A108299 -id:A108299 - cp's OEIS frontend

A077416 Chebyshev S-sequence with Diophantine property.

1, 13, 155, 1847, 22009, 262261, 3125123, 37239215, 443745457, 5287706269, 63008729771, 750817050983, 8946795882025, 106610733533317, 1270382006517779, 15137973344680031, 180385298129642593

Offset: 0

Wolfdieter Lang, Nov 29 2002

7*b(n)^2 - 5*a(n)^2 = 2 with companion sequence b(n) = A077417(n), n>=0.
a(n) = L(n,-12)*(-1)^n, where L is defined as in A108299; see also A077417 for L(n,+12). - Reinhard Zumkeller, Jun 01 2005
The aerated sequence (b(n))n>=1 = [1, 0, 13, 0, 155, 0, 1857, 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 = -10, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, May 12 2025

Ivan Panchenko, Table of n, a(n) for n = 0..200
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
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 (12,-1).

Cf. A054320(n-1) with companion A072256(n), n>=1.

I:=[1, 13]; [n le 2 select I[n] else 12*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 18 2018

LinearRecurrence[{12,-1},{1,13},30] (* Harvey P. Dale, Apr 03 2013 *)

x='x+O('x^30); Vec((1+x)/(1-12*x+x^2)) \\ G. C. Greubel, Jan 18 2018

[(lucas_number2(n,12,1)-lucas_number2(n-1,12,1))/10 for n in range(1, 18)] # Zerinvary Lajos, Nov 10 2009

a(n) = 12*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 12) + S(n-1, 12) = S(2*n, sqrt(14)) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0, S(n, 12) = A004191(n).
G.f.: (1+x)/(1-12*x+x^2).
a(n) = (ap^(2*n+1) - am^(2*n+1))/(ap - am) with ap := (sqrt(7)+sqrt(5))/sqrt(2) and am := (sqrt(7)-sqrt(5))/sqrt(2).
a(n) = Sum_{k=0..n} (-1)^k * binomial(2*n-k,k) * 14^(n-k).
a(n) = sqrt((7*A077417(n)^2 - 2)/5).
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 6), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 12*a(n)*a(n+1) + a(n+1)^2 = 14.
More generally, for real x, a(n+x)^2 - 12*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 14, where a(n) := (ap^(2*n+1) - am^(2*n+1))/(ap - am), ap := sqrt(7/2) + sqrt(5/2) and am := sqrt(7/2) - sqrt(5/2), as given above.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/14 (telescoping series).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(7/5) (telescoping product). (End)

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

A085260 Ratio-determined insertion sequence I(0.0833344) (see the link below).

Original entry on oeis.org

1, 12, 155, 2003, 25884, 334489, 4322473, 55857660, 721827107, 9327894731, 120540804396, 1557702562417, 20129592507025, 260127000028908, 3361521407868779, 43439651302265219, 561353945521579068

Offset: 1

John W. Layman, Jun 23 2003

nonn

This sequence is the ratio-determined insertion sequence (RDIS) "twin" to A078362 (see the link for an explanation of "twin"). See A082630 or A082981 for recent examples of RDIS sequences.
a(n) = L(n,13), where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(11)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Seems to be positive values of x (or y) satisfying x^2 - 13xy + y^2 + 11 = 0. - Colin Barker, Feb 10 2014
It appears that the b-file, formulas and programs are based on the conjectured, so far apparently unproved recurrence relation. - M. F. Hasler, Nov 05 2018
Nonnegative y values in solutions to the Diophantine equation 11*x^2 - 15*y^2 = -4. The corresponding x values are in A126866. Note that a(n+1)^2 - a(n)*a(n+2) = -11. - Klaus Purath, Mar 21 2025

Vincenzo Librandi, Table of n, a(n) for n = 1..900
A. Fink, R. K. Guy and M. Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types [local copy, corrected]
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv:1403.5962 [math.CO], 2014.
Index entries for linear recurrences with constant coefficients, signature (13,-1).

Cf. A078362, A082630, A082981.
Row 13 of array A094954.
Cf. similar sequences listed in A238379.

I:=[1,12]; [n le 2 select I[n] else 13*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 18 2018

CoefficientList[Series[(1 - x)/(1 - 13 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
LinearRecurrence[{13,-1}, {1,12}, 30] (* G. C. Greubel, Jan 18 2018 *)

my(x='x+O('x^30)); Vec(x*(1-x)/(1-13*x+x^2)) \\ G. C. Greubel, Jan 18 2018

It appears that the sequence satisfies a(n+1) = 13*a(n) - a(n-1). [Corrected by M. F. Hasler, Nov 05 2018]
If the recurrence a(n+2) = 13*a(n+1) - a(n) holds then for n > 0, a(n)*a(n+3) = 143 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
G.f.: x*(1-x)/(1 - 13*x + x^2). - Philippe Deléham, Nov 17 2008
For n>1, a(n) is the numerator of the continued fraction [1,11,1,11,...,1,11] with (n-1) repetitions of 1,11. - Greg Dresden, Sep 10 2019

A374439 Triangle read by rows: the coefficients of the Lucas-Fibonacci polynomials. T(n, k) = T(n - 1, k) + T(n - 2, k - 2) with initial values T(n, k) = k + 1 for k < 2.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 3, 4, 1, 1, 2, 4, 6, 3, 2, 1, 2, 5, 8, 6, 6, 1, 1, 2, 6, 10, 10, 12, 4, 2, 1, 2, 7, 12, 15, 20, 10, 8, 1, 1, 2, 8, 14, 21, 30, 20, 20, 5, 2, 1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1, 1, 2, 10, 18, 36, 56, 56, 70, 35, 30, 6, 2

Offset: 0

Peter Luschny, Jul 22 2024

There are several versions of Lucas and Fibonacci polynomials in this database. Our naming follows the convention of calling polynomials after the values of the polynomials at x = 1. This assumes a regular sequence of polynomials, that is, a sequence of polynomials where degree(p(n)) = n. This view makes the coefficients of the polynomials (the terms of a row) a refinement of the values at the unity.
A remarkable property of the polynomials under consideration is that they are dual in this respect. This means they give the Lucas numbers at x = 1 and the Fibonacci numbers at x = -1 (except for the sign). See the example section.
The Pell numbers and the dual Pell numbers are also values of the polynomials, at the points x = -1/2 and x = 1/2 (up to the normalization factor 2^n). This suggests a harmonized terminology: To call 2^n*P(n, -1/2) = 1, 0, 1, 2, 5, ... the Pell numbers (A000129) and 2^n*P(n, 1/2) = 1, 4, 9, 22, ... the dual Pell numbers (A048654).
Based on our naming convention one could call A162515 (without the prepended 0) the Fibonacci polynomials. In the definition above only the initial values would change to: T(n, k) = k + 1 for k < 1. To extend this line of thought we introduce A374438 as the third triangle of this family.
The triangle is closely related to the qStirling2 numbers at q = -1. For the definition of these numbers see A333143. This relates the triangle to A065941 and A103631.

			Triangle starts:
  [ 0] [1]
  [ 1] [1, 2]
  [ 2] [1, 2, 1]
  [ 3] [1, 2, 2,  2]
  [ 4] [1, 2, 3,  4,  1]
  [ 5] [1, 2, 4,  6,  3,  2]
  [ 6] [1, 2, 5,  8,  6,  6,  1]
  [ 7] [1, 2, 6, 10, 10, 12,  4,  2]
  [ 8] [1, 2, 7, 12, 15, 20, 10,  8,  1]
  [ 9] [1, 2, 8, 14, 21, 30, 20, 20,  5,  2]
  [10] [1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1]
.
Table of interpolated sequences:
  |  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |
  |  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|
  |    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |
  |  0 |        -1         |     1   |       1     |       1    |
  |  1 |         1         |     3   |       0     |       4    |
  |  2 |         0         |     4   |       1     |       9    |
  |  3 |         1         |     7   |       2     |      22    |
  |  4 |         1         |    11   |       5     |      53    |
  |  5 |         2         |    18   |      12     |     128    |
  |  6 |         3         |    29   |      29     |     309    |
  |  7 |         5         |    47   |      70     |     746    |
  |  8 |         8         |    76   |     169     |    1801    |
  |  9 |        13         |   123   |     408     |    4348    |

Paolo Xausa, Rows n = 0..150 of the triangle, flattened
Peter Luschny, Illustrating the polynomials.

Triangles related to Lucas polynomials: A034807, A114525, A122075, A061896, A352362.
Triangles related to Fibonacci polynomials: A162515, A053119, A168561, A049310, A374441.
Sums include: A000204 (Lucas numbers, row), A000045 & A212804 (even sums, Fibonacci numbers), A006355 (odd sums), A039834 (alternating sign row).
Type m^n*P(n, 1/m): A000129 & A048654 (Pell, m=2), A108300 & A003688 (m=3), A001077 & A048875 (m=4).
Adding and subtracting the values in a row of the table (plus halving the values obtained in this way): A022087, A055389, A118658, A052542, A163271, A371596, A324969, A212804, A077985, A069306, A215928.
Columns include: A040000 (k=1), A000027 (k=2), A005843 (k=3), A000217 (k=4), A002378 (k=5).
Diagonals include: A000034 (k=n), A029578 (k=n-1), abs(A131259) (k=n-2).
Cf. A029578 (subdiagonal), A124038 (row reversed triangle, signed).
Cf. A039961, A065941 (qStirling2), A103631, A108299, A131259. A133607, A194005, A333143, A374438 (m=3).

function T(n,k) // T = A374439
  if k lt 0 or k gt n then return 0;
  elif k le 1 then return k+1;
  else return T(n-1,k) + T(n-2,k-2);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 23 2025

A374439 := (n, k) -> ifelse(k::odd, 2, 1)*binomial(n - irem(k, 2) - iquo(k, 2), iquo(k, 2)):
# Alternative, using the function qStirling2 from A333143:
T := (n, k) -> 2^irem(k, 2)*qStirling2(n, k, -1):
seq(seq(T(n, k), k = 0..n), n = 0..10);

A374439[n_, k_] := (# + 1)*Binomial[n - (k + #)/2, (k - #)/2] & [Mod[k, 2]];
Table[A374439[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Paolo Xausa, Jul 24 2024 *)

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k > n: return 0
    if k < 2: return k + 1
    return T(n - 1, k) + T(n - 2, k - 2)

from math import comb as binomial
def T(n: int, k: int) -> int:
    o = k & 1
    return binomial(n - o - (k - o) // 2, (k - o) // 2) << o

def P(n, x):
    if n < 0: return P(n, x)
    return sum(T(n, k)*x**k for k in range(n + 1))
def sgn(x: int) -> int: return (x > 0) - (x < 0)
# Table of interpolated sequences
print("|  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |")
print("|  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|")
print("|    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |")
f = "| {0:2d} | {1:9d}         |  {2:4d}   |   {3:5d}     |    {4:4d}    |"
for n in range(10): print(f.format(n, -P(n, -1), P(n, 1), int(2**n*P(n, -1/2)), int(2**n*P(n, 1/2))))

from sage.combinat.q_analogues import q_stirling_number2
def A374439(n,k): return (-1)^((k+1)//2)*2^(k%2)*q_stirling_number2(n+1, k+1, -1)
print(flatten([[A374439(n, k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 23 2025

T(n, k) = 2^k' * binomial(n - k' - (k - k') / 2, (k - k') / 2) where k' = 1 if k is odd and otherwise 0.
T(n, k) = (1 + (k mod 2))*qStirling2(n, k, -1), see A333143.
2^n*P(n, -1/2) = A000129(n - 1), Pell numbers, P(-1) = 1.
2^n*P(n, 1/2) = A048654(n), dual Pell numbers.
T(2*n, n) = (1/2)*(-1)^n*( (1+(-1)^n)*A005809(n/2) - 2*(1-(-1)^n)*A045721((n-1)/2) ). - G. C. Greubel, Jan 23 2025

A164965 Cumulative sums of A010892.

Original entry on oeis.org

1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0

Offset: 0

Mark Dols, Sep 02 2009

Index entries for linear recurrences with constant coefficients, signature (2,-2,1).

Cf. A010892, A021823, A164925, A108299.

Table[-Floor[1/6 (-4 + n)] - Floor[1/6 (-3 + n)] + Floor[1/6 (-1 + n)] + Floor[n/6], {n, 0, 100}] (* John M. Campbell, Dec 23 2016 *)
LinearRecurrence[{2,-2,1},{1,2,2},100] (* Harvey P. Dale, Jul 17 2020 *)

G.f.: 1/((1 - x)*(1 - x + x^2)). - Philippe Deléham, Oct 11 2011
a(n) = a(n-1) - a(n-2) + 1. - Arkadiusz Wesolowski, Jun 08 2013
a(n) = -floor((n - 4)/6) - floor((n - 3)/6) + floor((n - 1)/6) + floor(n/6). - John M. Campbell, Dec 23 2016
E.g.f.: cosh(x) + 2*exp(x/2)*sin(sqrt(3)*x/2)/sqrt(3) + sinh(x). - Stefano Spezia, Feb 20 2023

Offset corrected by John M. Campbell, Dec 23 2016

A187660 Triangle read by rows: T(n,k) = (-1)^(floor(3k/2))binomial(floor((n+k)/2),k), 0 <= k <= n.

Original entry on oeis.org

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

Offset: 0

L. Edson Jeffery, Mar 12 2011

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

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

L. E. Jeffery, Danzer matrices
Guoce Xin and Yueming Zhong, Proving some conjectures on Kekulé numbers for certain benzenoids by using Chebyshev polynomials, arXiv:2201.02376 [math.CO], 2022.

Signed version of A046854.
Absolute values of a(n) form a reflected version of A065941, which is considered the main entry.
Cf. A046854, A066170, A130777, A267482.

A187660 := proc(n,k): (-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k) end: seq(seq(A187660(n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011

t[n_, k_] := (-1)^Floor[3 k/2] Binomial[Floor[(n + k)/2], k]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] (* L. Edson Jeffery, Oct 20 2017 *)

T(n,k) = (-1)^n*A066170(n,k).
abs(T(n,k)) = A046854(n,k) = abs(A066170(n,k)) = abs(A130777(n,k)).
abs(T(n,k)) = A065941(n,n-k) = abs(A108299(n,n-k)).

Edited and corrected by L. Edson Jeffery, Oct 20 2017

A191314 Triangle read by rows: T(n,k) is the number of dispersed Dyck paths (i.e., Motzkin paths with no (1,0) steps at positive heights) of length n and height k.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 7, 2, 1, 12, 6, 1, 1, 20, 12, 2, 1, 33, 27, 8, 1, 1, 54, 53, 16, 2, 1, 88, 108, 44, 10, 1, 1, 143, 208, 88, 20, 2, 1, 232, 405, 208, 65, 12, 1, 1, 376, 768, 415, 130, 24, 2, 1, 609, 1459, 908, 350, 90, 14, 1, 1, 986, 2734, 1804, 700, 180, 28, 2, 1, 1596, 5117, 3776, 1700, 544, 119, 16, 1

Offset: 0

Emeric Deutsch, May 31 2011

Row n has 1 + floor(n/2) entries.
Sum of entries in row n is binomial(n, floor(n/2)) = A001405(n).
T(n,1) = A000071(n+1) (Fibonacci numbers minus 1).
Sum_{k>=0} k * T(n,k) = A191315(n).
Extracting the even numbered rows, we obtain triangle A205946 with row sums A000984. The odd numbered rows yield triangle A205945 with row sums A001700. - Gary W. Adamson, Feb 01 2012

			T(5,2) = 2 because we have HUUDD and UUDDH, where U=(1,1), D=(1,-1), H=(1,0).
Triangle starts:
1;
1;
1,  1;
1,  2;
1,  4,  1;
1,  7,  2;
1, 12,  6, 1;
1, 20, 12, 2;
1, 33, 27, 8, 1;

Alois P. Heinz, Rows n = 0..200, flattened

Cf. A001405, A000071, A191315.

Cf. A205573, A205945, A001700, A205946, A000984.

F[0] := 1: F[1] := 1-z: for k from 2 to 12 do F[k] := sort(expand(F[k-1]-z^2*F[k-2])) end do: for k from 0 to 11 do h[k] := z^(2*k)/(F[k]*F[k+1]) end do: T := proc (n, k) options operator, arrow: coeff(series(h[k], z = 0, 20), z, n) end proc: for n from 0 to 16 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(x, y) option remember; `if`(y>x or y<0, 0, `if`(x=0, 1,
      (p->add(coeff(p, z, i)*z^max(i, y), i=0..degree(p,z)))
      (b(x-1, y-1))+ b(x-1, y+1)+`if`(y=0, b(x-1, y), 0)))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..14);  # Alois P. Heinz, Mar 12 2014

b[x_, y_] := b[x, y] = If[y>x || y<0, 0, If[x==0, 1, Function [{p}, Sum[ Coefficient[p, z, i]*z^Max[i, y], {i, 0, Exponent[p, z]}]][b[x-1, y-1]] + b[x-1, y+1] + If[y==0, b[x-1, y], 0]]]; T[n_] := Function[{p}, Table[Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][b[n, 0]]; Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Mar 31 2015, after Alois P. Heinz *)

G.f.: The g.f. of column k is z^{2k}/(F[k]*F[k+1]), where F[k] are polynomials in z defined by F[0]=1, F[1]=1-z, F[k]=F[k-1]-z^2*F[k-2] for k>=2. The coefficients of these polynomials form the triangle A108299.

Rows may be obtained by taking finite differences of A205573 columns from the top -> down. - Gary W. Adamson, Feb 01 2012

A205573 Array M read by antidiagonals in which successive rows evidently converge to A001405 (central binomial coefficients).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 3, 5, 1, 1, 1, 2, 3, 6, 8, 1, 1, 1, 2, 3, 6, 10, 13, 1, 1, 1, 2, 3, 6, 10, 19, 21, 1, 1, 1, 2, 3, 6, 10, 20, 33, 34, 1, 1, 1, 2, 3, 6, 10, 20, 35, 61, 55, 1, 1, 1, 2, 3, 6, 10, 20, 35, 69, 108, 89, 1

Offset: 0

L. Edson Jeffery, Jan 29 2012

CONJECTURE 1. Let M(n,k) (n,k >= 0) denote the entry in row n and column k of the array. For all n, M(n,j) = A001405(j), j=0,...,2*n+1; hence row n of M -> A001405 as n -> infinity.

Taking finite differences of even numbered columns from the top -> down yields triangle A205946 with row sums A000984, central binomial coefficients; while odd numbered columns yield triangle A205945 with row sums A001700. A205946 and A205945 represent the bisection of A191314. - Gary W. Adamson, Feb 01 2012

			Array begins:
  1, 1, 1, 1, 1,  1,  1,  1,  1,   1,   1,...
  1, 1, 2, 3, 5,  8, 13, 21, 34,  55,  89,...
  1, 1, 2, 3, 6, 10, 19, 33, 61, 108, 197,...
  1, 1, 2, 3, 6, 10, 20, 35, 69, 124, 241,...
  1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 251,...
  1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252,...
  ...
According to Conjecture 2, row n = 3 has g.f. F_3(x) = (1-2*x^2)/(1-x-3*x^2+2*x^3+x^4).

L. E. Jeffery, Unit-primitive matrices

Cf. A001405, A108299, A115139, A191314.

Cf. also A205945, A205946, A001700, A000984.

Let N=2*n+3. For each n>0, define the (n+1) X (n+1) tridiagonal unit-primitive matrix (see [Jeffery]) B_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], and put B_0 = [1]. Then, for all n, M(n,k)=[(B_n)^k]{n+1,n+1}, k=0,1,..., where X{n+1,n+1} denotes the lower right corner entry of X.
CONJECTURE 2 (Rows of M). Let S(n,i) denote term i in row n of A115139, i=0,...,floor(n/2), and let T(n,j) denote term j in row n of A108299, j=0,...,n.  The generating function for row n of M is of the form F_n(x) =sum[i=0,...,floor(n/2) S(n,i)*x^(2*i)]/sum[j=0,...,n T(n,j)*x^j].
CONJECTURE 3 (Columns of M). Let D(m,k) denote term m in column k of A191314, m=0,...,floor(k/2). The generating function for column k of M is of the form G_k(x)=sum[m=0,...,floor(k/2) D(m,k)*x^m]/(1-x).

A124645 Triangle T(n,k), 0<=k<=n, read by rows given by [1,-1,0,0,0,0,0,...] DELTA [ -1,2,-1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938 .

Original entry on oeis.org

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, 0, 0, 0, 0, 1, -5, 10, -10, 5, -1, 0, 0, 0, 0, 0, 1, -5, 10, -10, 5, -1, 0, 0, 0, 0, 0, 1, -6, 15, -20, 15, -6, 1, 0, 0, 0, 0, 0, 0, 1, -6, 15, -20, 15, -6, 1

Offset: 0

Philippe Deléham, Jun 13 2007, Aug 22 2007

Matrix inverse of A108299.

			Triangle begins:
  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;
  0,  0,  0,  0,  1, -5, 10, -10,   5, -1;
  0,  0,  0,  0,  0,  1, -5,  10, -10,  5, -1;
  0,  0,  0,  0,  0,  1, -6,  15, -20, 15, -6, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

[(-1)^(k+Floor(n/2))*Binomial(Floor((n+1)/2), k-Floor(n/2)): k in [0..n], n in [0..12]]; // G. C. Greubel, May 01 2021

Table[(-1)^(Floor[n/2]+k)*Binomial[Floor[(n+1)/2], k-Floor[n/2]], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 01 2021 *)

flatten([[(-1)^(k+(n//2))*binomial(((n+1)//2), k-(n//2)) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 01 2021

Row n has g.f.: x^[n/2]*(1-x)^(n-[n/2]).
G.f.: (1-x*y+x)/(1-x^2*y+x^2*y^2). - R. J. Mathar, Aug 11 2015
T(n, k) = (-1)^(k + floor(n/2)) * binomial(floor((n+1)/2), k - floor(n/2)). - G. C. Greubel, May 01 2021

A039961 Triangle of coefficients in a Fibonacci-like sequence of polynomials.

Original entry on oeis.org

1, 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

Offset: 1

N. J. A. Sloane

Essentially the same as A108299. - Philippe Deléham, Feb 27 2014

			Triangle starts:
  1
  1
  1 -1
  1 -1 -1
  1 -1 -2 1
  1 -1 -3 2 1
  ...

A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.

Cf. A065941, A108299.

q_{n+2}(x) = x*q_{n+1}(x)-q_n(x), q_1(x) = q_2(x) = 1.

More terms from Philippe Deléham, Feb 27 2014

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A191314 Triangle read by rows: T(n,k) is the number of dispersed Dyck paths (i.e., Motzkin paths with no (1,0) steps at positive heights) of length n and height k.

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