cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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

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Author

Keywords

Comments

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)

Examples

			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)
		

References

  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 148.
  • C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.
  • 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.

Crossrefs

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {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 */

Formula

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

Extensions

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

A111125 Triangle read by rows: T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.

Original entry on oeis.org

1, 3, 1, 5, 5, 1, 7, 14, 7, 1, 9, 30, 27, 9, 1, 11, 55, 77, 44, 11, 1, 13, 91, 182, 156, 65, 13, 1, 15, 140, 378, 450, 275, 90, 15, 1, 17, 204, 714, 1122, 935, 442, 119, 17, 1, 19, 285, 1254, 2508, 2717, 1729, 665, 152, 19, 1, 21, 385, 2079, 5148, 7007, 5733, 2940, 952, 189, 21, 1
Offset: 0

Views

Author

N. J. A. Sloane, Oct 16 2005

Keywords

Comments

Riordan array ((1+x)/(1-x)^2, x/(1-x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1-x), x/(1-x))(1+2*x, x*(1+x)). - Paul Barry, Oct 17 2005
Central coefficients of rows with odd numbers of term are A052227.
From Wolfdieter Lang, Jun 26 2011: (Start)
T(k,s) appears as T_s(k) in the Knuth reference, p. 285.
This triangle is related to triangle A156308(n,m), appearing in this reference as U_m(n) on p. 285, by T(k,s) - T(k-1,s) = A156308(k,s), k>=s>=1 (identity on p. 286). T(k,s) = A156308(k+1,s+1) - A156308(k,s+1), k>=s>=0 (identity on p. 286).
(End)
A111125 is jointly generated with A208513 as an array of coefficients of polynomials v(n,x): initially, u(1,x)= v(1,x)= 1; for n>1, u(n,x)= u(n-1,x) +x*(x+1)*v(n-1) and v(n,x)= u(n-1,x) +x*v(n-1,x) +1. See the Mathematica section. The columns of A111125 are identical to those of A208508. Here, however, the alternating row sums are periodic (with period 1,2,1,-1,-2,-1). - Clark Kimberling, Feb 28 2012
This triangle T(k,s) (with signs and columns scaled with powers of 5) appears in the expansion of Fibonacci numbers F=A000045 with multiples of odd numbers as indices in terms of odd powers of F-numbers. See the Jennings reference, p. 108, Theorem 1. Quoted as Lemma 3 in the Ozeki reference given in A111418. The formula is: F_{(2*k+1)*n} = Sum_{s=0..k} ( T(k,s)*(-1)^((k+s)*n)*5^s*F_{n}^(2*s+1) ), k >= 0, n >= 0. - Wolfdieter Lang, Aug 24 2012
From Wolfdieter Lang, Oct 18 2012: (Start)
This triangle T(k,s) appears in the formula x^(2*k+1) - x^(-(2*k+1)) = Sum_{s=0..k} ( T(k,s)*(x-x^(-1))^(2*s+1) ), k>=0. Prove the inverse formula (due to the Riordan property this will suffice) with the binomial theorem. Motivated to look into this by the quoted paper of Wang and Zhang, eq. (1.4).
Alternating row sums are A057079.
The Z-sequence of this Riordan array is A217477, and the A-sequence is (-1)^n*A115141(n). For the notion of A- and Z-sequences for Riordan triangles see a W. Lang link under A006232. (End)
The signed triangle ((-1)^(k-s))*T(k,s) gives the coefficients of (x^2)^s of the polynomials C(2*k+1,x)/x, with C the monic integer Chebyshev T-polynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1-x)/(1+x)^2, x/(1+x)^2). Proof by comparing the o.g.f. of the row polynomials where x is replaced by x^2 with the odd part of the bisection of the o.g.f. for C(n,x)/x. - Wolfdieter Lang, Oct 23 2012
From Wolfdieter Lang, Oct 04 2013: (Start)
The signed triangle S(k,s) := ((-1)^(k-s))*T(k,s) (see the preceding comment) is used to express in a (4*(k+1))-gon the length ratio s(4*(k+1)) = 2*sin(Pi/4*(k+1)) = 2*cos((2*k+1)*Pi/(4*(k+1))) of a side/radius as a polynomial in rho(4*(k+1)) = 2*cos(Pi/4*(k+1)), the length ratio (smallest diagonal)/side:
s(4*(k+1)) = Sum_{s=0..k} ( S(k,s)*rho(4*(k+1))^(2*s+1) ).
This is to be computed modulo C(4*(k+1), rho(4*(k+1)) = 0, the minimal polynomial (see A187360) in order to obtain s(4*(k+1)) as an integer in the algebraic number field Q(rho(4*(k+1))) of degree delta(4*(k+1)) (see A055034). Thanks go to Seppo Mustonen for asking me to look into the problem of the square of the total length in a regular n-gon, where this formula is used in the even n case. See A127677 for the formula in the (4*k+2)-gon. (End)
From Wolfdieter Lang, Aug 14 2014: (Start)
The row polynomials for the signed triangle (see the Oct 23 2012 comment above), call them todd(k,x) = Sum_{s=0..k} ( (-1)^(k-s)*T(k,s)*x^s ) = S(k, x-2) - S(k-1, x-2), k >= 0, with the Chebyshev S-polynomials (see their coefficient triangle (A049310) and S(-1, x) = 0), satisfy the recurrence todd(k, x) = (-1)^(k-1)*((x-4)/2)*todd(k-1, 4-x) + ((x-2)/2)*todd(k-1, x), k >= 1, todd(0, x) = 1. From the Aug 03 2014 comment on A130777.
This leads to a recurrence for the signed triangle, call it S(k,s) as in the Oct 04 2013 comment: S(k,s) = (1/2)*(1 + (-1)^(k-s))*S(k-1,s-1) + (2*(s+1)*(-1)^(k-s) - 1)*S(k-1,s) + (1/2)*(-1)^(k-s)*Sum_{j=0..k-s-2} ( binomial(j+s+2,s)*4^(j+2)* S(k-1, s+1+j) ) for k >= s >= 1, and S(k,s) = 0 if k < s and S(k,0) = (-1)^k*(2*k+1). Note that the recurrence derived from the Riordan A-sequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k-1,s-1+j), j=0..k-s), k >= s >=1.
(End)
From Tom Copeland, Nov 07 2015: (Start)
Rephrasing notes here: Append an initial column of zeros, except for a 1 at the top, to A111125 here. Then the partial sums of the columns of this modified entry are contained in A208513. Append an initial row of zeros to A208513. Then the difference of consecutive pairs of rows of the modified A208513 generates the modified A111125. Cf. A034807 and A127677.
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. 20 and 21) and Damianou and Evripidou (p. 7).
As suggested by the equations on p. 7 of Damianou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = (A(2*n+1, x) + 2)/x = (F(2*n+1, (2-x), 1, 0, 0, ... ) + 2)/x = F(2*n+1, -x, 2*x, -3*x, ..., (-1)^n n*x)/x = -F(2*n+1, x, 2*x, 3*x, ..., n*x)/x, where A(n,x) are the polynomials of A127677 and F(n, ...) are the Faber polynomials of A263196. Cf. A127672 and A127677.
(End)
The row polynomials P(k, x) of the signed triangle S(k, s) = ((-1)^(k-s))*T(k, s) are given from the row polynomials R(2*k+1, x) of triangle A127672 by
P(k, x) = R(2*k+1, sqrt(x))/sqrt(x). - Wolfdieter Lang, May 02 2021

Examples

			Triangle T(k,s) begins:
k\s  0    1     2     3     4     5     6    7    8   9 10
0:   1
1:   3    1
2:   5    5     1
3:   7   14     7     1
4:   9   30    27     9     1
5:  11   55    77    44    11     1
6:  13   91   182   156    65    13     1
7:  15  140   378   450   275    90    15    1
8:  17  204   714  1122   935   442   119   17    1
9:  19  285  1254  2508  2717  1729   665  152   19   1
10: 21  385  2079  5148  7007  5733  2940  952  189  21  1
... Extended and reformatted by _Wolfdieter Lang_, Oct 18 2012
Application for Fibonacci numbers F_{(2*k+1)*n}, row k=3:
F_{7*n} = 7*(-1)^(3*n)*F_n + 14*(-1)^(4*n)*5*F_n^3 + 7*(-1)^(5*n)*5^2*F_n^5 + 1*(-1)^(6*n)*5^3*F_n^7, n>=0. - _Wolfdieter Lang_, Aug 24 2012
Example for the  Z- and A-sequence recurrences  of this Riordan triangle: Z = A217477 = [3,-4,12,-40,...]; T(4,0) = 3*7 -4*14 +12*7 -40*1 = 9. A =  [1, 2, -1, 2, -5, 14, ..]; T(5,2) = 1*30 + 2*27 - 1*9 + 2*1= 77. _Wolfdieter Lang_, Oct 18 2012
Example for the (4*(k+1))-gon length ratio s(4*(k+1))(side/radius) as polynomial in the ratio rho(4*(k+1)) ((smallest diagonal)/side): k=0, s(4) = 1*rho(4) = sqrt(2); k=1, s(8) = -3*rho(8) + rho(8)^3 = sqrt(2-sqrt(2)); k=2, s(12) = 5*rho(12) - 5*rho(12)^3 + rho(12)^5, and C(12,x) = x^4 - 4*x^2 + 1, hence rho(12)^5 = 4*rho(12)^3 - rho(12), and s(12) = 4*rho(12) - rho(12)^3 = sqrt(2 - sqrt(3)). - _Wolfdieter Lang_, Oct 04 2013
Example for the recurrence for the signed triangle S(k,s)= ((-1)^(k-s))*T(k,s) (see the Aug 14 2014 comment above):
S(4,1) = 0 + (-2*2 - 1)*S(3,1) - (1/2)*(3*4^2*S(3,2) + 4*4^3*S(3,3)) = - 5*14 - 3*8*(-7) - 128*1 = -30. The recurrence from the Riordan A-sequence A115141 is S(4,1) = -7 -2*14 -(-7) -2*1 = -30. - _Wolfdieter Lang_, Aug 14 2014
		

Crossrefs

Mirror image of A082985, which see for further references, etc.
Also closely related to triangles in A098599 and A100218.

Programs

  • Magma
    [((2*n+1)/(n+k+1))*Binomial(n+k+1, 2*k+1): k in [0..n], n in [0..12]];  // G. C. Greubel, Feb 01 2022
  • Mathematica
    (* First program *)
    u[1, x_]:=1; v[1, x_]:=1; z=16;
    u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
    v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x] + 1;
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]  (* A208513 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]  (* A111125 *) (* Clark Kimberling, Feb 28 2012 *)
    (* Second program *)
    T[n_, k_]:= ((2*n+1)/(2*k+1))*Binomial[n+k, 2*k];
    Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 01 2022 *)
  • Sage
    @CachedFunction
    def T(n,k):
        if n< 0: return 0
        if n==0: return 1 if k == 0 else 0
        h = 3*T(n-1,k) if n==1 else 2*T(n-1,k)
        return T(n-1,k-1) - T(n-2,k) - h
    A111125 = lambda n,k: (-1)^(n-k)*T(n,k)
    for n in (0..9): [A111125(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012
    

Formula

T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.
From Peter Bala, Apr 30 2012: (Start)
T(n,k) = binomial(n+k,2*k) + 2*binomial(n+k,2*k+1).
The row generating polynomials P(n,x) are a generalization of the Morgan-Voyce polynomials b(n,x) and B(n,x). They satisfy the recurrence equation P(n,x) = (x+2)*P(n-1,x) - P(n-2,x) for n >= 2, with initial conditions P(0,x) = 1, P(1,x) = x+r+1 and with r = 2. The cases r = 0 and r = 1 give the Morgan-Voyce polynomials A085478 and A078812 respectively. Andre-Jeannin has considered the case of general r.
P(n,x) = U(n+1,1+x/2) + U(n,1+x/2), where U(n,x) denotes the Chebyshev polynomial of the second kind - see A053117. P(n,x) = (2/x)*(T(2*n+2,u)-T(2*n,u)), where u = sqrt((x+4)/4) and T(n,x) denotes the Chebyshev polynomial of the first kind - see A053120. P(n,x) = Product_{k = 1..n} ( x + 4*(sin(k*Pi/(2*n+1)))^2 ). P(n,x) = 1/x*(b(n+1,x) - b(n-1,x)) and P(n,x) = 1/x*{(b(2*n+2,x)+1)/b(n+1,x) - (b(2*n,x)+1)/b(n,x)}, where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478. Cf. A211957.
(End)
From Wolfdieter Lang, Oct 18 2012 (Start)
O.g.f. column No. s: ((1+x)/(1-x)^2)*(x/(1-x)^2)^s, s >= 0. (from the Riordan data given in a comment above).
O.g.f. of the row polynomials R(k,x):= Sum_{s=0..k} ( T(k,s)*x^s ), k>=0: (1+z)/(1-(2+x)*z+z^2) (from the Riordan property).
(End)
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 12 2013

Extensions

More terms from Paul Barry, Oct 17 2005

A263916 Coefficients of the Faber partition polynomials.

Original entry on oeis.org

-1, -2, 1, -3, 3, -1, -4, 4, 2, -4, 1, -5, 5, 5, -5, -5, 5, -1, -6, 6, 6, -6, 3, -12, 6, -2, 9, -6, 1, -7, 7, 7, -7, 7, -14, 7, -7, -7, 21, -7, 7, -14, 7, -1, -8, 8, 8, -8, 8, -16, 8, 4, -16, -8, 24, -8, -8, 12, 24, -32, 8, 2, -16, 20, -8, 1
Offset: 1

Views

Author

Tom Copeland, Oct 29 2015

Keywords

Comments

The coefficients of the Faber polynomials F(n,b(1),b(2),...,b(n)) (Bouali, p. 52) in the order of the partitions of Abramowitz and Stegun. Compare with A115131 and A210258.
These polynomials occur in discussions of the Virasoro algebra, univalent function spaces and the Schwarzian derivative, symmetric functions, and free probability theory. They are intimately related to symmetric functions, free probability, and Appell sequences through the raising operator R = x - d log(H(D))/dD for the Appell sequence inverse pair associated to the e.g.f.s H(t)e^(xt) (cf. A094587) and (1/H(t))e^(xt) with H(0)=1.
Instances of the Faber polynomials occur in discussions of modular invariants and modular functions in the papers by Asai, Kaneko, and Ninomiya, by Ono and Rolen, and by Zagier. - Tom Copeland, Aug 13 2019
The Faber polynomials, denoted by s_n(a(t)) where a(t) is a formal power series defined by a product formula, are implicitly defined by equation 13.4 on p. 62 of Hazewinkel so as to extract the power sums of the reciprocals of the zeros of a(t). This is the Newton identity expressing the power sum symmetric polynomials in terms of the elementary symmetric polynomials/functions. - Tom Copeland, Jun 06 2020
From Tom Copeland, Oct 15 2020: (Start)
With a_n = n! * b_n = (n-1)! * c_n for n > 0, represent a function with f(0) = a_0 = b_0 = 1 as an
A) exponential generating function (e.g.f), or formal Taylor series: f(x) = e^{a.x} = 1 + Sum_{n > 0} a_n * x^n/n!
B) ordinary generating function (o.g.f.), or formal power series: f(x) = 1/(1-b.x) = 1 + Sum_{n > 0} b_n * x^n
C) logarithmic generating function (l.g.f): f(x) = 1 - log(1 - c.x) = 1 + Sum_{n > 0} c_n * x^n /n.
Expansions of log(f(x)) are given in
I) A127671 and A263634 for the e.g.f: log[ e^{a.*x} ] = e^{L.(a_1,a_2,...)x} = Sum_{n > 0} L_n(a_1,...,a_n) * x^n/n!, the logarithmic polynomials, cumulant expansion polynomials
II) A263916 for the o.g.f.: log[ 1/(1-b.x) ] = log[ 1 - F.(b_1,b_2,...)x ] = -Sum_{n > 0} F_n(b_1,...,b_n) * x^n/n, the Faber polynomials.
Expansions of exp(f(x)-1) are given in
III) A036040 for an e.g.f: exp[ e^{a.x} - 1 ] = e^{BELL.(a_1,...)x}, the Bell/Touchard/exponential partition polynomials, a.k.a. the Stirling partition polynomials of the second kind
IV) A130561 for an o.g.f.: exp[ b.x/(1-b.x) ] = e^{LAH.(b.,...)x}, the Lah partition polynomials
V) A036039 for an l.g.f.: exp[ -log(1-c.x) ] = e^{CIP.(c_1,...)x}, the cycle index polynomials of the symmetric groups S_n, a.k.a. the Stirling partition polynomials of the first kind.
Since exp and log are a compositional inverse pair, one can extract the indeterminates of the log set of partition polynomials from the exp set and vice versa. For a discussion of the relations among these polynomials and the combinatorics of connected and disconnected graphs/maps, see Novak and LaCroix on classical moments and cumulants and the two books on statistical mechanics referenced in A036040. (End)

Examples

			F(1,b1) = - b1
F(2,b1,b2) = -2 b2 + b1^2
F(3,b1,b2,b3) = -3 b3 + 3 b1 b2 - b1^3
F(4,b1,...) = -4 b4 + 4 b1 b3 + 2 b2^2  - 4 b1^2 b2 + b1^4
F(5,...) = -5 b5 + 5 b1 b4 + 5 b2 b3 - 5 b1^2 b3 - 5 b1 b2^2 + 5 b1^3 b2 - b1^5
------------------------------
IF(1,b1) = -b1
IF(2,b1,,b2) = -b2 + b1^2
IF(3,b1,b2,b3) = -2 b3 + 3 b1 b2 - b1^3
IF(4,b1,...) = -6 b4 + 8 b1 b3 + 3 b2^2  - 6 b1^2 b2 + b1^4
IF(5,...) = -24 b5 + 30 b1 b4 + 20 b2 b3 - 20 b1^2 b3 - 15 b1 b2^2 + 10 b1^3 b2 - b1^5
------------------------------
For 1/(1+x)^2 = 1- 2x + 3x^2 - 4x^3 + 5x^4 - ..., F(n,-2,3,-4,...) = (-1)^(n+1) 2.
------------------------------
F(n,x,2x,...,nx), F(n,-x,2x,-3x,...,(-1)^n n*x), and F(n,(2-x),1,0,0,...) are related to the Chebyshev polynomials through A127677 and A111125. See also A110162, A156308, A208513, A217476, and A220668.
------------------------------
For b1 = p, b2 = q, and all other indeterminates 0, see A113279 and A034807.
For b1 = -y, b2 = 1 and all other indeterminates 0, see A127672.
		

References

  • H. Airault, "Symmetric sums associated to the factorization of Grunsky coefficients," in Groups and Symmetries: From Neolithic Scots to John McKay, CRM Proceedings and Lecture Notes: Vol. 47, edited by J. Harnad and P. Winternitz, American Mathematical Society, 2009.
  • D. Bleeker and B. Booss, Index Theory with Applications to Mathematics and Physics, International Press, 2013, (see section 16.7 Characteristic Classes and Curvature).
  • M. Hazewinkel, Formal Groups and Applications, Academic Press, New York San Francisco London, 1978, p. 120.
  • F. Hirzebruch, Topological methods in algebraic geometry. Second, corrected printing of the third edition. Die Grundlehren der Mathematischen Wissenschaften, Band 131 Springer-Verlag, Berlin Heidelberg New York, 1978, p. 11 and 92.
  • D. Knutson, λ-Rings and the Representation Theory of the Symmetric Group, Lect. Notes in Math. 308, Springer-Verlag, 1973, p. 35.
  • D. Yau, Lambda-Rings, World Scientific Publishing Co., Singapore, 2010, p. 45.

Crossrefs

Programs

  • Mathematica
    F[0] = 1; F[1] = -b[1]; F[2] = b[1]^2 - 2 b[2]; F[n_] := F[n] = -b[1] F[n - 1] - Sum[b[n - k] F[k], {k, 1, n - 2}] - n b[n] // Expand;
    row[n_] := (List @@ F[n]) /. b[_] -> 1 // Reverse;
    Table[row[n], {n, 1, 8}] // Flatten // Rest (* Jean-François Alcover, Jun 12 2017 *)

Formula

-log(1 + b(1) x + b(2) x^2 + ...) = Sum_{n>=1} F(n,b(1),...,b(n)) * x^n/n.
-d(1 + b(1) x + b(2) x^2 + ...)/dx / (1 + b(1) x + b(2) x^2 + ...) = Sum_{n>=1} F(n,b(1),...,b(n)) x^(n-1).
F(n,b(1),...,b(n)) = -n*b(n) - Sum_{k=1..n-1} b(n-k)*F(k,b(1),...,b(k)).
Umbrally, with B(x) = 1 + b(1) x + b(2) x^2 + ..., B(x) = exp[log(1-F.x)] and 1/B(x) = exp[-log(1-F.x)], establishing a connection to the e.g.f. of A036039 and the symmetric polynomials.
The Stirling partition polynomials of the first kind St1(n,b1,b2,...,bn;-1) = IF(n,b1,b2,...,bn) (cf. the Copeland link Lagrange a la Lah, signed A036039, and p. 184 of Airault and Bouali), i.e., the cyclic partition polynomials for the symmetric groups, and the Faber polynomials form an inverse pair for isolating the indeterminates in their definition, for example, F(3,IF(1,b1),IF(2,b1,b2)/2!,IF(3,b1,b2,b3)/3!)= b3, with bk = b(k), and IF(3,F(1,b1),F(2,b1,b2),F(3,b1,b2,b3))/3!= b3.
The polynomials specialize to F(n,t,t,...) = (1-t)^n - 1.
See Newton Identities on Wikipedia on relation between the power sum symmetric polynomials and the complete homogeneous and elementary symmetric polynomials for an expression in multinomials for the coefficients of the Faber polynomials.
(n-1)! F(n,x[1],x[2]/2!,...,x[n]/n!) = - p_n(x[1],...,x[n]), where p_n are the cumulants of A127671 expressed in terms of the moments x[n]. - Tom Copeland, Nov 17 2015
-(n-1)! F(n,B(1,x[1]),B(2,x[1],x[2])/2!,...,B(n,x[1],...,x[n])/n!) = x[n] provides an extraction of the indeterminates of the complete Bell partition polynomials B(n,x[1],...,x[n]) of A036040. Conversely, IF(n,-x[1],-x[2],-x[3]/2!,...,-x[n]/(n-1)!) = B(n,x[1],...,x[n]). - Tom Copeland, Nov 29 2015
For a square matrix M, determinant(I - x M) = exp[-Sum_{k>0} (trace(M^k) x^k / k)] = Sum_{n>0} [ P_n(-trace(M),-trace(M^2),...,-trace(M^n)) x^n/n! ] = 1 + Sum_{n>0} (d[n] x^n), where P_n(x[1],...,x[n]) are the cycle index partition polynomials of A036039 and d[n] = P_n(-trace(M),-trace(M^2),...,-trace(M^n)) / n!. Umbrally, det(I - x M)= exp[log(1 - b. x)] = exp[P.(-b_1,..,-b_n)x] = 1 / (1-d.x), where b_k = tr(M^k). Then F(n,d[1],...,d[n]) = tr[M^n]. - Tom Copeland, Dec 04 2015
Given f(x) = -log(g(x)) = -log(1 + b(1) x + b(2) x^2 + ...) = Sum_{n>=1} F(n,b(1),...,b(n)) * x^n/n, action on u_n = F(n,b(1),...,b(n)) with A133932 gives the compositional inverse finv(x) of f(x), with F(1,b(1)) not equal to zero, and f(g(finv(x))) = f(e^(-x)). Note also that exp(f(x)) = 1 / g(x) = exp[Sum_{n>=1} F(n,b(1),...,b(n)) * x^n/n] implies relations among A036040, A133314, A036039, and the Faber polynomials. - Tom Copeland, Dec 16 2015
The Dress and Siebeneicher paper gives combinatorial interpretations and various relations that the Faber polynomials must satisfy for integral values of its arguments. E.g., Eqn. (1.2) p. 2 implies [2 * F(1,-1) + F(2,-1,b2) + F(4,-1,b2,b3,b4)] mod(4) = 0. This equation implies that [F(n,b1,b2,...,bn)-(-b1)^n] mod(n) = 0 for n prime. - Tom Copeland, Feb 01 2016
With the elementary Schur polynomials S(n,a_1,a_2,...,a_n) = Lah(n,a_1,a_2,...,a_n) / n!, where Lah(n,...) are the refined Lah polynomials of A130561, F(n,S(1,a_1),S(2,a_1,a_2),...,S(n,a_1,...,a_n)) = -n * a_n since sum_{n > 0} a_n x^n = log[sum{n >= 0} S(n,a_1,...,a_n) x^n]. Conversely, S(n,-F(1,a_1),-F(2,a_1,a_2)/2,...,-F(n,a_1,...,a_n)/n) = a_n. - Tom Copeland, Sep 07 2016
See Corollary 3.1.3 on p. 38 of Ardila and Copeland's two MathOverflow links to relate the Faber polynomials, with arguments being the signed elementary symmetric polynomials, to the logarithm of determinants, traces of powers of an adjacency matrix, and number of walks on graphs. - Tom Copeland, Jan 02 2017
The umbral inverse polynomials IF appear on p. 19 of Konopelchenko as partial differential operators. - Tom Copeland, Nov 19 2018

Extensions

More terms from Jean-François Alcover, Jun 12 2017

A173958 Number A(n,k) of spanning trees in C_k X P_n; square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 2, 1, 3, 12, 1, 4, 75, 70, 1, 5, 384, 1728, 408, 1, 6, 1805, 31500, 39675, 2378, 1, 7, 8100, 508805, 2558976, 910803, 13860, 1, 8, 35287, 7741440, 140503005, 207746836, 20908800, 80782, 1, 9, 150528, 113742727, 7138643400, 38720000000, 16864848000, 479991603, 470832, 1
Offset: 1

Views

Author

Alois P. Heinz, Nov 26 2010

Keywords

Comments

Every row and every column of the array is a divisibility sequence, i.e., the terms satisfy the property that if n divides m then a(n) divides a(m) provided a(n) > 0. This follows from the representation of the elements of the array as a resultant. - Peter Bala, May 01 2014

Examples

			Square array A(n,k) begins:
  1,    2,      3,         4,           5,  ...
  1,   12,     75,       384,        1805,  ...
  1,   70,   1728,     31500,      508805,  ...
  1,  408,  39675,   2558976,   140503005,  ...
  1, 2378, 910803, 207746836, 38720000000,  ...
		

Crossrefs

Rows n=1-2 give: A000027, A006235.
Main diagonal gives A252767.
Cf. A156308.

Programs

  • Maple
    with(LinearAlgebra):
    A:= proc(n, m) local M, i, j;
         if m=1 then 1 else
          M:= Matrix(n*m, shape=symmetric);
          for i to n do
            for j to m-1 do M[m*(i-1)+j, m*(i-1)+j+1]:=-1 od;
            M[m*(i-1)+1, m*i]:= M[m*(i-1)+1, m*i]-1
          od;
          for i to n-1 do
            for j to m do M[m*(i-1)+j, m*i+j]:=-1 od
          od;
          for i to n*m do
            M[i,i]:= -add(M[i,j], j=1..n*m)
          od;
          Determinant(DeleteColumn(DeleteRow(M, 1), 1))
         fi
        end:
    seq(seq(A(n, 1+d-n), n=1..d), d=1..9);
    # Crude Maple program from N. J. A. Sloane, May 27 2012:
    Digits:=200;
    T:=(m,n)->round(Re(evalf(simplify(expand(
    m*mul(mul( 4*sin(h*Pi/m)^2+4*sin(k*Pi/(2*n))^2, h=1..m-1), k=1..n-1))))));
    # Alternative program using the resultant:
    for n from 1 to 10 do seq(k*resultant(simplify((2*(ChebyshevT(k,(x + 2)/2) - 1))/x), simplify(ChebyshevU(n-1,1 - x/2)), x), k = 1 .. 10) end do; # Peter Bala, May 01 2014
  • Mathematica
    t[m_, n_] := m*Product[Product[4*Sin[h*Pi/m]^2 + 4*Sin[k*Pi/(2*n)]^2, {h, 1, m-1}], {k, 1, n-1}]; Table[t[m, n-m+1] // Round, {n, 1, 9}, {m, n, 1, -1}] // Flatten (* Jean-François Alcover, Dec 05 2013, after N. J. A. Sloane *)

Formula

A(n,k) = m*Prod(Prod( 4*sin(h*Pi/m)^2+4*sin(k*Pi/(2*n))^2, h=1..m-1), k=1..n-1) [Kreweras]. - From N. J. A. Sloane, May 27 2012
Let T(n,x) and U(n,x) denote the Chebyshev polynomials of the first and second kind respectively. Let R(n,x) = 2*( T(n,(x + 2)/2) - 1 )/x (the row polynomials of A156308). Then the (n,k)-th element of the array equals k times the resultant (R(k,x), U(n-1,(2 - x)/2)). - Peter Bala, May 01 2014 [Corrected by Pontus von Brömssen, Apr 08 2025]

A132460 Irregular triangle read by rows of the initial floor(n/2) + 1 coefficients of 1/C(x)^n, where C(x) is the g.f. of the Catalan sequence (A000108).

Original entry on oeis.org

1, 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
Offset: 0

Views

Author

Paul D. Hanna, Aug 21 2007

Keywords

Comments

The length of row n is A008619(n).
Essentially equals a signed version of A034807, the triangle of Lucas polynomials. The initial n coefficients of 1/C(x)^n consist of row n followed by floor((n-1)/2) zeros for n > 0.
For the following formula for 1/C(x)^n see the W. Lang reference, proposition 1 on p. 411:
1/C(x)^n = (sqrt(x))^n*(S(n,1/sqrt(x)) - sqrt(x)*S(n-1,1/sqrt(x))*C(x)), n >= 0, with the Chebyshev polynomials S(n,x) with coefficients given in A049310. See also the coefficient array A115139 for P(n,x) = (sqrt(x)^(n-1))*S(n-1, 1/sqrt(x)). - Wolfdieter Lang, Sep 14 2013
This triangular array is composed of interleaved rows of reversed, A127677 (cf. A156308, A217476, A263916) and reversed, signed A111125. - Tom Copeland, Nov 07 2015
It seems that the n-th row lists the coefficients of the HOMFLYPT (HOMFLY) polynomial reduced to one variable for link family n, see Jablan's slide 38. - Andrey Zabolotskiy, Jan 16 2018
For n >= 1 row n gives the coefficients of the Girard-Waring formula for the sum of x1^n + x2^n in terms of the elementary symmetric functions e_1(x1,x2) = x1 + x2 and e_2(x1,x2) = x1*x2. This is an array using the partitions of n, in the reverse Abramowitz-Stegun order, with all partitions with parts larger than 2 eliminated. E.g., n = 4: x1^4 + x2^4 = 1*e1^4 - 4*e1^3*e2 + 2*e1*e2^2. See also A115131, row n = 4, with the mentioned partitions omitted. - Wolfdieter Lang, May 03 2019
Row n lists the coefficients of the n-th Faber polynomial for the replicable function given in A154272 with offset -1. - Ben Toomey, May 12 2020

Examples

			The irregular triangle T(n,k) begins:
n\k 0    1    2    3    4    5    6   7 ...
-------------------------------------------------
0:  1
1:  1
2:  1   -2
3:  1   -3
4:  1   -4    2
5:  1   -5    5
6:  1   -6    9   -2
7:  1   -7   14   -7
8:  1   -8   20  -16    2
9:  1   -9   27  -30    9
10: 1  -10   35  -50   25   -2
11: 1  -11   44  -77   55  -11
12: 1  -12   54 -112  105  -36    2
13: 1  -13   65 -156  182  -91   13
14: 1  -14   77 -210  294 -196   49  -2
... (reformatted - _Wolfdieter Lang_, May 03 2019)
		

Crossrefs

Cf. A000108, A008619, A034807 (Lucas polynomials), A111125, A115131 (Waring numbers), A127677, A132461 (row squared sums), A156308, A217476, A263916.

Programs

  • Mathematica
    T[0, 0] = 1; T[n_, k_] := (-1)^k (Binomial[n-k, k] + Binomial[n-k-1, k-1]);
    Table[T[n, k], {n, 0, 14}, {k, 0, n/2}] // Flatten (* Jean-François Alcover, Jun 04 2018 *)
  • PARI
    {T(n,k)=if(k>n\2,0,(-1)^k*(binomial(n-k, k)+binomial(n-k-1, k-1)))}

Formula

T(n,k) = (-1)^k*( C(n-k,k) + C(n-k-1,k-1) ) for n >= 0, 0 <= k <= floor(n/2).
T(0,0) = 1; T(n,k) = (-1)^k*n*binomial(n-k,k)/(n-k), k = 0..floor(n/2). - Wolfdieter Lang, May 03 2019

A208513 Triangle of coefficients of polynomials u(n,x) jointly generated with A111125; see the Formula section.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 6, 1, 1, 16, 20, 8, 1, 1, 25, 50, 35, 10, 1, 1, 36, 105, 112, 54, 12, 1, 1, 49, 196, 294, 210, 77, 14, 1, 1, 64, 336, 672, 660, 352, 104, 16, 1, 1, 81, 540, 1386, 1782, 1287, 546, 135, 18, 1, 1, 100, 825, 2640, 4290, 4004, 2275, 800, 170, 20, 1
Offset: 1

Views

Author

Clark Kimberling, Feb 28 2012

Keywords

Comments

The columns of A208513 are identical to those of A208509. Here, however, the alternating row sums are periodic (with period 1,0,-2,-3,-2,0).
From Tom Copeland, Nov 07 2015: (Start)
These polynomials may be expressed in terms of the Faber polynomials of A263916, similar to A127677.
Rephrasing notes in A111125: Append an initial column of zeros except for a 1 at the top to A111125. Then the rows of this entry contain the partial sums of the column sequences of modified A111125; therefore, the difference of consecutive pairs of rows of this entry, modified by appending an initial row of zeros to it, generates the modified A111125. (End)

Examples

			First five rows:
  1;
  1,  1;
  1,  4,  1;
  1,  9,  6, 1;
  1, 16, 20, 8, 1;
First five polynomials u(n,x):
  u(1,x) = 1;
  u(2,x) = 1 +    x;
  u(3,x) = 1 +  4*x +    x^2;
  u(4,x) = 1 +  9*x +  6*x^2 +   x^3;
  u(5,x) = 1 + 16*x + 20*x^2 + 8*x^3 + x^4;
		

Crossrefs

Programs

  • Magma
    A208513:= func< n,k | k eq 1 select 1 else (2*(n-1)/(n+k-2))*Binomial(n+k-2, 2*k-2) >;
    [A208513(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 02 2022
    
  • Mathematica
    (* First program *)
    u[1, x_]:=1; v[1, x_]:=1; z=16;
    u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
    v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x] + 1;
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n,z}];
    TableForm[cu]
    Flatten[%]  (* A208513 *)
    Table[Expand[v[n, x]], {n,z}]
    cv = Table[CoefficientList[v[n, x], x], {n,z}];
    TableForm[cv]
    Flatten[%]  (* A111125 *)
    (* Second program *)
    T[n_, k_]:= If[k==1, 1, ((n-1)/(k-1))*Binomial[n+k-3, 2*k-3]];
    Table[T[n, k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Feb 02 2022 *)
  • Sage
    def A208513(n,k): return 1 if (k==1) else ((n-1)/(k-1))*binomial(n+k-3, 2*k-3)
    flatten([[A208513(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 02 2022

Formula

Coefficients of u(n, x) from the mixed recurrence relations:
u(n,x) = u(n-1,x) + x*v(n-1,x),
v(n,x) = u(n-1,x) + (x+1)*v(n-1,x) + 1,
where u(1,x) = 1, u(2,x) = 1+x, v(1,x) = 1, v(2,x) = 3+x.
From Peter Bala, May 01 2012: (Start)
Working with an offset of 0: T(n,0) = 1; T(n,k) = (n/k)*binomial(n+k-1,2*k-1) = (n/k)*A078812(n,k) for k > 0. Cf. A156308.
O.g.f.: ((1-t)^2 + t^2*x)/((1-t)*((1-t)^2-t*x)) = 1 + (1+x)*t + (1+4*x+x^2)*t^2 + ....
u(n+1,x) = -1 + (b(2*n,x) + 1)/b(n,x), where b(n,x) = Sum_{k = 0..n} binomial(n+k, 2*k)*x^k are the Morgan-Voyce polynomials of A085478.
This triangle is formed from the even numbered rows of A211956 with a factor of 2^(k-1) removed from the k-th column entries.
(End)
T(n, k) = (2*(n-1)/(n+k-2))*binomial(n+k-2, 2*k-2). - G. C. Greubel, Feb 02 2022

A084534 Triangle read by rows: row #n has n+1 terms. T(n,0)=1, T(n,n)=2, T(n,m) = T(n-1,m-1) + Sum_{k=0..m} T(n-1-k,m-k).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 6, 9, 2, 1, 8, 20, 16, 2, 1, 10, 35, 50, 25, 2, 1, 12, 54, 112, 105, 36, 2, 1, 14, 77, 210, 294, 196, 49, 2, 1, 16, 104, 352, 660, 672, 336, 64, 2, 1, 18, 135, 546, 1287, 1782, 1386, 540, 81, 2, 1, 20, 170, 800, 2275, 4004, 4290, 2640, 825, 100, 2
Offset: 0

Views

Author

Gary W. Adamson, May 29 2003

Keywords

Comments

Sum of row #n = A000204(2n). (But sum of row #0 = 1.)
Row #n has the unsigned coefficients of the monic polynomial whose roots are 2 cos(Pi*(2k-1)/(4n)) for k=1..2n. [Comment corrected by Barry Brent, Jan 03 2006]
The positive roots are some diagonal lengths of a regular (4n)-gon, inscribed in the unit circle.
Polynomial of row #n = Sum_{m=0..n} (-1)^m * T(n,m) x^(2*n-2*m).
This is the unsigned version of the coefficient table for scaled Chebyshev T(2*n,x) polynomials. - Wolfdieter Lang, Mar 07 2007
Reversed A127677 (cf. A156308, A217476, A263916). - Tom Copeland, Nov 07 2015

Examples

			First few Chebyshev T(2*n,x) polynomials:
  T(2*0,x) = 1;
  T(2*1,x) = x^2 -   2;
  T(2*2,x) = x^4 -   4*x^2 +  2;
  T(2*3,x) = x^6 -   6*x^4 +  9*x^2 -  2;
  T(2*4,x) = x^8 -   8*x^6 + 20*x^4 - 16*x^2 +  2;
  T(2*5,x) = x^10 - 10*x^8 + 35*x^6 - 50*x^4 + 25*x^2 - 2;
Triangle begins as:
  1;
  1,  2;
  1,  4,  2;
  1,  6,  9,   2;
  1,  8, 20,  16,   2;
  1, 10, 35,  50,  25,  2;
  1, 12, 54, 112, 105, 36, 2;
		

References

  • I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 118.
  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990. p. 37, eq.(1.96) and p. 4. eq.(1.10).

Crossrefs

Row sums are A005248 for n > 0.
Companion triangle A082985.
Cf. A082985 (unsigned scaled coefficient table for Chebyshev's T(2*n+1, x) polynomials).

Programs

  • Magma
    A084534:= func< n,k | k eq 0 select 1 else 2*(n/k)*Binomial(2*n-k-1, k-1) >;
    [A084534(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 02 2022
    
  • Maple
    T := proc(n, m): if n=0 then 1 else binomial(2*n-m, m)*2*n/(2*n-m) fi: end: seq(seq(T(n,m),m=0..n),n=0..10); # Johannes W. Meijer, May 31 2018
  • Mathematica
    a[n_, m_] := Binomial[2n-m, m]*2n/(2n-m); a[0, 0] = 1; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Apr 12 2016, after Wolfdieter Lang *)
  • PARI
    T(n,m) = if(n==0, m==0, binomial(2*n-m, m)*2*n/(2*n-m)) \\ Andrew Howroyd, Dec 18 2017
    
  • Sage
    def A084534(n,k): return 1 if (k==0) else 2*(n/k)*binomial(2*n-k-1, k-1)
    flatten([[A084534(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 02 2022

Formula

T(n,m) = binomial(2*n-m, m)*2*n/(2*n-m) for n > 0. - Andrew Howroyd, Dec 18 2017
Signed version from Wolfdieter Lang, Mar 07 2007: (Start)
a(n,m)=0 if n
a(n,m)=0 if n
a(n,m)=0 if nA127674(n,n-m)/2^(2*(n-m)-1) (scaled coefficients of Chebyshev's T(2*n,x), decreasing even powers). [corrected by Johannes W. Meijer, May 31 2018] (End)

Extensions

Edited by Don Reble, Nov 12 2005

A152060 Triangle read by rows, characteristic polynomials of Cartan ring matrices.

Original entry on oeis.org

1, 1, -2, 1, -4, 3, 1, -6, 9, -4, 1, -8, 20, -16, 4, 1, -10, 35, -50, 25, -4, 1, -12, 54, -112, 105, -36, 4, 1, -14, 77, -210, 294, -196, 49, -4, 1, -16, 104, -352, 660, -672, 336, -64, 4, 1, -18, 135, -546, 1287, -1782, 1386, -540, 81, -4, 1, -20, 170, -800, 2275, -4004, 4290, -2640, 825, -100, 4
Offset: 0

Author

Keywords

Examples

			Triangle begins:
1;
1, -2;
1, -4, 3;
1, -6, 9, -4;
1, -8, 20, -16, 4;
1, -10, 35, -50, 25, -4;
1, -12, 54, -112, 105, -36, 4;
1, -14, 77, -210, 294, -196, 49, -4;
1, -16, 104, -352, 660, -672, 336, -64, 4;
1, -18, 135, -546, 1287, -1782, 1386, -540, 81, -4;
1, -20, 170, -800, 2275, -4004, 4290, -2640, 825, -100, 4;
...
Example: x^5 -10x^4 + 35x^3 -50x^2 + 25x - 4 = (x - 4) * (x^2 - 3x + 1)^2 is the characteristic polynomial of the matrix
[ 2,-1, 0, 0, 1]
[-1, 2,-1, 0, 0]
[ 0,-1, 2,-1, 0]
[ 0, 0,-1, 2,-1]
[ 1, 0, 0,-1, 2].
		

References

  • William G. Harter, University of Arkansas; personal communication

Crossrefs

Programs

  • Mathematica
    M[n_] := SparseArray[{Band[{1, 1}] -> 2, Band[{1, 2}] -> -1, Band[{2, 1}] -> -1, {1, n} -> 1, {n, 1} -> 1}, {n, n}];
    row[0] = {1}; row[1] = {1, -2};
    row[n_] := (-1)^n CharacteristicPolynomial[M[n], x] // CoefficientList[#, x]& // Reverse;
    Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Aug 08 2018 *)

Formula

Triangle read by rows, n-th row = characteristic polynomial of n X n Cartan ring matrix, defined as a Cartan matrix with 1's in the upper right and lower left corners, i.e., positions (1,n) and (n,1).
The coefficients of characteristic polynomials of matrices C_n, defined by
C_n=
(2 -1 0 ... 0 1)
(-1 2 -1 0 ... 0)
(0 -1 2 -1 0 ... 0)
...
(0 ... 0 -1 2 -1)
(1 0 ... 0 -1 2),
give the same triangle T(n,k), for n>0, k=0,...,n, with T(0,0)=1. - L. Edson Jeffery, Mar 27 2011
It appears that for n >= 3 the n-th row polynomial equals 2*T(2*n,sqrt(x)/2) + 2*(-1)^n, where T(n,x) denotes the Chebyshev polynomial of the first kind (A008310). Checked for n = 3 through n = 12. - Peter Bala, May 04 2014
Apparently, omitting the diagonal here, this triangular array is signed, reversed A156308 (cf. A127677, A217476, A263916). 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). - Tom Copeland, Nov 07 2015

Extensions

Edited by L. Edson Jeffery, Mar 26 2011
Some terms corrected from Peter Bala, May 04 2014

A156290 Triangle read by rows: alternating binomial coefficients with signs.

Original entry on oeis.org

1, -4, 1, 15, -6, 1, -56, 28, -8, 1, 210, -120, 45, -10, 1, -792, 495, -220, 66, -12, 1, 3003, -2002, 1001, -364, 91, -14, 1, -11440, 8008, -4368, 1820, -560, 120, -16, 1, 43758, -31824, 18564, -8568, 3060, -816, 153, -18, 1, -167960, 125970, -77520
Offset: 1

Author

Hartmut F. W. Hoft, Feb 07 2009

Keywords

Comments

Alternating binomial coefficients in the closed form expression for sequence A156289.
The Example lines below show the connection with Pascal's triangle A007318.

Examples

			R(2,1)=-4, R(3,3)=1, R(4,2)=28.
Here is Pascal's triangle with the entries in the present triangle preceded by a *:
......................1
.....................1, 1
...................1, 2,*1
.................1, 3, 3, 1
................1, 4, 6,*4,*1
..............1, 5, 10, 10, 5, 1
............1, 6, 15, 20,*15,*6,*1
..........1, 7, 21, 35, 35, 21, 7, 1
........1, 8, 28, 56, 70,*56,*28,*8,*1
...
		

References

  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Crossrefs

Coefficient factor in elements of sequence A156289, the inverse of lower triangular matrix A156308.
Cf. A007318.

Programs

  • Mathematica
    R[m_] := Flatten[Table[(-1)^(k + j) Binomial[2 k, k + j], {k, 1, m}, {j, 1, k}]]

Formula

R(k,j)=(-1)^(k+j)*Binomial(2k,k+j), for 1<= j<=k, and 0 otherwise.

Extensions

Edited by N. J. A. Sloane, Apr 05 2011

A220669 Coefficient array for powers of x^2 of the square of Chebyshev's C(2*n+1,x)/x =: tau(n,x) polynomials.

Original entry on oeis.org

1, 9, -6, 1, 25, -50, 35, -10, 1, 49, -196, 294, -210, 77, -14, 1, 81, -540, 1386, -1782, 1287, -546, 135, -18, 1, 121, -1210, 4719, -9438, 11011, -8008, 3740, -1122, 209, -22, 1, 169, -2366, 13013, -37180, 63206, -68952, 50388, -25194, 8645, -2002, 299, -26, 1
Offset: 0

Author

Wolfdieter Lang, Jan 04 2013

Keywords

Comments

The row lengths sequence of this irregular triangle is 2*n+1 = A005408(n).
For the coefficient array of powers of x^2 of the monic integer Chebyshev C(2*n+1,x)/x = :tau(n,x) polynomials see the signed triangle ((-1)^(n-m))*A111125(n,m). See the comment from Oct 23 2012.
The o.g.f. of the row polynomials p(n,x) := sum(a(n,m)*x^m, m=0..2*n), n>=0, is G2(x,z) = sum(p(n,x)*z^n,n=0..infinity) = (1+ (6-2*x)*z+z^2)/((1-z)*((z+1)^2-z*(x-2)^2)). Derived from the odd part of the bisection of the o.g.f. for the C(n,x)^2 polynomials. Note that p(n,x) = (tau(n,sqrt(x)))^2.

Examples

			The array begins:
n\m  0     1    2     3     4     5    6     7    8   9  10
0:   1
1:   9    -6    1
2:  25   -50   35   -10     1
3:  49  -196  294  -210    77   -14    1
4:  81  -540 1386 -1782  1287  -546  135   -18    1
5: 121 -1210 4719 -9438 11011 -8008 3740 -1122  209 -22   1
...
Row polynomial for n=1: p(1,x) = (tau(1,sqrt(x)))^2 = (3-1*x)^2 = 9 - 6*x +1*x^2.
Row polynomial for n=2: p(2,x) = (tau(2,sqrt(x)))^2 = (5 - 5*x + 1*x^2)^2 = 25 - 50*x + 35*x^2 - 10*x^3 + 1*x^4.
		

Crossrefs

Odd rows of A156308 with alternating signs of elements.

Formula

a(n,m) = [x^m] (p(n,x)), n>=0, 0 <= m <= 2*n, with p(n,x) = (C(2*n+1,sqrt(x))/sqrt(x))^2 = (tau(n,sqrt(x)))^2, For Chebyshev's C and tau polynomials see a comment above.
For n >= 0, 0 <= m <= 2*n, a(n,m) = [x^m*z^n] G2(x,z), where the o.g.f. G2(x,z) given in a comment above.
a(n,m) = (-1)^m * A156308(2*n+1,m+1). - Max Alekseyev, Mar 06 2018
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