A157000 -id:A157000 - cp's OEIS frontend

A029635 The (1,2)-Pascal triangle (or Lucas triangle) read by rows.

2, 1, 2, 1, 3, 2, 1, 4, 5, 2, 1, 5, 9, 7, 2, 1, 6, 14, 16, 9, 2, 1, 7, 20, 30, 25, 11, 2, 1, 8, 27, 50, 55, 36, 13, 2, 1, 9, 35, 77, 105, 91, 49, 15, 2, 1, 10, 44, 112, 182, 196, 140, 64, 17, 2, 1, 11, 54, 156, 294, 378, 336, 204, 81, 19, 2, 1, 12, 65, 210, 450, 672, 714, 540, 285, 100

Offset: 0

Mohammad K. Azarian

This is also called Vieta's array. - N. J. A. Sloane, Nov 22 2017
Dropping the first term and changing the boundary conditions to T(n,1)=n, T(n,n-1)=2 (n>=2), T(n,n)=1 yields the number of nonterminal symbols (which generate strings of length k) in a certain context-free grammar in Chomsky normal form that generates all permutations of n symbols. Summation over k (1<=k<=n) results in A003945. For the number of productions of this grammar: see A090327. Example: 1; 2, 1; 3, 2, 1; 4, 5, 2, 1; 5, 9, 7, 2, 1; 6, 14, 16, 9, 2, 1; In addition to the example of A090327 we have T(3,3)=#{S}=1, T(3,2)=#{D,E}=2 and T(3,1)=#{A,B,C}=3. - Peter R. J. Asveld, Jan 29 2004
Much as the original Pascal triangle gives the Fibonacci numbers as sums of its diagonals, this triangle gives the Lucas numbers (A000032) as sums of its diagonals; see Posamentier & Lehmann (2007). - Alonso del Arte, Apr 09 2012
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
It appears that for the infinite set of (1,N) Pascal's triangles, the binomial transform of the n-th row (n>0), followed by zeros, is equal to the n-th partial sum of (1, N, N, N, ...). Example: for the (1,2) Pascal's triangle, the binomial transform of the second row followed by zeros, i.e., of (1, 3, 2, 0, 0, 0, ...), is equal to the second partial sum of (1, 2, 2, 2, ...) = (1, 4, 9, 16, ...). - Gary W. Adamson, Aug 11 2015
Given any (1,N) Pascal triangle, let the binomial transform of the n-th row (n>1) followed by zeros be Q(x). It appears that the binomial transform of the (n-1)-th row prefaced by a zero is Q(n-1). Example: In the (1,2) Pascal triangle the binomial transform of row 3: (1, 4, 5, 2, 0, 0, 0, ...) is A000330 starting with 1: (1, 5, 14, 30, 55, 91, ...). The binomial transform of row 2 prefaced by a zero and followed by zeros, i.e., of (0, 1, 3, 2, 0, 0, 0, ...) is (0, 1, 5, 14, 30, 55, ...). - Gary W. Adamson, Sep 28 2015
It appears that in the array accompanying each (1,N) Pascal triangle (diagonals of the triangle), the binomial transform of (..., 1, N, 0, 0, 0, ...) preceded by (n-1) zeros generates the n-th row of the array (n>0). Then delete the zeros in the result. Example: in the (1,2) Pascal triangle, row 3 (1, 5, 14, 30, ...) is the binomial transform of (0, 0, 1, 2, 0, 0, 0, ...) with the resulting zeros deleted. - Gary W. Adamson, Oct 11 2015
Read as a square array (similar to the Example section Sq(m,j), but with Sq(0,0)=0 and Sq(m,j)=P(m+1,j) otherwise), P(n,k) are the multiplicities of the eigenvalues, lambda_n = n(n+k-1), of the Laplacians on the unit k-hypersphere, given by Teo (and Choi) as P(n,k) = (2n-k+1)(n+k-2)!/(n!(k-1)!). P(n,k) is also the numerator of a Dirichlet series for the Minakashisundarum-Pleijel zeta function for the sphere. Also P(n,k) is the dimension of the space of homogeneous, harmonic polynomials of degree k in n variables (Shubin, p. 169). For relations to Chebyshev polynomials and simple Lie algebras, see A034807. - Tom Copeland, Jan 10 2016
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Jan 15 2016

			Triangle begins:
  [0] [2]
  [1] [1, 2]
  [2] [1, 3,  2]
  [3] [1, 4,  5,  2]
  [4] [1, 5,  9,  7,   2]
  [5] [1, 6, 14, 16,   9,  2]
  [6] [1, 7, 20, 30,  25, 11,  2]
  [7] [1, 8, 27, 50,  55, 36, 13,  2]
  [8] [1, 9, 35, 77, 105, 91, 49, 15, 2]
.
Read as a square, the array begins:
  n\k| 0  1   2    3    4    5
  --------------------------------------
  0 |  2  2   2    2    2    2   A040000
  1 |  1  3   5    7    9   11   A005408
  2 |  1  4   9   16   25   36   A000290
  3 |  1  5  14   30   55   91   A000330
  4 |  1  6  20   50  105  196   A002415
  5 |  1  7  27   77  182  378   A005585
  6 |  1  8  35  112  294  672   A040977

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers. New York: Prometheus Books (2007): 97 - 105.
M. Shubin and S. Andersson, Pseudodifferential Operators and Spectral Theory, Springer Series in Soviet Mathematics, 1987.

Vincenzo Librandi, Rows n = 0..102, flattened
P. R. J. Asveld, Generating all permutations by context-free grammars in Chomsky normal form, Theoretical Computer Science 354 (2006) 118-130.
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Arthur T. Benjamin, The Lucas triangle recounted
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 25.
Junesang Choi, Determinants of the Laplacians on the n-dimensional unit sphere S^n , Advances in Difference Equations, 236, 2013.
N. Durov, S. Meljanac, A. Samsarov, Z. Skoda, A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra, arXiv preprint arXiv:math/0604096 [math.RT], 2006.
S. Falcon, On the Lucas triangle and its relationship with the k-Lucas numbers, Journal of Mathematical and Computational Science, 2 (2012), No. 3, 425-434. - _N. J. A. Sloane_, Sep 20 2012
Mark Feinberg, A Lucas triangle, Fibonacci Quart. 5 (1967), 486-490.
Hans-Christian Herbig, Daniel Herden, Christopher Seaton, An algebra generated by x-2, arXiv:1605.01572 [math.CO], 2016.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014 and J. Int. Seq. 18 (2015) 15.10.6.
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (1998), 98-109. [Here the initial term is 1 not 2]
C. M. Ringel, The Catalan combinatorics of the hereditary artin algebras, arXiv preprint arXiv:1502.06553 [math.RT], 2015.
Neville Robbins, The Lucas triangle revisited, Fibonacci Quarterly 43, May 2005, pp. 142-148.
Lee-Peng Teo, Zeta function of spheres and real projective spaces, arXiv:1412.0758v1 [math.NT], 2014.
Xu Wang, Xuxu Zhao, Haiyuan Yao, Structure and enumeration results of matchable Lucas cubes, arXiv:1810.07329 [math.CO], 2018. See Table 1 p. 3.
Y. Yang and J. Leida, Pascal decompositions of geometric arrays in matrices, The Fibonacci Quarterly 42.3 (2004).

Cf. A003945 (row sums), A007318, A034807, A061896, A029653 (row-reversed), A157000.
Sums along ascending antidiagonals give Lucas numbers, n>0.
Cf. A090327, A003945, A029638, A228196, A228576, A000330, A000217, A293600.

a029635 n k = a029635_tabl !! n !! k
a029635_row n = a029635_tabl !! n
a029635_tabl = [2] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,2]
-- Reinhard Zumkeller, Mar 12 2012, Feb 23 2012

T := proc(n, k) option remember;
if n = k then 2 elif k = 0 then 1 else T(n-1, k-1) + T(n-1, k) fi end:
for n from 0 to 8 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Dec 22 2024

t[0, 0] = 2; t[n_, k_] := If[k < 0 || k > n, 0, Binomial[n, k] + Binomial[n-1, k-1]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, May 03 2011 *)
(* The next program cogenerates A029635 and A029638. *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A029638  *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A029635 *)
(* Clark Kimberling, Feb 20 2012 *)
Table[Binomial[n,k]+Binomial[n-1,k-1],{n,0,20},{k,0,n}]//Flatten (* Harvey P. Dale, Feb 08 2024 *)

{T(n, k) = if( k<0 || k>n, 0, (n==0) + binomial(n, k) + binomial(n-1, k-1))}; /* Michael Somos, Jul 15 2003 */

# uses[riordan_array from A256893]
riordan_array((2-x)/(1-x), x/(1-x), 8) # Peter Luschny, Nov 09 2019

From Henry Bottomley, Apr 26 2002;  (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k).
T(n, k) = C(n, k) + C(n-1, k-1).
T(n, k) = C(n, k)*(n + k)/n.
T(n, k) = A007318(n, k) + A007318(n-1, k-1).
T(n, k) = A061896(n + k, k) but with T(0, 0) = 1 and T(1, 1) = 2.
Row sum is floor(3^2(n-1)) i.e., A003945. (End)
G.f.: 1 + (1 + x*y) / (1 - x - x*y). - Michael Somos, Jul 15 2003
G.f. for n-th row: (x+2*y)*(x+y)^(n-1).
O.g.f. for row n: (1+x)/(1-x)^(n+1). The entries in row n are the nonzero entries in column n of A053120 divided by 2^(n-1). - Peter Bala, Aug 14 2008
T(2n, n) - T(2n, n+1)= Catalan(n)= A000108(n). - Philippe Deléham, Mar 19 2009
With T(0, 0) = 1 : Triangle T(n, k), read by rows, given by [1,0,0,0,0,0,...] DELTA [2,-1,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011
With T(0, 0) = 1, as in the Example section below, this is known as Vieta's array. The LU factorization of the square array is given by Yang and Leida, equation 20. - Peter Bala, Feb 11 2012
For n > 0: T(n, k) = A097207(n-1, k), 0 <= k < n. - Reinhard Zumkeller, Mar 12 2012
For n > 0: T(n, k) = A029600(n, k) - A007318(n, k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Riordan array ((2-x)/(1-x), x/(1-x)). - Philippe Deléham, Mar 15 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 4*x + 5*x^2/2! + 2*x^3/3!) = 1 + 5*x + 14*x^2/2! + 30*x^3/3! + 55*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
For n>=1: T(n, 0) + T(n, 1) + T(n, 2) = A000217(n+1). T(n, n-2) = (n-1)^2. - Bob Selcoe, Mar 29 2016:

More terms from David W. Wilson

a(0) changed to 2 (was 1) by Daniel Forgues, Jul 06 2010

A034807 Triangle T(n,k) of coefficients of Lucas (or Cardan) polynomials.

Original entry on oeis.org

2, 1, 1, 2, 1, 3, 1, 4, 2, 1, 5, 5, 1, 6, 9, 2, 1, 7, 14, 7, 1, 8, 20, 16, 2, 1, 9, 27, 30, 9, 1, 10, 35, 50, 25, 2, 1, 11, 44, 77, 55, 11, 1, 12, 54, 112, 105, 36, 2, 1, 13, 65, 156, 182, 91, 13, 1, 14, 77, 210, 294, 196, 49, 2, 1, 15, 90, 275, 450, 378, 140, 15, 1, 16, 104

Offset: 0

N. J. A. Sloane

These polynomials arise in the following setup. Suppose G and H are power series satisfying G + H = G*H = 1/x. Then G^n + H^n = (1/x^n)*L_n(-x).
Apart from signs, triangle of coefficients when 2*cos(nt) is expanded in terms of x = 2*cos(t). For example, 2*cos(2t) = x^2 - 2, 2*cos(3t) = x^3 - 3x and 2*cos(4t) = x^4 - 4x^2 + 2. - Anthony C Robin, Jun 02 2004
Triangle of coefficients of expansion of Z_{nk} in terms of Z_k.
Row n has 1 + floor(n/2) terms. - Emeric Deutsch, Dec 25 2004
T(n,k) = number of k-matchings of the cycle C_n (n > 1). Example: T(6,2)=9 because the 2-matchings of the hexagon with edges a, b, c, d, e, f are ac, ad, ae, bd, be, bf, ce, cf and df. - Emeric Deutsch, Dec 25 2004
An example for the first comment: G=c(x), H=1/(x*c(x)) with c(x) the o.g.f. Catalan numbers A000108: (x*c(x))^n + (1/c(x))^n = L(n,-x)= Sum_{k=0..floor(n/2)} T(n,k)*(-x)^k.
This triangle also supplies the absolute values of the coefficients in the multiplication formulas for the Lucas numbers A000032.
From L. Edson Jeffery, Mar 19 2011: (Start)
This sequence is related to rhombus substitution tilings. A signed version of it (see A132460), formed as a triangle with interlaced zeros extending each row to n terms, begins as
  {2}
  {1,  0}
  {1,  0, -2}
  {1,  0, -3,  0}
  {1,  0, -4,  0,  2}
  {1,  0, -5,  0,  5,  0}
  ....
For the n X n tridiagonal unit-primitive matrix G_(n,1) (n >= 2) (see the L. E. Jeffery link below), defined by
G_(n,1) =
  (0 1 0 ... 0)
  (1 0 1 0 ... 0)
  (0 1 0 1 0 ... 0)
  ...
  (0 ... 0 1 0 1)
  (0 ... 0 2 0),
Row n (i.e., {T(n,k)}, k=0..n) of the signed table gives the coefficients of its characteristic function: c_n(x) = Sum_{k=0..n} T(n,k)*x^(n-k) = 0. For example, let n=3. Then
G_(3,1) =
  (0 1 0)
  (1 0 1)
  (0 2 0),
and row 3 of the table is {1,0,-3,0}. Hence c_3(x) = x^3 - 3*x = 0. G_(n,1) has n distinct eigenvalues (the solutions of c_n(x) = 0), given by w_j = 2*cos((2*j-1)*Pi/(2*n)), j=1..n. (End)
For n > 0, T(n,k) is the number of k-subsets of {1,2,...,n} which contain neither consecutive integers nor both 1 and n. Equivalently, T(n,k) is the number of k-subsets without neighbors of a set of n points on a circle. - José H. Nieto S., Jan 17 2012
With the first column omitted, this gives A157000. - Philippe Deléham, Mar 17 2013
The number of necklaces of k black and n - k white beads with no adjacent black beads (Kaplansky 1943). Coefficients of the Dickson polynomials D(n,x,-a). - Peter Bala, Mar 09 2014
From Tom Copeland, Nov 07 2015: (Start)
This triangular array is composed of interleaved rows of reversed, unsigned A127677 (cf. A156308, A217476, A263916) and reversed A111125 (cf. A127672).
See also A113279 for another connection to symmetric and Faber polynomials.
The difference of consecutive rows gives the previous row shifted.
For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 12, 20, and 21) and Damianou and Evripidou (p. 7). (End)
Diagonals are related to multiplicities of eigenvalues of the Laplacian on hyperspheres through A029635. - Tom Copeland, Jan 10 2016
For n>=3, also the independence and matching polynomials of the n-cycle graph C_n. See also A284966. - Eric W. Weisstein, Apr 06 2017
Apparently, with the rows aerated and then the 2s on the diagonal removed, this matrix becomes the reverse, or mirror, of unsigned A117179. See also A114525 - Tom Copeland, May 30 2017
Briggs's (1633) table with an additional column of 2s on the right can be used to generate this table. See p. 69 of the Newton reference. - Tom Copeland, Jun 03 2017
From Liam Solus, Aug 23 2018: (Start)
For n>3 and k>0, T(n,k) equals the number of Markov equivalence classes with skeleton the cycle on n nodes having exactly k immoralities. See Theorem 2.1 of the article by A. Radhakrishnan et al. below.
For n>2 odd and r = floor(n/2)-1, the n-th row is the coefficient vector of the Ehrhart h*-polynomial of the r-stable (n,2)-hypersimplex. See Theorem 4.14 in the article by B. Braun and L. Solus below.
(End)
Conjecture: If a(n) = H(a,b,c,d,n) is a second-order linear recurrence with constant coefficients defined as a(0) = a, a(1)= b, a(n) = c*a(n-1) + d*a(n-2) then a(m*n) = H(a, H(a,b,c,d,m), Sum_{k=0..floor(m/2)} T(m,k)*c^(m-2*k)*d^k, (-1)^(m+1)*d^m, n) (Wolfdieter Lang). - Gary Detlefs, Feb 06 2023
For the proof of the preceding conjecture see the Detlefs and Lang link. There also proofs for several properties of this table are found. - Wolfdieter Lang, Apr 25 2023
From Mohammed Yaseen, Nov 09 2024: (Start)
Let m   - 1/m   = x, then
    m^2 + 1/m^2 = x^2 + 2,
    m^3 - 1/m^3 = x^3 + 3*x,
    m^4 + 1/m^4 = x^4 + 4*x^2 + 2,
    m^5 - 1/m^5 = x^5 + 5*x^3 + 5*x,
    m^6 + 1/m^6 = x^6 + 6*x^4 + 9*x^2 + 2,
    m^7 - 1/m^7 = x^7 + 7*x^5 + 14*x^3 + 7*x, etc. (End)

			I have seen two versions of these polynomials: One version begins L_0 = 2, L_1 = 1, L_2 = 1 + 2*x, L_3 = 1 + 3*x, L_4 = 1 + 4*x + 2*x^2, L_5 = 1 + 5*x + 5*x^2, L_6 = 1 + 6*x + 9*x^2 + 2*x^3, L_7 = 1 + 7*x + 14*x^2 + 7*x^3, L_8 = 1 + 8*x + 20*x^2 + 16*x^3 + 2*x^4, L_9 = 1 + 9*x + 27*x^2 + 30*x^3 + 9*x^4, ...
The other version (probably the more official one) begins L_0(x) = 2, L_1(x) = x, L_2(x) = 2 + x^2, L_3(x) = 3*x + x^3, L_4(x) = 2 + 4*x^2 + x^4, L_5(x) = 5*x + 5*x^3 + x^5, L_6(x) = 2 + 9*x^2 + 6*x^4 + x^6, L_7(x) = 7*x + 14*x^3 + 7*x^5 + x^7, L_8(x) = 2 + 16*x^2 + 20*x^4 + 8*x^6 + x^8, L_9(x) = 9*x + 30*x^3 + 27*x^5 + 9*x^7 + x^9.
From _John Blythe Dobson_, Oct 11 2007: (Start)
Triangle begins:
  2;
  1;
  1,  2;
  1,  3;
  1,  4,  2;
  1,  5,  5;
  1,  6,  9,   2;
  1,  7, 14,   7;
  1,  8, 20,  16,   2;
  1,  9, 27,  30,   9;
  1, 10, 35,  50,  25,   2;
  1, 11, 44,  77,  55,  11;
  1, 12, 54, 112, 105,  36,   2;
  1, 13, 65, 156, 182,  91,  13;
  1, 14, 77, 210, 294, 196,  49,  2;
  1, 15, 90, 275, 450, 378, 140, 15;
(End)
From _Peter Bala_, Mar 20 2025: (Start)
Let S = x + y and M = -x*y. Then the triangle gives the coefficients when expressing the symmetric polynomial x^n + y^n as a polynomial in S and M. For example,
x^2 + y^2 = S^2 + 2*M; x^3 + y^3 = S^3 + 3*S*M; x^4 + y^4 = S^4 + 4*(S^2)*M + 2*M^2;
x^5 + y^5 = S^5 + 5*(S^3)*M + 5*S*M^2; x^6 + y^6 = S^6 + 6*(S^4)*M + 9*(S^2)*M^2 + 2*M^3. See Woko. In general x^n + y^n = 2*(-i)^n *(sqrt(M))^n * T(n, i*S/(2*sqrt(M))), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. (End)

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Thomas Koshy, Fibonacci and Lucas Numbers with Applications. New York, etc.: John Wiley & Sons, 2001. (Chapter 13, "Pascal-like Triangles," is devoted to the present triangle.)
The Royal Society Newton Tercentenary Celebrations, Cambridge Univ. Press, 1947.

T. D. Noe, Rows n = 0..100 of triangle, flattened
Moussa Benoumhani, A Sequence of Binomial Coefficients Related to Lucas and Fibonacci Numbers, J. Integer Seqs., Vol. 6, 2003.
Benjamin Braun and Liam Solus, r-stable hypersimplices, arXiv:1408.4713 [math.CO], 2014-2016; Journal of Combinatorial Theory, Series A 157 (2018): 349-388.
Johann Cigler and Hans-Christian Herbig, Factorization of spread polynomials, arXiv:2412.18958 [math.NT], 2024. See p. 2.
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Pantelis A. Damianou and Charalampos A. Evripidou, Characteristic and Coxeter polynomials for affine Lie algebras, arXiv preprint arXiv:1409.3956 [math.RT], 2014.
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Rcurrence Sequence<, arXiv:2304.12937[nath.CO], 2023.
Joseph Doolittle, Lukas Katthän, Benjamin Nill, and Francisco Santos, Empty simplices of large width, arXiv:2103.14925 [math.CO], 2021.
Sergio Falcon, On the Lucas triangle and its relationship with the k-Lucas numbers, Journal of Mathematical and Computational Science, 2 (2012), No. 3, 425-434.
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Feihu Liu, Ying Wang, Yingrui Zhang, and Zihao Zhang,Hankel Determinants for Convolution of Power Series: An Extension of Cigler's Results, arXiv:2503.17187 [math.CO], 2025. See p.5.
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Lorenzo Venturello, Koszul Gorenstein algebras from Cohen-Macaulay simplicial complexes, arXiv:2106.05051 [math.AC], 2021.
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Wikipedia, Metallic mean.
Justin Eze Woko, A Pascal-like Triangle for alpha^n + beta^n, The Mathematical Gazette, Vol. 81, No. 490 (Mar., 1997), pp. 75-79.

Cf. A029635, A061896, A111125, A113279, A114525, A127672, A127677, A156308, A217476, A263916.

Cf. A117179.

T:= proc(n,k) if n=0 and k=0 then 2 elif k>floor(n/2) then 0 else n*binomial(n-k,k)/(n-k) fi end: for n from 0 to 15 do seq(T(n,k), k=0..floor(n/2)) od; # yields sequence in triangular form # Emeric Deutsch, Dec 25 2004

t[0, 0] = 2; t[n_, k_] := Binomial[n-k, k] + Binomial[n-k-1, k-1]; Table[t[n, k], {n, 0, 16}, {k, 0, Floor[n/2]}] // Flatten (* Jean-François Alcover, Dec 30 2013 *)
CoefficientList[Table[x^(n/2) LucasL[n, 1/Sqrt[x]], {n, 0, 15}], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
Table[Select[Reverse[CoefficientList[LucasL[n, x], x]], 0 < # &], {n, 0, 16}] // Flatten (* Robert G. Wilson v, May 03 2017 *)
CoefficientList[FunctionExpand @ Table[2 (-x)^(n/2) Cos[n ArcSec[2 Sqrt[-x]]], {n, 0, 15}], x] // Flatten (* Eric W. Weisstein, Apr 03 2018 *)
CoefficientList[Table[2 (-x)^(n/2) ChebyshevT[n, 1/(2 Sqrt[-x])], {n, 0, 15}], x] // Flatten (* Eric W. Weisstein, Apr 03 2018 *)

{T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k) + binomial(n-k-1, k-1) + (n==0))}; /* Michael Somos, Jul 15 2003 */

Row sums = A000032. T(2n, n-1) = A000290(n), T(2n+1, n-1) = A000330(n), T(2n, n-2) = A002415(n). T(n, k) = A029635(n-k, k), if n>0. - Michael Somos, Apr 02 1999
Lucas polynomial coefficients: 1, -n, n*(n-3)/2!, -n*(n-4)*(n-5)/3!, n*(n-5)*(n-6)*(n-7)/4!, - n*(n-6)*(n-7)*(n-8)*(n-9)/5!, ... - Herb Conn and Gary W. Adamson, May 28 2003
G.f.: (2-x)/(1-x-x^2*y). - Vladeta Jovovic, May 31 2003
T(n, k) = T(n-1, k) + T(n-2, k-1), n>1. T(n, 0) = 1, n>0. T(n, k) = binomial(n-k, k) + binomial(n-k-1, k-1) = n*binomial(n-k-1, k-1)/k, 0 <= 2*k <= n except T(0, 0) = 2. - Michael Somos, Apr 02 1999
T(n,k) = (n*(n-1-k)!)/(k!*(n-2*k)!), n>0, k>=0. - Alexander Elkins (alexander_elkins(AT)hotmail.com), Jun 09 2007
O.g.f.: 2-(2xt+1)xt/(-t+xt+(xt)^2). (Cf. A113279.) - Tom Copeland, Nov 07 2015
T(n,k) = A011973(n-1,k) + A011973(n-3,k-1) = A011973(n,k) - A011973(n-4,k-2) except for T(0,0)=T(2,1)=2. - Xiangyu Chen, Dec 24 2020
L_n(x) = ((x+sqrt(x^2+4))/2)^n + (-((x+sqrt(x^2+4))/2))^(-n). See metallic means. - William Krier, Sep 01 2023

Improved description, more terms, etc., from Michael Somos

cp's OEIS Frontend

A029635 The (1,2)-Pascal triangle (or Lucas triangle) read by rows.

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A034807 Triangle T(n,k) of coefficients of Lucas (or Cardan) polynomials.

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A174625 Table T(n,k) with the coefficients of the polynomial P_n(x) = P_{n-1}(x) + x*P_{n-2}(x) + 1 in row n, by decreasing exponent of x.

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