cp's OEIS Frontend

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020

Row sums give the Fibonacci sequence. So do the alternating row sums.

Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]

Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009

The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011

Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011

From Gary W. Adamson, Aug 25 2019: (Start)

Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.

For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.

Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:

3...............1

5...............1, 2

7...............1, 2, 3

9...............1, 2, 4

...............3 (singleton terms reduce to "1") (9 has two rows)

11...............1, 2, 4, 3, 5

13...............1, 2, 4, 5, 3, 6

15...............1, 2, 4, 7

................3, 6 (dividing through by the gcd gives (1, 2))

................5. (singleton terms reduce to "1")

The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)

From Gary W. Adamson, Sep 30 2019: (Start)

The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)

From Gary W. Adamson, Aug 21 2019: (Start)

Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1, 0]; the f(x) trajectory is:

....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,

....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)

Received from Herb Conn, Jan 2004: (Start)

Let x = 2*cos(2A) (A = Angle); then

sin(A)/sin A = 1

sin(3A)/sin A = x + 1

sin(5A)/sin A = x^2 + x - 1

sin(7A)/sin A = x^3 + x - 2x - 1

sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1

... (signed ++--++...). (End)

Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012

The matrix inverse is given by

1;

1, 1;

0, -1, 1;

0, 1, -2, 1;

0, 0, 1, -2, 1;

0, 0, -1, 3, -3, 1;

0, 0, 0, -1, 3, -3, 1;

0, 0, 0, 1, -4, 6, -4, 1;

0, 0, 0, 0, 1, -4, 6, -4, 1;

... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004

Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005

As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008

From Emeric Deutsch, Jun 18 2010: (Start)

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)

From L. Edson Jeffery, Mar 01 2011: (Start)

This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table

1

1

1 1

1 1

1 2 1

1 2 1

1 3 3 1

1 3 3 1

...,

then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with

1;

1, -1;

1, -1, -1;

1, -2, -1, 1;

1, -2, -3, 1, 1;

...

(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.

For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)

Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011

Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012

Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014

From Jeremy Dover, Jun 07 2016: (Start)

T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.

T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

Matrix inverse of A124645.

Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:

L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));

Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);

Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);

abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);

T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;

T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;

T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;

T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;

T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;

T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;

T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;

T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;

T(n,n) = L(n,0) = (-1)^floor((n+1)/2);

L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);

L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;

L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;

L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;

L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;

L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;

L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;

L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;

L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;

L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;

L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;

L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;

L(n,13) = A085260(n);

L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;

L(n,n) = A108366(n); L(n,-n) = A108367(n).

Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005

From L. Edson Jeffery, Mar 12 2011: (Start)

Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form

G_N=A_{N,1}=

(0 1 0 ... 0)

(1 0 1 0 ... 0)

(0 1 0 1 0 ... 0)

...

(0 ... 0 1 0 1)

(0 ... 0 1 1),

with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,

G_7=A_{7,1}=

(0 1 0)

(1 0 1)

(0 1 1).

We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)

The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011

The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

Equivalently, a triangle read by rows, where the rows are obtained by sorting the elements of the rows of Pascal's triangle (A007318) into descending order. - Philippe Deléham, May 21 2005

Equivalently, as a triangle read by rows, this is T(n,k)=binomial(n,floor((n-k)/2)); column k then has e.g.f. Bessel_I(k,2x)+Bessel_I(k+1,2x). - Paul Barry, Feb 28 2006

Antidiagonal sums are A037952(n+1) = C(n+1,[n/2]). Matrix inverse is the row reversal of triangle A066170. Eigensequence is A125094(n) = Sum_{k=0..n-1} A125093(n-1,k)*A125094(k). - Paul D. Hanna, Nov 20 2006

Riordan array (1/(1-x-x^2*c(x^2)),x*c(x^2)); where c(x)=g.f.for Catalan numbers A000108. - Philippe Deléham, Mar 17 2007

Triangle T(n,k), 0<=k<=n, read by rows given by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 27 2007

This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007

T(n,k) is the number of paths from (0,k) to some (n,m) which never dip below y=0, touch y=0 at least once and are made up only of the steps (1,1) and (1,-1). This can be proved using the recurrence supplied by Deléham. - Gerald McGarvey, Oct 15 2008

Triangle read by rows = partial sums of A053121 terms starting from the right. - Gary W. Adamson, Oct 24 2008

As a subset of the "family of triangles" (Deleham comment of Sep 25 2007), beginning with a signed variant of A061554, M = (-1,0) = (1; -1, 1; 2, -1, 1; -3, 3, -1, 1; ...) successive binomial transforms of M yield (0,1) - A089942; (1,2) - A039599; (2,3) - A124733; (3,4) - A124574; (4,5) - A126331; ... such that the binomial transform of the triangle generated from (n,n+1) = the triangle generated from (n+1,n+2). Similarly, another subset beginning with A053121 - (0,0), and taking successive binomial transforms yields (1,1) - A064189; (2,2) - A039598; (3,3) - A091965, ... By rows, the triangle generated from (n,n) can be obtained by taking pairwise sums from the (n-1,n) triangle starting from the right. For example, row 2 of (1,2) - A039599 = (2, 3, 1); and taking pairwise sums from the right we obtain (5, 4, 1) = row 2 of (2,2) - A039598. - Gary W. Adamson, Aug 04 2011

The triangle by rows (n) with alternating signs (+-+...) from the top as a set of simultaneous equations solves for diagonal lengths of odd N (N = 2n+1) regular polygons. The constants in each case are powers of c = 2*cos(2*Pi/N). By way of example, the first 3 rows relate to the heptagon and the simultaneous equations are (1,0,0) = 1; (-1,1,0) = c = 1.24697...; and (2,-1,1) = c^2. The answers are 1, 2.24697..., and 1.801...; the 3 distinct diagonal lengths of the heptagon with edge = 1. - Gary W. Adamson, Sep 07 2011

From L. Edson Jeffery, Mar 22 2011: (Start)

Let A be the unit-primitive matrix (see [Jeffery])

A=A_(7,2)=

(0 0 1)

(0 1 1)

(1 1 1).

Let B={b(n)} be this sequence shifted to the right one place and setting b(0)=3. Then B=(3,2,6,11,26,...) with generating function (3-4*x-x^2)/(1-2*x-x^2+x^3) and b(n)=Trace(A^n). (End)

The following identity hold true (a(n)^2 - a(2n+2))/2 = A094648(n+1) = (-1)^(n+1)*A096975(n+1) - for the proof see Witula et al.'s papers - Roman Witula, Jul 25 2012

We note that the joined sequences (-1)^(n+1)*a(n) and A094648(n) form a two-sided sequence defined either by the recurrence formula x(n+3) + x(n+2) - 2x(n+1) - x(n) = 0, n in Z, x(0)=3, x(-1)=-2, x(1)=-1, or by the following trigonometric identities: x(n) = (c(1))^n + (c(2))^n + (c(4))^n = (c(1)c(2))^(-n) + (c(1)c(4))^(-n) + (c(2)c(4))^(-n) = (s(2)/s(1))^n + (s(4)/s(2))^n + (s(1)/s(4))^n, for n in Z, where c(j) := 2*cos(2Pi*j/7) and s(j) := sin(2*Pi*j/7) - for the proof see Witula's and Witula et al.'s papers. - Roman Witula, Jul 25 2012

We have 4*a(n+2) - a(n) = 7*A077998(n+2). - Roman Witula, Aug 13 2012

Two very intriguing identities of trigonometric nature hold: (-1)^n*(a(n)-a(n-1)) = c(1)*c(2)^(-n) + c(2)*c(4)^(-n) + c(4)*c(1)^(-n), and (-1)^(n+1)*(a(n-1)-a(n+1)) = c(1)*c(4)^(-n-1) + c(2)*c(1)^(-n-1) + c(4)*c(2)^(-n-1), where a(-1):=3 and c(j) is defined as above. For the proof see Remark 6 in the first Witula's paper. - Roman Witula, Aug 14 2012

With respect to the form of the trigonometric formulas describing a(n), we call this sequence the Berndt-type sequence number 20 for the argument 2Pi/7. The A-numbers of other Berndt-type sequences numbers are given in below. - Roman Witula, Sep 30 2012

Inverse of triangle in A061554.

Signed version of A046854.

From Paul Barry, May 21 2009: (Start)

Riordan array ((1-x)/(1+x^2),x/(1+x^2)).

This triangle is the coefficient triangle for the Hankel transforms of the family of generalized Catalan numbers that satisfy a(n;r)=r*a(n-1;r)+sum{k=1..n-2, a(k)*a(n-1-k;r)}, a(0;r)=a(1;r)=1. The Hankel transform of a(n;r) is h(n)=sum{k=0..n, T(n,k)*r^k} with g.f. (1-x)/(1-r*x+x^2). These sequences include A086246, A000108, A002212. (End)

From Wolfdieter Lang, Jun 11 2011: (Start)

The Riordan array ((1+x)/(1+x^2),x/(1+x^2)) with entries Phat(n,k)= ((-1)^(n-k))*T(n,k) and o.g.f. Phat(x,z)=(1+z)/(1-x*z+z^2) for the row polynomials Phat(n,x) is related to Chebyshev C and S polynomials as follows.

Phat(n,x) = (R(n+1,x)-R(n,x))/(x+2) = S(2*n,sqrt(2+x))

with R(n,x)=C_n(x) in the Abramowitz and Stegun notation, p. 778, 22.5.11. See A049310 for the S polynomials. Proof from the o.g.f.s.

Recurrence for the row polynomials Phat(n,x):

Phat(n,x) = x*Phat(n-1,x) - Phat(n-2,x) for n>=1; Phat(-1,x)=-1, Phat(0,x)=1.

The A-sequence for this Riordan array Phat (see the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices) is given by 1, 0, -1, 0, -1, 0, -2, 0, -5,.., starting with 1 and interlacing the negated A000108 with zeros (o.g.f. 1/c(x^2) = 1-c(x^2)*x^2, with the o.g.f. c(x) of A000108).

The Z-sequence has o.g.f. sqrt((1-2*x)/(1+2*x)), and it is given by A063886(n)*(-1)^n.

The A-sequence of the Riordan array T(n,k) is identical with the one for the Riordan array Phat, and the Z-sequence is -A063886(n).

(End)

The row polynomials P(n,x) are the characteristic polynomials of the adjacency matrices of the graphs which look like P_n (n vertices (nodes), n-1 lines (edges)), but vertex no. 1 has a loop. - Wolfdieter Lang, Nov 17 2011

From Wolfdieter Lang, Dec 14 2013: (Start)

The zeros of P(n,x) are x(n,j) = -2*cos(2*Pi*j/(2*n+1)), j=1..n. From P(n,x) = (-1)^n*S(2*n,sqrt(2-x)) (see, e.g., the Lemma 6 of the W. Lang link).

The discriminants of the P-polynomials are given in A052750. (End)

Pairwise sums of A006356. Cf. A033303, A077850. - Ralf Stephan, Jul 06 2003

Number of (3412, P)-avoiding involutions in S_{n+1}, where P={1342, 1423, 2314, 3142, 2431, 4132, 3241, 4213, 21543, 32154, 43215, 15432, 53241, 52431, 42315, 15342, 54321}. - Ralf Stephan, Jul 06 2003

Number of 31- and 22-avoiding words of length n on alphabet {1,2,3} which do not end in 3 (e.g., at n=3, we have 111, 112, 121, 132, 211, 212, 232, 321 and 332). See A028859, A001519. - Jon Perry, Aug 04 2003

Form the graph with matrix A=[1, 1, 1; 1, 0, 0; 1, 0, 1]. Then the sequence 1,1,2,4,... with g.f. (1-x-x^2)/(1-2x-x^2+x^3) counts closed walks of length n at the degree 3 vertex. - Paul Barry, Oct 02 2004

a(n) is the number of Motzkin (n+1)-sequences whose flatsteps all occur at level <=1 and whose height is <=2. For example, a(5)=45 counts all 51 Motzkin 6-paths except FUUFDD, UFUFDD, UUFDDF, UUFDFD, UUFFDD, UUUDDD (the first five violate the flatstep restriction and the last violates the height restriction). - David Callan, Dec 09 2004

From Paul Barry, Nov 03 2010: (Start)

The g.f. of 1,1,2,4,9,... can be expressed as 1/(1-x/(1-x/(1-x^2))) and as 1/(1-x-x^2/(1-x-x^2)).

The second expression shows the link to the Motzkin numbers. (End)

From Emeric Deutsch, Oct 31 2010: (Start)

a(n) is the number of compositions of n into odd summands when we have two kinds of 1's. Proof: the g.f. of the set S={1,1',3,5,7,...} is g=2x+x^3/(1-x^2) and the g.f. of finite sequences of elements of S is 1/(1-g). Example: a(4)=20 because we have 1+3, 1'+3, 3+1, 3+1', and 2^4=16 of sums x+y+z+u, where x,y,z,u are taken from {1,1'}.

(End)

a(n-1) is the top left entry of the n-th power of any of the six 3 X 3 matrices [1, 1, 0; 1, 1, 1; 0, 1, 0] or [1, 1, 1; 0, 1, 1; 1, 1, 0] or [1, 0, 1; 1, 1, 1; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 0, 1, 1] or [1, 0, 1; 0, 0, 1; 1, 1, 1] or [1, 1, 0; 1, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

Conjecture: (i) Let n > 1 and N=2*n+1. Row n of T gives the coefficients of the characteristic polynomial p_N(x)=Sum_{k=0..n} T(n,k)*x^(n-k) of the n X n Danzer matrix D_{N,n-1} = {{0,...,0,1}, {0,...,0,1,1}, ..., {0,1,...,1}, {1,...,1}}. (ii) Let S_0(t)=1, S_1(t)=t and S_r(t)=t*S_(r-1)(t)-S_(r-2)(t), r > 1 (cf. A049310). Then p_N(x)=0 has solutions w_{N,j}=S_(n-1)(phi_{N,j}), where phi_{N,j}=2*(-1)^(j+1)*cos(j*Pi/N), j = 1..n. - L. Edson Jeffery, Dec 18 2011

The entry a(n,k), n >= 0, k = 0,1,...,g, where g=n, of this irregular triangle is the coefficient of (1+q^2)^k*q^(g-k) in the representation of the Gaussian polynomial [2n+1,1]q = Sum{k=0..g) a(n,k)*(1+q^2)^k*q^(g-k).

The sequence arises in the formal derivation of the stability polynomial B(x) = Sum_{i=0..N} d_i T(iM,x) of rank N, and degree L, where T(iM,x) denotes the Chebyshev polynomial of the first kind of degree iM. The coefficients d_i are determined by order conditions on the stability polynomial.

Conjecture: More generally, the Gaussian polynomial [2*n+m+1-(m mod 2),m]q = Sum{k=0..g(m;n)} a(m;n,k)*(1+q^2)^k*q^(g(m;n)-k), for m >= 0, n >= 0, where g(m;n) = m*n if m is odd and (2*n+1)*m/2 if m is even, and the tabf array entries a(m;n,k) are the coefficients of the g.f. for the row n polynomials G(m;n,x) = (d^m/dt^m)G(m;n,t,x)/m!|{t=0}, with G(m;n,t,x) = (1+t)*Product{k=1..n+(m - m (mod 2))/2}(1 + t^2 + 2*t*T(k,x/2) (Chebyshev's T-polynomials). Hence a(m;n,k) = [x^k]G(m;n,x), for k=0..g(m;n). The present entry is the instance m = 2. (Thanks to Wolfdieter Lang for clarifying the text on the general prescription of a(m;n,k).)

Signed version of A046854, A130777.

Conjecture: row n is U(n, x/2) + U(n-1, x/2) where U is the sequence of Chebyshev polynomials of the second kind. - Thomas Baruchel, Jun 03 2018 [For a proof see the following comment.]

From Wolfdieter Lang, Oct 19 2019: (Start)

The row polynomial R(n, x) = Sum_{k=0..n} a(n, k)*x^k = [2*n+1]_q / q^n with the q-number [2*n+1]_q := (1 - q^n)/(1 - q), which for q = 1 becomes 2*n+1, and x = x(q) = q + q^(-1). See the simplified Name and the first comment. In terms of Chebyshev S polynomials (A049310) this q-number is written as [2*n+1]_q = q^n*S(2*n, q^(1/2) + q^(-1/2)), hence R(n, x) = S(2*n, sqrt(2+x)) = S(n, x) + S(n-1, x) (which proves the conjecture of the previous comment).

For the o.g.f. of R(n, x) see the formula section.

My motivation for looking at this sequence came from the Brändli and Beyne paper's recurrence for the polynomial P_m(s) which coincides with R(n, x), with m -> n and s -> x. (End)

A294099(n, k) = Sum_{j=0..k} n^j * T(n, j) for all n, k in Z. - Michael Somos, Jun 19 2023

Middle line of 5-wave sequence A038201.

Let M denotes the 5 X 5 matrix = row by row (1,1,1,1,1)(1,1,1,1,0)(1,1,1,0,0)(1,1,0,0,0)(1,0,0,0,0) and A(n) the vector (x(n),y(n),z(n),t(n),u(n))=M^n*A where A is the vector (1,1,1,1,1) then a(n)=z(n). - Benoit Cloitre, Apr 02 2002

a(n) appears in the formula for 1/rho(11)^n, with rho(11) := 2*cos(Pi/11) (length ratio (smallest diagonal/side) in the regular 11-gom) when written in the power basis of the degree 5 number field Q(rho(11)): 1/rho(11)^n = a(n)*1 + A230080(n)*rho(11) - A230081(n)*rho(11)^2 - A069006(n-1)* rho(11)^3 + a(n-1)*rho(11)^4, n >= 0, with A069006(-1) = 0 = a(-1). See A230080 with the example for n=4. - Wolfdieter Lang, Nov 04 2013

From Wolfdieter Lang, Nov 20 2013: (Start)

The limit a(n+1)/a(n) for n -> infinity is omega(11) := S(4, x) = 1 - 3*x^2 + x^4 with x = rho(11). omega(11) = 1/(2*cos(Pi*5/11)), approx. 3.51333709. For the Chebyshev S-polynomial see A049310. For rho(11) see the preceding comment. The decimal expansion of omega(11) is given in A231186. omega(11) is an integer in Q(rho(11)) with power basis coefficients [1,0,-3,0,1]. It is known to be the length ratio (longest diagonal)/side in the regular 11-gon.

This limit follows from the a(n)-recurrence and the solutions of X^5 - 3*X^4 - 3*X^3 + 4*X^2 + X - 1 = 0, which are given by the inverse of the known solutions of the minimal polynomial C(11, x) of rho(11) (see A187360). The other four X solutions are 1/rho(11), with coefficients [3,3,-4,-1,1] in the power basis of Q(rho(11)), approx. 0.52110856, 1/(2*cos(Pi*3/11)) with coefficients [-1,-1,1,0,0], approx. 0.763521119, 1/(2*cos(Pi*7/11)) with coefficients [0,-3,3,1,-1], approx. -1.20361562, and 1/(2*cos(Pi*9/11)) with coefficients [0,1,3,0,-1], approx. -0.59435114. These solutions for X are therefore irrelevant for this sequence.

The same limit omega(11) is therefore obtained for the sequences A069006, A230080 and A230081. See the Nov 04 2013 comment.

(End)