cp's OEIS Frontend

For the definition of Q-residue, see A193649.

The row polynomials R(n,x) := Sum_{m=0..n} a(n,m)*x^m have been called Chebyshev C_n(x) polynomials in the Abramowitz-Stegun handbook, p. 778, 22.5.11 (see A049310 for the reference, and note that on p. 774 the S and C polynomials have been mixed up in older printings). - Wolfdieter Lang, Jun 03 2011

This is a signed version of triangle A114525.

The unsigned column sequences (without zeros) are, for m=1..11: A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.

The row polynomials R(n,x) := Sum_{m=0..n} a(n,m)*x*m, give for n=2,3,...,floor(N/2) the positive zeros of the Chebyshev S(N-1,x)-polynomial (see A049310) in terms of its largest zero rho(N):= 2*cos(Pi/N) by putting x=rho(N). The order of the positive zeros is falling: n=1 corresponds to the largest zero rho(N) and n=floor(N/2) to the smallest positive zero. Example N=5: rho(5)=phi (golden section), R(2,phi)= phi^2-2 = phi-1, the second largest (and smallest) positive zero of S(4,x). - Wolfdieter Lang, Dec 01 2010

The row polynomial R(n,x), for n >= 1, factorizes into minimal polynomials of 2*cos(Pi/k), called C(k,x), with coefficients given in A187360, as follows.

R(n,x) = Product_{d|oddpart(n)} C(2*n/d,x)

= Product_{d|oddpart(n)} C(2^(k+1)*d,x),

with oddpart(n)=A000265(n), and 2^k is the largest power of 2 dividing n, where k=0,1,2,...

(Proof: R and C are monic, the degree on both sides coincides, and the zeros of R(n,x) appear all on the r.h.s.) - Wolfdieter Lang, Jul 31 2011 [Theorem 1B, eq. (43) in the W. Lang link. - Wolfdieter Lang, Apr 13 2018]

The zeros of the row polynomials R(n,x) are 2*cos(Pi*(2*k+1)/(2*n)), k=0,1, ..., n-1; n>=1 (from those of the Chebyshev T-polynomials). - Wolfdieter Lang, Sep 17 2011

The discriminants of the row polynomials R(n,x) are found under A193678. - Wolfdieter Lang, Aug 27 2011

The determinant of the N X N matrix M(N) with entries M(N;n,m) = R(m-1,x[n]), 1 <= n,m <= N, N>=1, and any x[n], is identical with twice the Vandermondian Det(V(N)) with matrix entries V(N;n,m) = x[n]^(m-1). This is an instance of the general theorem given in the Vein-Dale reference on p. 59. Note that R(0,x) = 2 (not 1). See also the comments from Aug 26 2013 under A049310 and from Aug 27 2013 under A000178. - Wolfdieter Lang, Aug 27 2013

This triangle a(n,m) is also used to express in the regular (2*(n+1))-gon, inscribed in a circle of radius R, the length ratio side/R, called s(2*(n+1)), as a polynomial in rho(2*(n+1)), the length ratio (smallest diagonal)/side. See the bisections ((-1)^(k-s))*A111125(k,s) and A127677 for comments and examples. - Wolfdieter Lang, Oct 05 2013

From Tom Copeland, Nov 08 2015: (Start)

These are the characteristic polynomials a_n(x) = 2*T_n(x/2) for the adjacency matrix of the Coxeter simple Lie algebra B_n, related to the Cheybshev polynomials of the first kind, T_n(x) = cos(n*q) with x = cos(q) (see p. 20 of Damianou). Given the polynomial (x - t)*(x - 1/t) = 1 - (t + 1/t)*x + x^2 = e2 - e1*x + x^2, the symmetric power sums p_n(t,1/t) = t^n + t^(-n) of the zeros of this polynomial may be expressed in terms of the elementary symmetric polynomials e1 = t + 1/t = y and e2 = t*1/t = 1 as p_n(t,1/t) = a_n(y) = F(n,-y,1,0,0,...), where F(n,b1,b2,...,bn) are the Faber polynomials of A263916.

The partial sum of the first n+1 rows given t and y = t + 1/t is PS(n,t) = Sum_{k=0..n} a_n(y) = (t^(n/2) + t^(-n/2))*(t^((n+1)/2) - t^(-(n+1)/2)) / (t^(1/2) - t^(-1/2)). (For n prime, this is related simply to the cyclotomic polynomials.)

Then a_n(y) = PS(n,t) - PS(n-1,t), and for t = e^(iq), y = 2*cos(q), and, therefore, a_n(2*cos(q)) = PS(n,e^(iq)) - PS(n-1,e^(iq)) = 2*cos(nq) = 2*T_n(cos(q)) with PS(n,e^(iq)) = 2*cos(nq/2)*sin((n+1)q/2) / sin(q/2).

(End)

R(45, x) is the famous polynomial used by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593 to pose four problems, solved by Viète. See, e.g., the Havil reference, pp. 69-74. - Wolfdieter Lang, Apr 28 2018

From Wolfdieter Lang, May 05 2018: (Start)

Some identities for the row polynomials R(n, x) following from the known ones for Chebyshev T-polynomials (A053120) are:

(1) R(-n, x) = R(n, x).

(2) R(n*m, x) = R(n, R(m, x)) = R(m, R(n, x)).

(3) R(2*k+1, x) = (-1)^k*x*S(2*k, sqrt(4-x^2)), k >= 0, with the S row polynomials of A049310.

(4) R(2*k, x) = R(k, x^2-2), k >= 0.

(End)

For y = z^n + z^(-n) and x = z + z^(-1), Hirzebruch notes that y(z) = R(n,x) for the row polynomial of this entry. - Tom Copeland, Nov 09 2019

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)

This is the root congruent to 2 mod 5.

Or, residues modulo 5^n of a 5-adic solution of x^2+1=0.

The radix-5 expansion of a(n) is obtained from the n rightmost digits in the expansion of the following pentadic integer:

...422331102414131141421404340423140223032431212 = u

The residues modulo 5^n of the other 5-adic solution of x^2+1=0 are given by A048899 which corresponds to the pentadic integer -u:

...022113342030313303023040104021304221412013233 = -u

The digits of u and -u are given in A210850 and A210851, respectively. - Wolfdieter Lang, May 02 2012

For approximations for p-adic square roots see also the W. Lang link under A268922. - Wolfdieter Lang, Apr 03 2016

From Jianing Song, Sep 06 2022: (Start)

For n > 0, a(n)-1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 1 modulo 5.

For n > 0, a(n)+1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 3 modulo 5. (End)

This is the root congruent to 3 (mod 5) for n>0.

The other case with the 2 (mod 5) numbers (except for n=0) is given in A048898. - Wolfdieter Lang, Feb 19 2016

From Jianing Song, Sep 06 2022: (Start)

For n > 0, a(n)-1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 2 modulo 5.

For n > 0, a(n)+1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 4 modulo 5. (End)

See A048898 for the successive approximations to this 5-adic integer, called there u.

The digits of -u, the other 5-adic integer sqrt(-1), are given in A210851.

a(n) is the unique solution of the linear congruence 2*A048898(n)*a(n) + A210848(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A210848, eq. (6) on p. 86 adapted to this case. a(0)=2 follows from the formula given below.

If n>0, a(n) == A210848(n) (mod 5), since A048898(n) == 2 (mod 5). - Álvar Ibeas, Feb 21 2017

If a(n)=0 then A048899(n+1) and A048899(n) coincide.

a(n) + A210851(n) = 4 for n >= 1. - Robert Israel, Mar 04 2016

From Jianing Song, Sep 06 2022: (Start)

With a(0) = 1, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 1 modulo 5.

With a(0) = 3, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 3 modulo 5. (End)

This square root of -1 in the 5-adic integers is equal to the 5-adic limit of the sequence {L(5^n,2)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 02 2022

See A048899 for the successive approximations to this 5-adic integer, called -u in a comment on A048898.

The digits of u, the other 5-adic integer sqrt(-1), are given in A210850.

a(n) is the (unique) solution of the linear congruence 2*A048899(n)*a(n) + A210849(n) == 0 (mod 5), n >= 1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A210848, eq. (6) on p. 86 adapted to this case. a(0)=3 follows from the formula given below.

If n>0, a(n) == -(A210849(n)) (mod 5), since A048899(n) == 3 (mod 5). - Álvar Ibeas, Feb 21 2017

If a(n)=0 then A048899(n+1) and A048899(n) coincide.

From Jianing Song, Sep 06 2022: (Start)

With a(0) = 2, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 2 modulo 5.

With a(0) = 4, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 4 modulo 5. (End)

This square root of -1 in the 5-adic integers is equal to the 5-adic limit of the sequence {L(5^n,3)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 02 2022

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.