A084844 -id:A084844 - cp's OEIS frontend

A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 10, 12, 5, 0, 1, 5, 17, 33, 29, 8, 0, 1, 6, 26, 72, 109, 70, 13, 0, 1, 7, 37, 135, 305, 360, 169, 21, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 0, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89, 0, 1, 11, 101, 747, 4289, 18200, 53353, 98145, 98209, 42837, 5741, 144

Offset: 1

Roger L. Bagula, Jan 29 2010

Equals A073133 with an additional column A(.,0).
If the first column and top row are deleted, antidiagonal reading yields A118243.
Adding a top row of 1's and antidiagonal reading downwards yields A157103.
Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....
From Jianing Song, Jul 14 2018: (Start)
All rows have strong divisibility, that is, gcd(A(n,k_1), A(n,k_2)) = A(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0.
Let E(n,m) be the smallest number l such that m divides A(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime.
E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.
Let pi(n,m) be the Pisano period of A(n, k) modulo m, i.e, the smallest number l such that A(n, k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n.
pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n, p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1), pi(n,m_2)) if gcd(m_1,m_2) = 1.
If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e.
Let z(n,m) be the number of zeros in a period of A(n, k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1.
Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:
((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4
......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*
......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*
......1.........3,7.............1...................0...................0
.....-1.........1,5.............0...................0...................1
.....-1.........3,7.............0...................1...................0
* The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019]
z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.
(End)
From Michael A. Allen, Mar 06 2023: (Start)
Removing the first (n=0) row of A352361 gives this sequence.
Row n is the n-metallonacci sequence.
A(n,k) is (for k>0) the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			The array, A(n, k), starts in row n = 1 with columns k >= 0 as
  0      1      1      2      3      5      8
  0      1      2      5     12     29     70
  0      1      3     10     33    109    360
  0      1      4     17     72    305   1292
  0      1      5     26    135    701   3640
  0      1      6     37    228   1405   8658
  0      1      7     50    357   2549  18200
  0      1      8     65    528   4289  34840
  0      1      9     82    747   6805  61992
  0      1     10    101   1020  10301 104030
  0      1     11    122   1353  15005 166408
Antidiagonal triangle, T(n, k), begins as:
  0;
  0, 1;
  0, 1, 1;
  0, 1, 2,  2;
  0, 1, 3,  5,   3;
  0, 1, 4, 10,  12,   5;
  0, 1, 5, 17,  33,  29,    8;
  0, 1, 6, 26,  72, 109,   70,   13;
  0, 1, 7, 37, 135, 305,  360,  169,  21;
  0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34;

Jianing Song, Antidiagonals n = 1..100, flattened (the first 30 antidiagonals from Vincenzo Librandi)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Rows n include: A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9), A041041 (n=10), A049666 (n=11), A041061 (n=12), A140455 (n=13), A041085 (n=14), A154597 (n=15), A041113 (n=16), A178765 (n=17), A041145 (n=18), A243399 (n=19), A041181 (n=20). (Note that there are offset shifts for rows n = 5, 7, 8, 10, 12, 14, 16..20.)
Columns k include: A000004 (k=0), A000012 (k=1), A000027 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Entry points for A(n,k) modulo m: A001177 (n=1), A214028 (n=2), A322907 (n=3).
Pisano period for A(n,k) modulo m: A001175 (n=1), A175181 (n=2), A175182 (n=3), A175183 (n=4), A175184 (n=5), A175185 (n=6).
Number of zeros in a period for A(n,k) modulo m: A001176 (n=1), A214027 (n=2), A322906 (n=3).
Cf. A084844, A084845, A118243, A157103, A271782, A316269.
Sums include: A304357, A304359.
Similar to: A073133.

A172236:= func< n,k | k le 1 select k else Evaluate(DicksonSecond(k-1,-1), n-k) >;
[A172236(n,k): k in [0..n-1], n in [1..13]]; // G. C. Greubel, Sep 29 2024

A172236[n_,k_]:=Fibonacci[k, n-k];
Table[A172236[n, k], {n,15}, {k,0,n-1}]//Flatten

A(n, k) = if (k==0, 0, if (k==1, 1, n*A(n, k-1) + A(n, k-2)));
tabl(nn) = for(n=1, nn, for (k=0, nn, print1(A(n, k), ", ")); print); \\ Jianing Song, Jul 14 2018 (program from Michel Marcus; see also A316269)

A(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018

def A172236(n,k): return sum(binomial(k-j-1,j)*(n-k)^(k-2*j-1) for j in range(1+(k-1)//2))
flatten([[A172236(n,k) for k in range(n)] for n in range(1,14)]) # G. C. Greubel, Sep 29 2024

A(n,k) = (((n + sqrt(n^2 + 4))/2)^k - ((n-sqrt(n^2 + 4))/2)^k)/sqrt(n^2 + 4), n >= 1, k >= 0. - Jianing Song, Jun 27 2018
For n >= 1, Sum_{i=1..k} 1/A(n,2^i) = ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/A(n,2^k)), where u = (n + sqrt(n^2 + 4))/2, v = (n - sqrt(n^2 + 4))/2 are the two roots of the polynomial x^2 - n*x - 1. As a result, Sum_{i>=1} 1/A(n,2^i) = (n^2 + 4 - n*sqrt(n^2 + 4))/(2*n). - Jianing Song, Apr 21 2019
From G. C. Greubel, Sep 29 2024: (Start)
A(n, k) = F_{k}(n) (Fibonacci polynomials F_{n}(x)) (array).
T(n, k) = F_{k}(n-k) (antidiagonal triangle).
Sum_{k=0..n-1} T(n, k) = A304357(n) - (1-(-1)^n)/2.
Sum_{k=0..n-1} (-1)^k*T(n, k) = (-1)*A304359(n) + (1-(-1)^n)/2.
T(2*n, n) = A084844(n).
T(2*n+1, n+1) = A084845(n). (End)

More terms from Jianing Song, Jul 14 2018

A352361 Array read by ascending antidiagonals. A(n, k) = Fibonacci(k, n), where Fibonacci(n, x) are the Fibonacci polynomials.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 2, 2, 0, 0, 1, 3, 5, 3, 1, 0, 1, 4, 10, 12, 5, 0, 0, 1, 5, 17, 33, 29, 8, 1, 0, 1, 6, 26, 72, 109, 70, 13, 0, 0, 1, 7, 37, 135, 305, 360, 169, 21, 1, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 1

Offset: 0

Peter Luschny, Mar 18 2022

From Michael A. Allen, Mar 26 2023: (Start)
Row n is the n-metallonacci sequence for n>0.
A(n,k), for n > 0 and k > 0, is the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			Array, A(n,k), starts:
  n\k 0, 1, 2,  3,   4,    5,     6,      7,       8,        9, ...
  -------------------------------------------------------------------------
  [0] 0, 1, 0,  1,   0,    1,     0,      1,       0,        1, ... A000035;
  [1] 0, 1, 1,  2,   3,    5,     8,     13,      21,       34, ... A000045;
  [2] 0, 1, 2,  5,  12,   29,    70,    169,     408,      985, ... A000129;
  [3] 0, 1, 3, 10,  33,  109,   360,   1189,    3927,    12970, ... A006190;
  [4] 0, 1, 4, 17,  72,  305,  1292,   5473,   23184,    98209, ... A001076;
  [5] 0, 1, 5, 26, 135,  701,  3640,  18901,   98145,   509626, ... A052918;
  [6] 0, 1, 6, 37, 228, 1405,  8658,  53353,  328776,  2026009, ... A005668;
  [7] 0, 1, 7, 50, 357, 2549, 18200, 129949,  927843,  6624850, ... A054413;
  [8] 0, 1, 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, ... A041025;
  [9] 0, 1, 9, 82, 747, 6805, 61992, 564733, 5144589, 46866034, ... A099371;
      |  |  |  | A054602 |   A124152;
      |  |  |  A002522   A057721;
      |  |  A001477;
      |  A000012;
      A000004;
Antidiagonals, T(n, k), begin as:
  0;
  0, 1;
  0, 1, 0;
  0, 1, 1,  1;
  0, 1, 2,  2,   0;
  0, 1, 3,  5,   3,   1;
  0, 1, 4, 10,  12,   5,   0;
  0, 1, 5, 17,  33,  29,   8,   1;
  0, 1, 6, 26,  72, 109,  70,  13,  0;
  0, 1, 7, 37, 135, 305, 360, 169, 21, 1;

G. C. Greubel, Antidiagonals n = 0..50, flattened
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Other versions of this array are A073133, A157103, A172236.
Rows n: A000035 (n=0), A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9).
Columns k: A000004 (k=0), A000012 (k=1), A001477 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Cf. A084844 (main diagonal), A352362 (Lucas polynomials), A350470 (Jacobsthal polynomials).
Sums include: A304357 (row sums), A304359.
Cf. A084845.

A352361:= func< n, k | k le 1 select k else Evaluate(DicksonSecond(k-1, -1), n-k) >;
[A352361(n, k): k in [0..n], n in [0..13]]; // G. C. Greubel, Sep 29 2024

seq(seq(combinat:-fibonacci(k, n - k), k = 0..n), n = 0..11);

Table[Fibonacci[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten
(* or *)
A[n_, k_] := With[{s = Sqrt[n^2 + 4]}, ((n + s)^k - (n - s)^k) / (2^k*s)];
Table[Simplify[A[n, k]], {n, 0, 9}, {k, 0, 9}] // TableForm

A(n, k) = ([1, k; 1, k-1]^n)[2, 1] ;
export(A)
for(k = 0, 9, print(parvector(10, n, A(n - 1, k))))

def A352361(n, k): return lucas_number1(k,n-k,-1)
flatten([[A352361(n, k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Sep 29 2024

A(n, k) = Sum_{j=0..floor((k-1)/2)} binomial(k-j-1, j)*n^(k-2*j-1).
A(n, k) = ((n + s)^k - (n - s)^k) / (2^k*s) where s = sqrt(n^2 + 4).
A(n, k) = [x^k] (x / (1 - n*x - x^2)).
A(n, k) = n^(k-1)*hypergeom([1 - k/2, 1/2 - k/2], [1 - k], -4/n^2) for n,k >= 1.
A(n, n) = T(2*n, n) = A084844(n).
From G. C. Greubel, Sep 29 2024: (Start)
T(n, k) = A(n-k, k) (antidiagonal triangle).
T(2*n+1, n+1) = A084845(n).
Sum_{k=0..n} T(n, k) = A304357(n) (row sums).
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)*A304359(n). (End)

A157103 Array A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 3, 1, 1, 5, 12, 10, 4, 1, 1, 8, 29, 33, 17, 5, 1, 1, 13, 70, 109, 72, 26, 6, 1, 1, 21, 169, 360, 305, 135, 37, 7, 1, 1, 34, 408, 1189, 1292, 701, 228, 50, 8, 1, 1, 55, 985, 3927, 5473, 3640, 1405, 357, 65, 9, 1

Offset: 0

Gary W. Adamson, Feb 22 2009

From Michael A. Allen, Mar 30 2023: (Start)
Column k is the k-metallonacci sequence for k > 0.
T(n,k) is, for n > 0 and k > 0, the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are k kinds of squares available. (End)

			Array begins:
  1,  1,   1,    1,     1,     1,      1,      1, ... (A000012);
  1,  1,   2,    3,     4,     5,      6,      7, ... (A000027);
  1,  2,   5,   10,    17,    26,     37,     50, ... (A002522);
  1,  3,  12,   33,    72,   135,    228,    357, ...;
  1,  5,  29,  109,   305,   701,   1405,   2549, ...;
  1,  8,  70,  360,  1292,  3640,   8658,  18200, ...;
  1, 13, 169, 1189,  5473, 18901,  53353, 129949, ...;
  1, 21, 408, 3927, 23184, 98145, 328776, 927843, ...;
  ...
First few rows of the triangle:
  1;
  1,   1;
  1,   1,    1;
  1,   2,    2,     1;
  1,   3,    5,     3,     1;
  1,   5,   12,    10,     4,     1;
  1,   8,   29,    33,    17,     5,     1;
  1,  13,   70,   109,    72,    26,     6,     1;
  1,  21,  169,   360,   305,   135,    37,     7,    1;
  1,  34,  408,  1189,  1292,   701,   228,    50,    8,   1;
  1,  55,  985,  3927,  5473,  3640,  1405,   357,   65,   9,   1;
  1,  89, 2378, 12970, 23184, 18901,  8658,  2549,  528,  82,  10,  1;
  1, 144, 5741, 42837, 98209, 98145, 53353, 18200, 4289, 747, 101, 11, 1;
  ...
Example: Column 3 = (1, 3, 10, 33, 109, 360, ...) = A006190.

G. C. Greubel, Antidiagonals n = 0..50, flattened
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894 [math.NT], 2015. See Table 3.

Columns k=1 to 9: A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371.
Cf. A084844, A084845.
Essentially the transpose of A073133, A172236, A352361.

A157103:= func< n,k | k eq 0 or k eq n select 1 else Evaluate(DicksonSecond(n, -1), k) >;
[A157103(n-k, k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jan 11 2022

A157103 := proc(n,k)
    if k = 0 then
        1;
    else
        mul(k-2*I*cos(l*Pi/(n+1)),l=1..n) ;
        combine(%,trig) ;
        round(%) ;
    end if;
end proc:
seq( seq(A157103(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Feb 27 2023

(* First program *)
T[, 0]=1; T[n, n_]=1; T[, ]=0;
T[n_, k_] /; 0 <= k <= n := k T[n-1, k] + T[n-2, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Aug 07 2018 *)
(* Second program *)
T[n_, k_]:= If[k==0 || k==n, 1, Fibonacci[n-k+1, k]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 11 2022 *)

def A157103(n,k): return 1 if (k==0 or k==n) else lucas_number1(n+1, k, -1)
flatten([[A157103(n-k, k) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Jan 11 2022

A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1 (array).
A(n, 1) = A000045(n+1).
T(n, k) = k*T(n-1, k) + T(n-2, k) with T(n, 0) = T(n, n) = 1 (triangle).
From G. C. Greubel, Jan 11 2022: (Start)
T(n, k) = Fibonacci(n-k+1, k), with T(n, 0) = T(n, n) = 1.
T(2*n, n) = A084845(n) for n >= 1, with T(0, 0) = 1.
T(2*n+1, n+1) = A084844(n). (End)

Edited by G. C. Greubel, Jan 11 2022

A084845 Numerators of the continued fraction n+1/(n+1/...) [n times].

Original entry on oeis.org

1, 5, 33, 305, 3640, 53353, 927843, 18674305, 426938895, 10928351501, 309601751184, 9616792908241, 324971855514293, 11868363584907985, 465823816409224245, 19553538801258341377, 874091571490181406680

Offset: 1

Hollie L. Buchanan II, Jun 08 2003

The n-th term of the Lucas sequence U(n,-1). The denominator is the (n-1)-th term. Adjacent terms of the sequence U(n,-1) are relatively prime. - T. D. Noe, Aug 19 2004

			a(4) = 305 since 4+1/(4+1/(4+1/4)) = 305/72.

Alois P. Heinz, Table of n, a(n) for n = 1..386
Eric Weisstein's World of Mathematics, Lucas Sequence

Cf. A084844 (denominators).

Cf. A097690, A097691, A117715.

A084845 := proc(n)
    fibonacci(n+1,n) ;
end proc:
seq(A084845(n),n=1..20) ; # Zerinvary Lajos, Dec 01 2006

myList[n_] := Module[{ex = {n}}, Do[ex = {ex, n}, {n - 1}]; Flatten[ex]] Table[Numerator[FromContinuedFraction[myList[n]]], {n, 1, 20}]
Table[s=n; Do[s=n+1/s, {n-1}]; Numerator[s], {n, 20}] (* T. D. Noe, Aug 19 2004 *)

{a(n)=polcoeff(1/(1-n*x-x^2+x*O(x^n)),n)} \\ Paul D. Hanna, Dec 27 2012

from sympy import fibonacci
def a117715(n, m): return 0 if n==0 else fibonacci(n, m)
def a(n): return a117715(n + 1, n)
print([a(n) for n in range(1, 31)]) # Indranil Ghosh, Aug 12 2017

a(n) = Sum_{k=0..floor(n/2)}* binomial(n-k, k)*n^(n-2k). - Michel Lagneau
a(n) = [x^n] 1/(1 - n*x - x^2). - Paul D. Hanna, Dec 27 2012
a(n) = (s^(n+1) - (-s)^(-n-1))/(2*s - n), where s = (n + sqrt(n^2 + 4))/2. - Vladimir Reshetnikov, May 07 2016
a(n) = A117715(n+1,n). - Alois P. Heinz, Aug 12 2017

A097690 Numerators of the continued fraction n-1/(n-1/...) [n times].

Original entry on oeis.org

1, 3, 21, 209, 2640, 40391, 726103, 15003009, 350382231, 9127651499, 262424759520, 8254109243953, 281944946167261, 10393834843080975, 411313439034311505, 17391182043967249409, 782469083251377707328

Offset: 1

T. D. Noe, Aug 19 2004

The n-th term of the Lucas sequence U(n,1). The denominator is the (n-1)-th term. Adjacent terms of the sequence U(n,1) are relatively prime.

			a(4) = 209 because 4-1/(4-1/(4-1/4)) = 209/56.

Alois P. Heinz, Table of n, a(n) for n = 1..386
Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, Metered Parking Functions, arXiv:2406.12941 [math.CO], 2024. See pp. 3, 8, 10, 22.
Pascual Jara and Miguel L. Rodríguez, Solving quadratic congruences, Arhimede Math. J. (2020) Vol. 7, No. 2, 105-120.
Eric Weisstein's World of Mathematics, Lucas Sequence
Wikipedia, Chebyshev polynomials.

Cf. A084844, A084845, A097691 (denominators), A179943, A323118.

Table[s=n; Do[s=n-1/s, {n-1}]; Numerator[s], {n, 20}]
Table[DifferenceRoot[Function[{y, m}, {y[1 + m] == n*y[m] - y[m - 1], y[0] == 1, y[1] == n}]][n], {n, 1, 20}] (* Benedict W. J. Irwin, Nov 05 2016 *)

{a(n)=polcoeff(1/(1-n*x+x^2+x*O(x^n)), n)} \\ Paul D. Hanna, Dec 27 2012

a(n) = polchebyshev(n, 2, n/2); \\ Seiichi Manyama, Mar 03 2021

a(n) = sum(k=0, n, (n-2)^k*binomial(n+1+k, 2*k+1)); \\ Seiichi Manyama, Mar 03 2021

[lucas_number1(n,n-1,1) for n in range(19)] # Zerinvary Lajos, Jun 25 2008

a(n) = [x^n] 1/(1 - n*x + x^2). - Paul D. Hanna, Dec 27 2012
a(n) = y(n,n), where y(m+1,n) = n*y(m,n) - y(m-1,n) with y(0,n)=1, y(1,n)=n. - Benedict W. J. Irwin, Nov 05 2016
From Seiichi Manyama, Mar 03 2021: (Start)
a(n) = U(n,n/2) where U(n,x) is a Chebyshev polynomial of the second kind.
a(n) = Sum_{k=0..n} (n-2)^(n-k) * binomial(2*n+1-k,k) = Sum_{k=0..n} (n-2)^k * binomial(n+1+k,2*k+1). (End)

A097691 Denominators of the continued fraction n-1/(n-1/...) [n times].

Original entry on oeis.org

1, 2, 8, 56, 551, 6930, 105937, 1905632, 39424240, 922080050, 24057287759, 692686638072, 21817946138353, 746243766783074, 27543862067299424, 1091228270370045824, 46187969968474139807, 2080128468827570457762, 99318726126650358502921, 5011361251329169946919800

Offset: 1

T. D. Noe, Aug 19 2004

The (n-1)-th term of the Lucas sequence U(n,1). The numerator is the n-th term. Adjacent terms of the sequence U(n,1) are relatively prime.

			a(4) = 56 because 4-1/(4-1/(4-1/4)) = 209/56.

Alois P. Heinz, Table of n, a(n) for n = 1..387
Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, Metered Parking Functions, arXiv:2406.12941 [math.CO], 2024. See pp. 8, 22.
Pascual Jara and Miguel L. Rodríguez, Solving quadratic congruences, Arhimede Math. J. (2020) Vol. 7, No. 2, 105-120.
Eric Weisstein's World of Mathematics, Lucas Sequence

Cf. A084844, A084845, A097690 (numerators).

Table[s=n; Do[s=n-1/s, {n-1}]; Denominator[s], {n, 20}]
Table[Abs[Fibonacci[n, I n]], {n, 20}] (* Vladimir Reshetnikov, Oct 16 2018 *)

[lucas_number1(n,n,1) for n in range(1,19)] # Zerinvary Lajos, Jul 16 2008

a(n) = ChebyshevU(n-1,n/2). - Gary Detlefs, Oct 15 2011
a(n) = abs((2^(-n) * (sqrt(4 - n^2) + i*n)^n - 2^n*(-sqrt(4 - n^2) - i*n)^(-n))/sqrt(4 - n^2)), where i is the imaginary unit, for n > 2. - Daniel Suteu, May 31 2017
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jun 03 2017

A117715 Triangle, read by rows, T(n, k) = Fibonacci(n, k), where Fibonacci(n, x) is the Fibonacci polynomial.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 2, 5, 10, 0, 3, 12, 33, 72, 1, 5, 29, 109, 305, 701, 0, 8, 70, 360, 1292, 3640, 8658, 1, 13, 169, 1189, 5473, 18901, 53353, 129949, 0, 21, 408, 3927, 23184, 98145, 328776, 927843, 2298912, 1, 34, 985, 12970, 98209, 509626, 2026009, 6624850, 18674305, 46866034

Offset: 0

Roger L. Bagula, Apr 13 2006

			Triangle begins as:
  0;
  1,  1;
  0,  1,   2;
  1,  2,   5,   10;
  0,  3,  12,   33,   72;
  1,  5,  29,  109,  305,   701;
  0,  8,  70,  360, 1292,  3640,  8658;
  1, 13, 169, 1189, 5473, 18901, 53353, 129949;

Steven Wolfram, The Mathematica Book, Cambridge University Press, 3rd ed. 1996, page 728

Alois P. Heinz, Rows n = 0..140, flattened
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
Wikipedia, Fibonacci Polynomial

Cf. A000045, A117716, A049310, A073133, A157103 (antidiagonals).
Main diagonal and first lower diagonal give: A084844, A084845.
Cf. A352361.

A117715:= func< n, k | k eq 0 select (n mod 2) else Evaluate(DicksonSecond(n-1, -1), k) >;
[A117715(n, k): k in [0..n], n in [0..13]]; // G. C. Greubel, Oct 01 2024

with(combinat):for n from 0 to 9 do seq(fibonacci(n,m), m = 0 .. n) od; # Zerinvary Lajos, Apr 09 2008

Table[Fibonacci[n, k], {n,0,12}, {k,0,n}]//Flatten

from sympy import fibonacci
def T(n, m): return 0 if n==0 else fibonacci(n, m)
for n in range(21): print([T(n, m) for m in range(n + 1)]) # Indranil Ghosh, Aug 12 2017

def A117715(n,k): return lucas_number1(n, k, -1)
flatten([[A117715(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 01 2024

T(n, 1) = A000045(n).
T(n, 3) = A006190(n).
T(n, 4) = A001076(n).
T(n, 5) = A052918(n-1).
T(5, k) = A057721(k).
T(6, k) = A124152(k).
T(n, k) = (-1)^(n-1)*A352361(n-k, n). - G. C. Greubel, Oct 01 2024

Definition simplified by the Assoc. Editors of the OEIS, Nov 17 2009

A304357 Antidiagonal sums of the first quadrant of array A(k,m) = F_k(m), F_k(m) being the k-th Fibonacci polynomial evaluated at m.

Original entry on oeis.org

0, 1, 1, 3, 5, 13, 32, 94, 297, 1036, 3911, 15918, 69350, 321779, 1582745, 8220349, 44925187, 257563819, 1544896976, 9671289892, 63051738167, 427254561854, 3003872526303, 21876513464296, 164790822258172, 1282198404741305, 10292007232817249, 85126350266370355

Offset: 0

Alois P. Heinz, May 11 2018

nonn

Equivalently, antidiagonal sums of the third quadrant of array A(k,m).

It seems that: a(n+1) is the sum of the n-th antidiagonal of triangle A101494; a(n)-(n mod 2) is the sum of the n-th antidiagonal of array A172236; and a(n+1)+(n mod 2) is the sum of row n of triangle A157103. - Mathew Englander, Feb 28 2021

Alois P. Heinz, Table of n, a(n) for n = 0..600
Wikipedia, Fibonacci polynomials
Wikipedia, Quadrant (plane geometry)

Cf. A000045, A084844, A304359, A101494, A172236, A157103.

F:= (n, k)-> (<<0|1>, <1|k>>^n)[1, 2]:
a:= n-> add(F(j, n-j), j=0..n):
seq(a(n), n=0..30);
# second Maple program:
F:= proc(n, k) option remember;
      `if`(n<2, n, k*F(n-1, k)+F(n-2, k))
    end:
a:= n-> add(F(j, n-j), j=0..n):
seq(a(n), n=0..30);
# third Maple program:
a:= n-> add(combinat[fibonacci](j, n-j), j=0..n):
seq(a(n), n=0..30);

a[n_] := Sum[Fibonacci[j, n - j], {j, 0, n}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 02 2018, from 3rd Maple program *)

a(n) = Sum_{j=0..n} F_j(n-j).

a(n+1) = Sum_{j = 0..n} Sum_{i = j..floor((n+j)/2)} binomial(i,j)*(n+j-2*i)^j (empirically). - Mathew Englander, Feb 28 2021

cp's OEIS Frontend

A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

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A352361 Array read by ascending antidiagonals. A(n, k) = Fibonacci(k, n), where Fibonacci(n, x) are the Fibonacci polynomials.

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A157103 Array A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1, read by antidiagonals.

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A084845 Numerators of the continued fraction n+1/(n+1/...) [n times].

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A097690 Numerators of the continued fraction n-1/(n-1/...) [n times].

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A097691 Denominators of the continued fraction n-1/(n-1/...) [n times].

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A320534 a(n) = ((1 + sqrt(4n^2 + 1))^n + (1 - sqrt(4n^2 + 1))^n)/2^n.