T(n,k) = Sum_{i = k..floor(n/2)} binomial(n,2*i) *binomial(i,k). (End)

Name changed, keyword:easy added, crossrefs A028297 and A039991 added, and g.f. corrected by Wolfdieter Lang, Aug 06 2014

A193723 Mirror of the fusion triangle A193722.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 18, 21, 8, 1, 54, 81, 45, 11, 1, 162, 297, 216, 78, 14, 1, 486, 1053, 945, 450, 120, 17, 1, 1458, 3645, 3888, 2295, 810, 171, 20, 1, 4374, 12393, 15309, 10773, 4725, 1323, 231, 23, 1, 13122, 41553, 58320, 47628, 24948, 8694, 2016, 300, 26, 1

Offset: 0

Clark Kimberling, Aug 04 2011

A193723 is obtained by reversing the rows of the triangle A193722.
Triangle T(n,k), read by rows, given by [2,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 04 2011
From Philippe Deléham, Nov 14 2011: (Start)
Riordan array ((1-x)/(1-3x), x/(1-3x)).
Product A200139*A007318 as infinite lower triangular arrays. (End)

			First six rows:
    1;
    2,   1;
    6,   5,   1;
   18,  21,   8,   1;
   54,  81,  45,  11,   1;
  162, 297, 216,  78,  14,   1;

Cf. A084938, A193722, A052924 (antidiagonal sums), Diagonals: A000012, A016789, A081266, Columns: A025192, A081038.

z = 9; a = 1; b = 1; c = 1; d = 2;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193722 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A193723 *)

Write w(n,k) for the triangle at A193722.  The triangle at A193723 is then given by w(n,n-k).
T(n,k) = T(n-1,k-1) + 3*T(n-1,k) with T(0,0)=T(1,1)=1 and T(1,0)=2. - Philippe Deléham, Oct 05 2011
From Philippe Deléham, Nov 14 2011: (Start)
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for x=-2,-1,0,1,2,3,4,5,6,7,8,9 respectively.
T(n,k) = Sum_{j>=0} T(n-1-j,k-1)*3^j.
G.f.: (1-x)/(1-(3+y)*x). (End)

Showing 1-8 of 8 results.

cp's OEIS Frontend

A097805 Number of compositions of n with k parts, T(n, k) = binomial(n-1, k-1) for n, k >= 1 and T(n, 0) = 0^n, triangle read by rows for n >= 0 and 0 <= k <= n.

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A028297 Coefficients of Chebyshev polynomials of the first kind: triangle of coefficients in expansion of cos(n*x) in descending powers of cos(x).

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A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

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A081038 3rd binomial transform of (1,2,0,0,0,0,0,0,...).

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A118800 Triangle read by rows: T satisfies the matrix products: C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle.

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A118801 Triangle T that satisfies the matrix products: C*[T^-1]*C = T and T*[C^-1]*T = C, where C is Pascal's triangle.

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A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.

A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.

A201701 Riordan triangle ((1-x)/(1-2x), x^2/(1-2x)).