A070998 -id:A070998 - cp's OEIS frontend

A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

1, 1, -1, 1, -1, -1, 1, -1, -2, 1, 1, -1, -3, 2, 1, 1, -1, -4, 3, 3, -1, 1, -1, -5, 4, 6, -3, -1, 1, -1, -6, 5, 10, -6, -4, 1, 1, -1, -7, 6, 15, -10, -10, 4, 1, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1, 1, -1, -11, 10, 45, -36, -84, 56, 70

Offset: 0

Reinhard Zumkeller, Jun 01 2005

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

			Triangle begins:
  1;
  1,  -1;
  1,  -1,  -1;
  1,  -1,  -2,   1;
  1,  -1,  -3,   2,   1;
  1,  -1,  -4,   3,   3,  -1;
  1,  -1,  -5,   4,   6,  -3,  -1;
  1,  -1,  -6,   5,  10,  -6,  -4,   1;
  1,  -1,  -7,   6,  15, -10, -10,   4,   1;
  1,  -1,  -8,   7,  21, -15, -20,  10,   5,  -1;
  1,  -1,  -9,   8,  28, -21, -35,  20,  15,  -5,  -1;
  1,  -1, -10,   9,  36, -28, -56,  35,  35, -15,  -6,   1;
  ...

Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
Jay Kappraff, S. Jablan, G. Adamson, & R. Sazdonovich: "Golden Fields, Generalized Fibonacci Sequences, & Chaotic Matrices"; FORMA, Vol 19, No 4, (2005).

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Henry W. Gould, A Variant of Pascal's Triangle, Corrections, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271.
L. Edson Jeffery, Unit-primitive matrices.
Ju, Hyeong-Kwan On the sequence generated by a certain type of matrices. Honam Math. J. 39, No. 4, 665-675 (2017).
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894 [math.NT], 2015.
Frank Ruskey and Carla Savage, Gray codes for set partitions and restricted growth tails, Australasian Journal of Combinatorics, Volume 10(1994), pp. 85-96. See Table 1 p. 95.

Cf. A049310, A039961, A124645 (matrix inverse).
Triangle sums (see the comments): A193884 (Kn11), A154955 (Kn21), A087960 (Kn22), A000007 (Kn3), A010892 (Fi1), A134668 (Fi2), A078031 (Ca2), A193669 (Gi1), A001519 (Gi3), A193885 (Ze1), A050935 (Ze3). - Johannes W. Meijer, Aug 08 2011
Cf. A003558.
Cf. A033999, A059841.

a108299 n k = a108299_tabl !! n !! k
a108299_row n = a108299_tabl !! n
a108299_tabl = [1] : iterate (\row ->
   zipWith (+) (zipWith (*) ([0] ++ row) a033999_list)
               (zipWith (*) (row ++ [0]) a059841_list)) [1,-1]
-- Reinhard Zumkeller, May 06 2012

A108299 := proc(n,k): binomial(n-floor((k+1)/2), floor(k/2))*(-1)^floor((k+1)/2) end: seq(seq(A108299 (n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011

t[n_, k_?EvenQ] := I^k*Binomial[n-k/2, k/2]; t[n_, k_?OddQ] := -I^(k-1)*Binomial[n+(1-k)/2-1, (k-1)/2]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 16 2013 *)

{T(n,k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)),n,x)+y*O(y^k),k,y)} (Hanna)

T(n,k) = binomial(n-floor((k+1)/2),floor(k/2))*(-1)^floor((k+1)/2).
T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0 < k < n, T(n, 0) = 1, T(n, n) = (-1)^floor((n+1)/2).
G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - Paul D. Hanna, Jun 12 2005
The generating polynomial (in z) of row n >= 0 is (u^(2*n+1) + v^(2*n+1))/(u + v), where u and v are defined by u^2 + v^2 = 1 and u*v = z. - Emeric Deutsch, Jun 16 2011
From Johannes W. Meijer, Aug 08 2011: (Start)
abs(T(n,k)) = A065941(n,k) = abs(A187660(n,n-k));
T(n,n-k) = A130777(n,k); abs(T(n,n-k)) = A046854(n,k) = abs(A066170(n,k)). (End)

Corrected and edited by Philippe Deléham, Oct 20 2008

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681

Offset: 1

Ralf Stephan, May 31 2004

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.

Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...

Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.
Elizabeth Wilmer, A note on Stephan's conjecture 87
Elizabeth Wilmer, A note on Stephan's conjecture 87 [cached copy]

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)

PARI
```
T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)
```

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449

Offset: 0

Bruno Berselli, Feb 25 2014

First bisection of A041611.

Bruno Berselli, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (36,-1).

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

[n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020

CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]

a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016

m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
print(f.coefficients())

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A018913 a(n) = 9*a(n - 1) - a(n - 2) for n>1, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 9, 80, 711, 6319, 56160, 499121, 4435929, 39424240, 350382231, 3114015839, 27675760320, 245967827041, 2186034683049, 19428344320400, 172669064200551, 1534593233484559, 13638670037160480, 121213437100959761

Offset: 0

R. K. Guy

Define the sequence L(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0. This is L(1,9).
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 9's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,8}. - Milan Janjic, Jan 25 2015
Not to be confused with the Pisot L(1,9) sequence, which is A001019. - R. J. Mathar, Feb 13 2016
Lim_{n->oo} a(n+1)/a(n) = (9 + sqrt(77))/2 = A092290 + 1 = 8.887482... - Wolfdieter Lang, Nov 16 2023

			G.f. = x + 9*x^2 + 80*x^3 + 711*x^4 + 6319*x^5 + 56160*x^6 + 499121*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 12.
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=9, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=11.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).
Index entries for Pisot sequences

Cf. A000027, A001906, A001353, A004254, A001109, A004187, A001090.
Cf. A056918(n)=sqrt{77*(a(n))^2 +4}, that is, a(n)=sqrt((A056918(n)^2 - 4)/77).
Cf. A092290 + 1.

I:=[0, 1]; [n le 2 select I[n] else 9*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

CoefficientList[Series[x/(1 - 9*x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 23 2012 *)

concat(0, Vec(x/(1-9*x+x^2) + O(x^30))) \\ Michel Marcus, Sep 06 2017

[lucas_number1(n,9,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

G.f.: x/(1-9*x+x^2).
a(n) = S(2*n-1, sqrt(11))/sqrt(11) = S(n-1, 9); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
a(n) = (((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n)/sqrt(77). - Barry E. Williams, Aug 21 2000
a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*8^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/7*(7 + sqrt(77)).
Product {n >= 2} (1 - 1/a(n)) = 1/18*(7 + sqrt(77)). (End)
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*7^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*11^k. - Peter Bala, Jul 17 2023
E.g.f.: 2*exp(9*x/2)*sinh(sqrt(77)*x/2)/sqrt(77). - Stefano Spezia, Feb 23 2025
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = sqrt(11/7) [telescoping product: ((a(2*n) + 1)/(a(2*n) - 1))^2 = (11 - 4/(a(n+1) - a(n))^2)/(11 - 4/(a(n) - a(n-1))^2), leading to 11 - 7*Product_{k = 1..n} ((a(2*k) + 1)/(a(2*k) - 1))^2 = 4/A070998(n)^2]. - Peter Bala, May 18 2025

G.f. adapted to the offset by Vincenzo Librandi, Dec 23 2012

A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

Original entry on oeis.org

1, 10, 89, 791, 7030, 62479, 555281, 4935050, 43860169, 389806471, 3464398070, 30789776159, 273643587361, 2432002510090, 21614379003449, 192097408520951, 1707262297685110, 15173263270645039, 134852107138120241, 1198495700972437130, 10651609201613813929

Offset: 0

Wolfdieter Lang, Aug 04 2000

This is the m=11 member of the m-family of sequences S(n,m-2)+S(n-1,m-2) = S(2*n,sqrt(m)) (for S(n,x) see Formula). The m=4..10 instances are A005408, A002878, A001834, A030221, A002315, A033890 and A057080, resp. The m=1..3 (signed) sequences are: A057078, A057077 and A057079, resp.
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim_{n->oo} a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 89, 389806471, 192097408520951, 7477414486269626733119, ... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 10, 0, 89, 0, 791, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -7, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=11.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A005408, A002878, A001834, A030221, A002315, A033890, A057080.
Cf. A057078, A057077, A057079.
Cf. A018913, A049310.
Cf. A100047.

A057081 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,10]);
    else
        9*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1 + x)/(1 - 9*x + x^2), {x,0,50}], x] (* or *) LinearRecurrence[{9,-1}, {1,10}, 50] (* G. C. Greubel, Apr 12 2017 *)

Vec((1+x)/(1-9*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,9,1)-lucas_number2(n-1,9,1))/7 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

a(n) = 9*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 9) + S(n-1, 9) = S(2*n, sqrt(11)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 9) = A018913(n).
G.f.: (1+x)/(1-9*x+x^2).
Let q(n, x) = Sum{i=0..n} x^(n-i)*binomial(2*n-i, i), a(n) = (-1)^n*q(n, -11). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-9)*(-1)^n, where L is defined as in A108299; see also A070998 for L(n,+9). - Reinhard Zumkeller, Jun 01 2005
From Peter Bala, Jun 08 2025: (Start)
a(n) = (1/sqrt(7)) * ( ((sqrt(11) + sqrt(7))/2)^(2*n+1) - ((sqrt(11) - sqrt(7))/2)^(2*n+1) ).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/11 (telescoping series: 11/(a(n) - 1/a(n)) = 1/A018913(n+1) + 1/A018913(n)).
Conjecture: for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(11/7) [telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (1 - 4/b(n+1))/(1 - 4/b(n)), where b(n) = 2 + A056918(n)]. (End)

A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, -1, -1, 1, -2, 1, 1, 1, -3, 5, -1, 0, 1, -4, 11, -13, 1, -1, 1, -5, 19, -41, 34, -1, 1, 1, -6, 29, -91, 153, -89, 1, 0, 1, -7, 41, -169, 436, -571, 233, -1, -1, 1, -8, 55, -281, 985, -2089, 2131, -610, 1, 1, 1, -9, 71, -433, 1926, -5741, 10009, -7953, 1597, -1, 0

Offset: 1

L. Edson Jeffery, Bob Selcoe and Andrew Hone, Feb 01 2018

This array is used to compute A269252: A269252(n) = least k such that |A(n,k)| is a prime, or -1 if no such k exists.
For detailed theory, see [Hone].
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

			Array begins:
1   0  -1     1     0      -1       1         0        -1           1
1  -1   1    -1     1      -1       1        -1         1          -1
1  -2   5   -13    34     -89     233      -610      1597       -4181
1  -3  11   -41   153    -571    2131     -7953     29681     -110771
1  -4  19   -91   436   -2089   10009    -47956    229771    -1100899
1  -5  29  -169   985   -5741   33461   -195025   1136689    -6625109
1  -6  41  -281  1926  -13201   90481   -620166   4250681   -29134601
1  -7  55  -433  3409  -26839  211303  -1663585  13097377  -103115431
1  -8  71  -631  5608  -49841  442961  -3936808  34988311  -310957991
1  -9  89  -881  8721  -86329  854569  -8459361  83739041  -828931049

Robert Price, Table of n, a(n) for n = 1..5050
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269251, A269252, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.
Cf. A094954 (unsigned version of this array, but missing the first row).
Cf. Rows: A057078, A033999, A099496, A079935 (or A001835), A004253, A001653, A049685, A070997, A070998, A138288 (or A072256), ...
Cf. Columns: A000012, A001477 (A000027), A110331 (A165900), A123972, A192398, ...

(* Array: *)
Grid[Table[LinearRecurrence[{-n, -1}, {1, 1 - n}, 10], {n, 10}]]
(*Array antidiagonals flattened (gives this sequence):*)
A299045[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] (-n)^(k - j), {j, 0, k}]; Flatten[Table[A299045[n - k, k], {n, 11}, {k, 0, n - 1}]]

{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*(-n)^(k-j))}; /* Michael Somos, Jun 19 2023 */

G.f. for row n: (1 + x)/(1 + n*x + x^2), n >= 1.

A(n, k) = B(-n, k) where B = A294099. - Michael Somos, Jun 19 2023

A165253 Triangle T(n,k), read by rows given by [1,0,1,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 5, 1, 0, 1, 10, 15, 7, 1, 0, 1, 15, 35, 28, 9, 1, 0, 1, 21, 70, 84, 45, 11, 1, 0, 1, 28, 126, 210, 165, 66, 13, 1, 0, 1, 36, 210, 462, 495, 286, 91, 15, 1, 0, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 0, 1, 55, 495, 1716, 3003, 3003, 1820, 680

Offset: 0

Philippe Deléham, Sep 10 2009

Mirror image of triangle in A121314.

			Triangle begins:
  1;
  1,    0;
  1,    1,    0;
  1,    3,    1,    0;
  1,    6,    5,    1,    0;
  1,   10,   15,    7,    1,    0;
  1,   15,   35,   28,    9,    1,    0;
  1,   21,   70,   84,   45,   11,    1,    0;
  1,   28,  126,  210,  165,   66,   13,    1,    0;
  1,   36,  210,  462,  495,  286,   91,   15,    1,    0,
  1,   45,  330,  924, 1287, 1001,  455,  120,   17,    1,    0;

Indranil Ghosh, Rows 0..125 of triangle, flattened

Cf. A054142, A085478, A121314.

m = 13;
(* DELTA is defined in A084938 *)
DELTA[Join[{1, 0, 1}, Table[0, {m}]], Join[{0, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)

T(0,0)=1, T(n,k) = binomial(n-1+k,2k) for n >= 1.
Sum {k=0..n} T(n,k)*x^k = A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000007(n), A001519(n), A047849(n), A165310(n), A165311(n), A165312(n), A165314(n), A165322(n), A165323(n), A165324(n) for x= 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 26 2009
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=T(1,0)=1, T(1,1)=0. - Philippe Deléham, Feb 18 2012
G.f.: (1-x-y*x)/((1-x)^2-y*x). - Philippe Deléham, Feb 19 2012

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A133607 Triangle read by rows: T(n, k) = qStirling2(n, k, q) for q = -1, with 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, -1, 0, 1, -1, -1, 0, 1, -1, -2, 1, 0, 1, -1, -3, 2, 1, 0, 1, -1, -4, 3, 3, -1, 0, 1, -1, -5, 4, 6, -3, -1, 0, 1, -1, -6, 5, 10, -6, -4, 1, 0, 1, -1, -7, 6, 15, -10, -10, 4, 1, 0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1

Offset: 0

Philippe Deléham, Dec 27 2007

Previous name: Triangle T(n,k), 0<=k<=n, read by rows given by [0, 1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, -2, 1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

			Triangle begins:
  1;
  0, 1;
  0, 1, -1;
  0, 1, -1, -1;
  0, 1, -1, -2, 1;
  0, 1, -1, -3, 2, 1;
  0, 1, -1, -4, 3, 3, -1;
  0, 1, -1, -5, 4, 6, -3, -1;
  0, 1, -1, -6, 5, 10, -6, -4, 1;
  0, 1, -1, -7, 6, 15, -10, -10, 4, 1;
  0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  0, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...
Triangle A103631 begins:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 1, 2, 1;
  0, 1, 1, 3, 2, 1;
  0, 1, 1, 4, 3, 3, 1;
  0, 1, 1, 5, 4, 6, 3, 1;
  0, 1, 1, 6, 5, 10, 6, 4, 1;
  0, 1, 1, 7, 6, 15, 10, 10, 4, 1;
  0, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1;
  0, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1;
  0, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1;
  ...
Triangle A108299 begins:
  1;
  1, -1;
  1, -1, -1;
  1, -1, -2, 1;
  1, -1, -3, 2, 1;
  1, -1, -4, 3, 3, -1;
  1, -1, -5, 4, 6, -3, -1;
  1, -1, -6, 5, 10, -6, -4, 1;
  1, -1, -7, 6, 15, -10, -10, 4, 1;
  1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...

Another version is A108299.

Unsigned version is A103631 (T(n,k) = A103631(n,k)*A057077(k)).

m = 13
(* DELTA is defined in A084938 *)
DELTA[Join[{0, 1}, Table[0, {m}]], Join[{1, -2, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)
qStirling2[n_, k_, q_] /; 1 <= k <= n := q^(k-1) qStirling2[n-1, k-1, q] + Sum[q^j, {j, 0, k-1}] qStirling2[n-1, k, q];
qStirling2[n_, 0, _] := KroneckerDelta[n, 0];
qStirling2[0, k_, _] := KroneckerDelta[0, k];
qStirling2[, , _] = 0;
Table[qStirling2[n, k, -1], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 10 2020 *)

from sage.combinat.q_analogues import q_stirling_number2
for n in (0..9):
    print([q_stirling_number2(n,k).substitute(q=-1) for k in [0..n]])
# Peter Luschny, Mar 09 2020

Sum_{k, 0<=k<=n}T(n,k)*x^(n-k)= A057077(n), A010892(n), A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n-1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 respectively .
Sum_{k, 0<=k<=n}T(n,k)*x^k = A000007(n), A010892(n), A133631(n), A133665(n), A133666(n), A133667(n), A133668(n), A133669(n), A133671(n), A133672(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively .
G.f.: (1-x+y*x)/(1-x+y^2*x^2). - Philippe Deléham, Mar 14 2012
T(n,k) = T(n-1,k) - T(n-2,k-2), T(0,0) = T(1,1) = T(2,1) = 1, T(1,0) = T(2,0) = 0, T(2,2) = -1 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 14 2012

New name from Peter Luschny, Mar 09 2020

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

Original entry on oeis.org

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024

Offset: 0

Wolfdieter Lang, Jul 29 2014

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
  The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
  [1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014

Wikipedia, Dirichlet kernel.

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).
Cf. A180870 (row reversed), A157751, A228565, A049310, A053120, A057077, A004526, A008805, A058187, A189976, A189980, A005408, A033999.

T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)

def T(n, k):
    if k == 0: return (-1)^(n//2)
    if (0 <= n and n < k) or (n == -1 and k == 1): return 0
    return 2*T(n-1, k-1) - T(n-2, k)
for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

cp's OEIS Frontend

A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

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A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

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A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

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A018913 a(n) = 9*a(n - 1) - a(n - 2) for n>1, a(0)=0, a(1)=1.

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A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

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A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)*binomial(k-floor((j+1)/2), floor(j/2))*(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

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A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.