A162515 -id:A162515 - cp's OEIS frontend

A168561 Riordan array (1/(1-x^2), x/(1-x^2)). Unsigned version of A049310.

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 1, 0, 3, 0, 1, 0, 3, 0, 4, 0, 1, 1, 0, 6, 0, 5, 0, 1, 0, 4, 0, 10, 0, 6, 0, 1, 1, 0, 10, 0, 15, 0, 7, 0, 1, 0, 5, 0, 20, 0, 21, 0, 8, 0, 1, 1, 0, 15, 0, 35, 0, 28, 0, 9, 0, 1, 0, 6, 0, 35, 0, 56, 0, 36, 0, 10, 0, 1, 1, 0, 21, 0, 70, 0, 84, 0, 45, 0, 11, 0, 1

Offset: 0

Philippe Deléham, Nov 29 2009

Row sums: A000045(n+1), Fibonacci numbers.
A168561*A007318 = A037027, as lower triangular matrices. Diagonal sums : A077957. - Philippe Deléham, Dec 02 2009
T(n,k) is the number of compositions of n+1 into k+1 odd parts. Example: T(4,2)=3 because we have 5 = 1+1+3 = 1+3+1 = 3+1+1.
Coefficients of monic Fibonacci polynomials (rising powers of x). Ftilde(n, x) = x*Ftilde(n-1, x) + Ftilde(n-2, x), n >=0, Ftilde(-1,x) = 0, Ftilde(0, x) = 1. G.f.: 1/(1 - x*z - z^2). Compare with Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jul 29 2014

			The triangle T(n,k) begins:
n\k 0  1   2   3   4    5    6    7    8    9  10  11  12  13 14 15 ...
0:  1
1:  0  1
2:  1  0   1
3:  0  2   0   1
4:  1  0   3   0   1
5:  0  3   0   4   0    1
6:  1  0   6   0   5    0    1
7:  0  4   0  10   0    6    0    1
8:  1  0  10   0  15    0    7    0    1
9:  0  5   0  20   0   21    0    8    0    1
10: 1  0  15   0  35    0   28    0    9    0   1
11: 0  6   0  35   0   56    0   36    0   10   0   1
12: 1  0  21   0  70    0   84    0   45    0  11   0   1
13: 0  7   0  56   0  126    0  120    0   55   0  12   0   1
14: 1  0  28   0 126    0  210    0  165    0  66   0  13   0  1
15: 0  8   0  84   0  252    0  330    0  220   0  78   0  14  0  1
... reformatted by _Wolfdieter Lang_, Jul 29 2014.
------------------------------------------------------------------------

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - From _N. J. A. Sloane_, May 10 2012
Tom Copeland, Addendum to Elliptic Lie Triad
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A162515 (rows reversed), A112552, A102426 (deflated).

A168561:=proc(n,k) if n-k mod 2 = 0 then binomial((n+k)/2,k) else 0 fi end proc:
seq(seq(A168561(n,k),k=0..n),n=0..12) ; # yields sequence in triangular form

Table[If[EvenQ[n + k], Binomial[(n + k)/2, k], 0], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 16 2017 *)

T(n,k) = if ((n+k) % 2, 0, binomial((n+k)/2,k));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print();); \\ Michel Marcus, Oct 09 2016

Sum_{k=0..n} T(n,k)*x^k = A059841(n), A000045(n+1), A000129(n+1), A006190(n+1), A001076(n+1), A052918(n), A005668(n+1), A054413(n), A041025(n), A099371(n+1), A041041(n), A049666(n+1), A041061(n), A140455(n+1), A041085(n), A154597(n+1), A041113(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 respectively. - Philippe Deléham, Dec 02 2009
T(2n,2k) = A085478(n,k). T(2n+1,2k+1) = A078812(n,k). Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000045(n+1), A006131(n), A015445(n), A168579(n), A122999(n) for x = 0,1,2,3,4,5 respectively. - Philippe Deléham, Dec 02 2009
T(n,k) = binomial((n+k)/2,k) if (n+k) is even; otherwise T(n,k)=0.
G.f.: (1-z^2)/(1-t*z-z^2) if offset is 1.
T(n,k) = T(n-1,k-1) + T(n-2,k), T(0,0) = 1, T(0,1) = 0. - Philippe Deléham, Feb 09 2012
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 09 2012
From R. J. Mathar, Feb 04 2022: (Start)
Sum_{k=0..n} T(n,k)*k = A001629(n+1).
Sum_{k=0..n} T(n,k)*k^2 = 0,1,4,11,... = 2*A055243(n)-A099920(n+1).
Sum_{k=0..n} T(n,k)*k^3 = 0,1,8,29,88,236,... = 12*A055243(n) -6*A001629(n+2) +A001629(n+1)-6*(A001872(n)-2*A001872(n-1)). (End)

Typo in name corrected (1(1-x^2) changed to 1/(1-x^2)) by Wolfdieter Lang, Nov 20 2010

A053119 Triangle of coefficients of Chebyshev's S(n,x) polynomials (exponents in decreasing order).

Original entry on oeis.org

1, 1, 0, 1, 0, -1, 1, 0, -2, 0, 1, 0, -3, 0, 1, 1, 0, -4, 0, 3, 0, 1, 0, -5, 0, 6, 0, -1, 1, 0, -6, 0, 10, 0, -4, 0, 1, 0, -7, 0, 15, 0, -10, 0, 1, 1, 0, -8, 0, 21, 0, -20, 0, 5, 0, 1, 0, -9, 0, 28, 0, -35, 0, 15, 0, -1, 1, 0, -10, 0, 36, 0, -56, 0, 35, 0, -6, 0, 1, 0, -11, 0, 45, 0, -84, 0, 70, 0, -21, 0, 1

Offset: 0

Wolfdieter Lang

These polynomials also give the determinant of the tridiagonal matrix having x on the diagonal and -1 next to these x. - M. F. Hasler, Oct 15 2019

The polynomial S(n,x) is the character of the irreducible (n+1) dimensional representation of the Lie algebra sl_2 when x is the character of irreducible 2-dimesional representation. - Leonid Bedratyuk, Oct 28 2023

			The triangle begins:
n\m 0  1   2  3   4  5   6  7   8  9  10 ...
0:  1
1:  1  0
2:  1  0  -1
3:  1  0  -2  0
4:  1  0  -3  0   1
5:  1  0  -4  0   3  0
6:  1  0  -5  0   6  0  -1
7:  1  0  -6  0  10  0  -4  0
8:  1  0  -7  0  15  0 -10  0   1
9:  1  0  -8  0  21  0 -20  0   5  0
10: 1  0  -9  0  28  0 -35  0  15  0  -1
... Reformatted. - _Wolfdieter Lang_, Dec 17 2013
E.g., fourth row (n=3) corresponds to polynomial S(3,x)= x^3-2*x.
Triangle of absolute values of coefficients (coefficients of Fibonacci polynomials) with exponents in increasing order begins:
[1]
[0, 1]
[1, 0, 1]
[0, 2, 0, 1]
[1, 0, 3, 0, 1]
[0, 3, 0, 4, 0, 1]
[1, 0, 6, 0, 5, 0, 1]
[0, 4, 0, 10, 0, 6, 0, 1]
[1, 0, 10, 0, 15, 0, 7, 0, 1]
[0, 5, 0, 20, 0, 21, 0, 8, 0, 1]
See A162515 for the Fibonacci polynomials with reversed row entries, starting there with row 1. - _Wolfdieter Lang_, Dec 16 2013

D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 232, Sect. 3.3.38.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

T. D. Noe, Rows n=0..100 of triangle, flattened
Tom Copeland, Addendum to Elliptic Lie Triad
Index entries for sequences related to Chebyshev polynomials.

Row sums give A000045. Reflection of A049310.

Cf. A162515. - Wolfdieter Lang, Dec 16 2013

A053119 := (n, k) -> if k::even then (-1)^binomial(k, 2)*binomial(n - k/2, k/2)
else 0 fi: seq(seq(A053119(n, k), k = 0..n), n = 0..11); # Peter Luschny, Jul 20 2024

ChebyshevS[n_, x_] := ChebyshevU[n, x/2]; Flatten[ Table[ Reverse[ CoefficientList[ ChebyshevS[n, x], x]], {n, 0, 12}]] (* Jean-François Alcover, Nov 25 2011 *)

tabl(nn) = for (n=0, nn, print(Vec(polchebyshev(n, 2, x/2)))); \\ Michel Marcus, Jan 14 2016

a(n,m) = A049310(n,n-m).
G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2).
Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with G.f. 1/(1-x*z-z^2).
a(n, m) := 0 if n < m or m odd, else ((-1)^(3*m/2))*binomial(n-m/2, n-m); a(n, m) = a(n-1, m) - a(n-2, m-2), a(n, -2) := 0 =: a(n, -1), a(0, 0) = 1, a(n, m) = 0 if n < m or m odd.
G.f. for m-th column (signed triangle): (-1)^(3*m/2)*x^m/(1-x)^(m/2+1) if m >= 0 is even else 0.
Recurrence for the (unsigned) Fibonacci polynomials: F[1]=1, F[2]=x; for n>2, F[n] = x*F[n-1]+F[n-2].
a = 2*A192011 - 3*A192174. - Thomas Baruchel, Jun 02 2018
Recurrence for the polynomials S(n) = x S(n-1) - S(n-2); S(0) = 1, S(1) = x. - M. F. Hasler, Oct 15 2019

A162517 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x + d)^n - (x - d)^n)/(2*d), where d = sqrt(x+4).

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 4, 4, 16, 0, 5, 10, 41, 8, 16, 6, 20, 86, 48, 96, 0, 7, 35, 161, 169, 348, 48, 64, 8, 56, 280, 456, 992, 384, 512, 0, 9, 84, 462, 1044, 2449, 1744, 2400, 256, 256, 10, 120, 732, 2136, 5482, 5920, 8640, 2560, 2560, 0, 11, 165, 1122, 4026, 11407, 16721, 26420, 14240, 14720, 1280, 1024

Offset: 1

Clark Kimberling, Jul 05 2009

			First six rows:
  0
  1
  2...0
  3...1...4
  4...4...16...0
  5...10..41...8...16

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened

Cf. A162514, A162515, A162516.

Cf. A002605, A063727.

m:=12;
Q:= func< n,x | ((x+Sqrt(x+4))^n - (x-Sqrt(x+4))^n)/(2*Sqrt(x+4)) >;
R:=PowerSeriesRing(Rationals(), m+1);
T:= func< n,k | Coefficient(R!( Q(n, x) ), n-k) >;
[0] cat [T(n,k): k in [1..n], n in [1..m]]; // G. C. Greubel, Jul 09 2023

Q[n_, x_]:= Q[n, x]= ((x+Sqrt[x+4])^n -(x-Sqrt[x+4])^n)/(2*Sqrt[x+4]);
T[n_, k_]:= Coefficient[Series[P[n,x], {x,0,n-k+1}], x, n-k];
Join[{0}, Table[T[n,k], {n,12}, {k,n}]//Flatten] (* G. C. Greubel, Jul 09 2023 *)

def Q(n,x): return ((x+sqrt(x+4))^n - (x-sqrt(x+4))^n)/(2*sqrt(x+4))
def T(n,k):
    P. = PowerSeriesRing(QQ)
    return P( Q(n,x) ).list()[n-k]
[0]+flatten([[T(n,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Jul 09 2023

Q(n,x) = (P(n+1, x) - x*P(n,x))/(x+4), where P(n, x) is the n-th polynomial of A162516.
Q(n, x) also has the recurrence Q(n, x) = 2*x*Q(n-1, x) - (x^2 - x - 4)*Q(n-2, x).
From G. C. Greubel, Jul 09 2023: (Start)
T(n, k) = [x^(n-k)](((x+sqrt(x+4))^n -(x-sqrt(x+4))^n)/(2*sqrt(x+4))).
Sum_{k=1..n-1} T(n, k) = A063727(n-2), n >= 2.
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = A002605(n-1). (End)

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

A162514 Triangle of coefficients of polynomials defined by the Binet form P(n,x) = U^n + L^n, where U = (x + d)/2, L = (x - d)/2, d = (4 + x^2)^(1/2). Decreasing powers of x.

Original entry on oeis.org

2, 1, 0, 1, 0, 2, 1, 0, 3, 0, 1, 0, 4, 0, 2, 1, 0, 5, 0, 5, 0, 1, 0, 6, 0, 9, 0, 2, 1, 0, 7, 0, 14, 0, 7, 0, 1, 0, 8, 0, 20, 0, 16, 0, 2, 1, 0, 9, 0, 27, 0, 30, 0, 9, 0, 1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2, 1, 0, 11, 0, 44, 0, 77, 0, 55, 0, 11, 0, 1, 0, 12, 0, 54, 0, 112, 0, 105, 0, 36, 0, 2, 1, 0, 13, 0

Offset: 0

Clark Kimberling, Jul 05 2009

For a signed version of this triangle corresponding to the row reversed version of the triangle A127672 see A244422. - Wolfdieter Lang, Aug 07 2014

The row reversed triangle is A114525. - Paolo Bonzini, Jun 23 2016

			Triangle begins
   2;  == 2
   1, 0;  == x + 0
   1, 0,  2;  == x^2 + 2
   1, 0,  3, 0;  == x^3 + 3*x + 0
   1, 0,  4, 0,  2;
   1, 0,  5, 0,  5, 0;
   1, 0,  6, 0,  9, 0,  2;
   1, 0,  7, 0, 14, 0,  7, 0;
   1, 0,  8, 0, 20, 0, 16, 0,  2;
   1, 0,  9, 0, 27, 0, 30, 0,  9, 0;
   1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2;
   ...
From _Wolfdieter Lang_, Aug 07 2014: (Start)
The row polynomials R(n, x) are:
  R(0, x) = 2, R(1, x) = 1 =   x*P(1,1/x),  R(2, x) = 1 + 2*x^2 = x^2*P(2,1/x), R(3, x) = 1 + 3*x^2 = x^3*P(3,1/x), ...
(End)

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A000032, A114525, A162515, A162516, A162517.

Table[Reverse[CoefficientList[LucasL[n, x], x]], {n, 0, 12}]//Flatten  (* G. C. Greubel, Nov 05 2018 *)

P(n)=
{
    local(U, L, d, r, x);
    if ( n<0, return(0) );
    x = 'x+O('x^(n+1));
    d=(4 + x^2)^(1/2);
    U=(x+d)/2;  L=(x-d)/2;
    r = U^n+L^n;
    r = truncate(r);
    return( r );
}
for (n=0, 10, print(Vec(P(n))) ); /* show triangle */
/* Joerg Arndt, Jul 24 2011 */

P(n,x) = x*P(n-1,x) + P(n-2,x) for n >= 2, P(0,x) = 2, P(1,x) = x.
From Wolfdieter Lang, Aug 07 2014: (Start)
T(n,m) = [x^(n-m)] P(n,x), m = 0, 1, ..., n and  n >= 0.
G.f. of polynomials P(n,x): (2 - x*z)/(1 - x*z - z^2).
G.f. of row polynomials R(n,x) = Sum_{m=0..n} T(n,m)*x^m:  (2 - z)/(1 - z - (x*z)^2) (rows for P(n,x) reversed).
(End)
For n > 0, T(n,2*m+1) = 0, T(n,2*m) = A034807(n,m). - Paolo Bonzini, Jun 23 2016

Name clarified by Wolfdieter Lang, Aug 07 2014

A162516 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x+d)^n + (x-d)^n)/2, where d=sqrt(x+4).

Original entry on oeis.org

1, 1, 0, 1, 1, 4, 1, 3, 12, 0, 1, 6, 25, 8, 16, 1, 10, 45, 40, 80, 0, 1, 15, 75, 121, 252, 48, 64, 1, 21, 119, 287, 644, 336, 448, 0, 1, 28, 182, 588, 1457, 1360, 1888, 256, 256, 1, 36, 270, 1092, 3033, 4176, 6240, 2304, 2304, 0, 1, 45, 390, 1890, 5925, 10801, 17780, 11680, 12160, 1280, 1024

Offset: 0

Clark Kimberling, Jul 05 2009

			First six rows:
  1;
  1,  0;
  1,  1,  4;
  1,  3, 12,  0;
  1,  6, 25,  8, 16;
  1, 10, 48, 40, 80, 0;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A162514, A162515, A162517.
Cf. A000217, A026150, A084057, A125818, A199572.
For fixed k, the sequences P(n,k), for n=1,2,3,4,5, are A084057, A084059, A146963, A081342, A081343, respectively.

m:=12;
p:= func< n,x | ((x+Sqrt(x+4))^n + (x-Sqrt(x+4))^n)/2 >;
R:=PowerSeriesRing(Rationals(), m+1);
T:= func< n,k | Coefficient(R!( p(n,x) ), n-k) >;
[T(n,k): k in [0..n], n in [0..m]]; // G. C. Greubel, Jul 09 2023

P[n_, x_]:= P[n, x]= ((x+Sqrt[x+4])^n + (x-Sqrt[x+4])^n)/2;
T[n_, k_]:= Coefficient[Series[P[n, x], {x,0,n-k+1}], x, n-k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 08 2020; Jul 09 2023 *)

def p(n,x): return ((x+sqrt(x+4))^n + (x-sqrt(x+4))^n)/2
def T(n,k):
    P. = PowerSeriesRing(QQ)
    return P( p(n,x) ).list()[n-k]
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 09 2023

P(n,x) = 2*x*P(n-1,x) - (x^2 -x -4)*P(n-2,x).
From G. C. Greubel, Jul 09 2023: (Start)
T(n, k) = [x^(n-k)] ( ((x+sqrt(x+4))^n + (x-sqrt(x+4))^n)/2 ).
T(n, 1) = A000217(n-1), n >= 1.
T(n, n) = A199572(n).
Sum_{k=0..n} T(n, k) = A084057(n).
Sum_{k=0..n} 2^k*T(n, k) = A125818(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A026150(n).
Sum_{k=0..n} (-2)^k*T(n, k) = A133343(n). (End)

A374441 Triangle read by rows: T(n, k) = binomial(n - floor(k/2), ceiling(k/2)) - binomial(n - ceiling(k/2), floor(k/2)).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 1, 0, 0, 4, 0, 3, 0, 0, 0, 5, 0, 6, 0, 1, 0, 0, 6, 0, 10, 0, 4, 0, 0, 0, 7, 0, 15, 0, 10, 0, 1, 0, 0, 8, 0, 21, 0, 20, 0, 5, 0, 0, 0, 9, 0, 28, 0, 35, 0, 15, 0, 1, 0, 0, 10, 0, 36, 0, 56, 0, 35, 0, 6, 0, 0, 0, 11, 0, 45, 0, 84, 0, 70, 0, 21, 0, 1, 0

Offset: 0

Peter Luschny, Jul 19 2024

Member of the family of Fibonacci polynomials (A011973, A162515, ...) and Chebyshev polynomials (A053119).

			Triangle starts:
  [ 0]  0;
  [ 1]  0, 0;
  [ 2]  0, 1, 0;
  [ 3]  0, 2, 0,  0;
  [ 4]  0, 3, 0,  1, 0;
  [ 5]  0, 4, 0,  3, 0,  0;
  [ 6]  0, 5, 0,  6, 0,  1, 0;
  [ 7]  0, 6, 0, 10, 0,  4, 0,  0;
  [ 8]  0, 7, 0, 15, 0, 10, 0,  1, 0;
  [ 9]  0, 8, 0, 21, 0, 20, 0,  5, 0, 0;
  [10]  0, 9, 0, 28, 0, 35, 0, 15, 0, 1, 0;

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
Henry W. Gould, A Variant of Pascal's Triangle, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271, with corrections.

Cf. A374440 (odd columns agree).
Cf. A000071 (row sums), A065941, A194005, A103631, A007318.
Cf. A011973, A162515, A053119.

T := (n, k) -> if k::even then 0 else binomial(n - (k + 1)/2, (k + 1)/2) fi:
# Or as a recurrence:
T := proc(n, k) option remember; if k::even or k > n then 0 elif k = 1 then n - 1 else T(n - 1, k) + T(n - 2, k - 2) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..12);

A374441[n_, k_] := If[OddQ[k], Binomial[n - (k + 1)/2, (k + 1)/2], 0];
Table[A374441[n, k], {n, 0, 15}, {k, 0, n}] (* Paolo Xausa, Nov 16 2024 *)

from math import isqrt, comb
def A374441(n):
    a = (m:=isqrt(k:=n+1<<1))-(k<=m*(m+1))
    b = n-comb(a+1,2)
    return comb(a-(b+1>>1),b+1>>1) if b&1 else 0 # Chai Wah Wu, Nov 14 2024

from math import comb as binomial
def row(n: int) -> list[int]:
    return [binomial(n - (k+1)//2, (k+1)//2) if k%2 else 0 for k in range(n+1)]
for n in range(11): print(row(n))  # Peter Luschny, Nov 21 2024

T(n, k) = [x^(n-k)][z^n] (x / (1 - x*z - z^2)).
T(n, k) = binomial(n - (k + 1)/2, (k + 1)/2) if k is odd, and otherwise 0.
Sum_{k=0..n} T(n, k) = Fibonacci(n + 1) - 1.
Columns with odd index agree with the odd indexed columns of A374440.

A374438 Triangle read by rows: T(n, k) = T(n - 1, k) + T(n - 2, k - 2), with initial values T(n, k) = k + 1 for k < 3.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 3, 2, 1, 2, 3, 4, 3, 1, 2, 3, 6, 6, 2, 1, 2, 3, 8, 9, 6, 3, 1, 2, 3, 10, 12, 12, 9, 2, 1, 2, 3, 12, 15, 20, 18, 8, 3, 1, 2, 3, 14, 18, 30, 30, 20, 12, 2, 1, 2, 3, 16, 21, 42, 45, 40, 30, 10, 3, 1, 2, 3, 18, 24, 56, 63, 70, 60, 30, 15, 2

Offset: 0

Peter Luschny, Jul 22 2024

See A374439 and the cross-references for comments about this family of triangles, where the recurrence is defined as in the name, but with an additional parameter m for the initial values: T(n, k) = k + 1 for k < m.

As m -> oo, the rows of the triangles become the initial segments of the integers.

			Triangle starts:
  [ 0] [1]
  [ 1] [1, 2]
  [ 2] [1, 2, 3]
  [ 3] [1, 2, 3,  2]
  [ 4] [1, 2, 3,  4,  3]
  [ 5] [1, 2, 3,  6,  6,  2]
  [ 6] [1, 2, 3,  8,  9,  6,  3]
  [ 7] [1, 2, 3, 10, 12, 12,  9,  2]
  [ 8] [1, 2, 3, 12, 15, 20, 18,  8,  3]
  [ 9] [1, 2, 3, 14, 18, 30, 30, 20, 12,  2]
  [10] [1, 2, 3, 16, 21, 42, 45, 40, 30, 10, 3]

Family of triangles: A162515 (m=1, Fibonacci), A374439 (m=2, Lucas), this triangle (m=3).
Row sums: A187890 (apart from initial terms), also A001060 + 1 (with 1 prepended).
Cf. A006355 (odd sums), A187893 (even sums).
Cf. related to deltas: A065220, A210673.

M := 3;  # family index
T := proc(n, k) option remember; if k > n then 0 elif k < M then k + 1 else
T(n - 1, k) + T(n - 2, k - 2) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..11);

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k > n: return 0
    if k < 3: return k + 1
    return T(n - 1, k) + T(n - 2, k - 2)

A290864 Numbers k such that the k-th Fibonacci polynomial evaluated at k is prime.

Original entry on oeis.org

2, 5, 71, 8419

Offset: 1

Bobby Jacobs, Aug 12 2017

Numbers k such that A084844(k) = A117715(k,k) is prime.
a(5) > 9200. - Giovanni Resta, Aug 13 2017
Except for a(1), all terms == 1 or 5 (mod 6). - Robert Israel, Aug 13 2017

			5 is in the sequence because A117715(5,5) = 701 is prime.

Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Wikipedia, Fibonacci polynomials

Cf. A000045, A011973, A084844, A117715, A162515, A168561.

select(t -> isprime(combinat:-fibonacci(t,t)), [2,seq(seq(6*i+j,j=[1,5]),i=0..100)]); # Robert Israel, Aug 13 2017

Select[Range[100], PrimeQ@ Fibonacci[#, #] &] (* Giovanni Resta, Aug 13 2017 *)

a(4) from Giovanni Resta, Aug 13 2017

A346038 Triangle read by rows T(n, k) such that Fib(n, x+1) = Sum_{k=1..n} T(n, k)*Fib(k, x) where Fib(n, x) is the n-th Fibonacci polynomial.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 0, 3, 3, 1, -2, 2, 6, 4, 1, -4, -3, 7, 10, 5, 1, -3, -12, 0, 16, 15, 6, 1, 5, -18, -21, 11, 30, 21, 7, 1, 20, -4, -50, -24, 35, 50, 28, 8, 1, 29, 48, -51, -98, -9, 78, 77, 36, 9, 1, 1, 124, 45, -164, -150, 42, 147, 112, 45, 10, 1, -94, 128, 282, -67, -365, -177, 154, 250, 156, 55, 11, 1

Offset: 1

Michel Marcus, Jul 02 2021

			Triangle begins:
   1;
   1,  1;
   1,  2, 1;
   0,  3, 3,  1;
  -2,  2, 6,  4, 1;
  -4, -3, 7, 10, 5, 1;
  ...
The first 3 Fibonacci polynomials are 1, x, x^2 + 1. So F3(n, x+1) = x^2 + 2*x + 2  = 1*1 + 2*x + 1*(x^2+1) = 1*F(1,x) + 2*F(2, x) + 1*F(3,x), so the 3rd row is [1, 2, 1].

Michel Marcus, Rows n=1..100 of triangle, flattened
Math StackExchange, Variable transformation x->x+1 in Fibonacci-polynomial Fn(x)
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

Cf. A000012, A000027, A000217, A005581: diagonals.

Cf. A162515 and A168561 (Fibonacci polynomials coefficients).

rowV(n) = my(v= if (n==0, [0], n--; vector(n+1, k, k--; if (k%2==0, binomial(n-k/2, k/2))))); Pol(v); \\ A162515
rowT(n, vfp, vfp1) = {my(vp1 = vfp1[n], vc = vector(n), i=n); forstep (k = poldegree(vp1), 0, -1, vc[i] = polcoef(vp1, k)/polcoef(vfp[k+1], k); vp1 -= vfp[k+1]*vc[i]; i--;); vc;}
tabl(nn) = {my(vfp = vector(nn, k, rowV(k))); my(vfp1 = vector(nn, k, subst(vfp[k], x, x+1))); for(n=1, nn, print((rowT(n, vfp, vfp1))););}

cp's OEIS Frontend

A168561 Riordan array (1/(1-x^2), x/(1-x^2)). Unsigned version of A049310.

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A053119 Triangle of coefficients of Chebyshev's S(n,x) polynomials (exponents in decreasing order).

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A162517 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x + d)^n - (x - d)^n)/(2*d), where d = sqrt(x+4).

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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.

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A162514 Triangle of coefficients of polynomials defined by the Binet form P(n,x) = U^n + L^n, where U = (x + d)/2, L = (x - d)/2, d = (4 + x^2)^(1/2). Decreasing powers of x.

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A162516 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x+d)^n + (x-d)^n)/2, where d=sqrt(x+4).

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