A054413 -id:A054413 - cp's OEIS frontend

A037027 Skew Fibonacci-Pascal triangle read by rows.

1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 5, 10, 9, 4, 1, 8, 20, 22, 14, 5, 1, 13, 38, 51, 40, 20, 6, 1, 21, 71, 111, 105, 65, 27, 7, 1, 34, 130, 233, 256, 190, 98, 35, 8, 1, 55, 235, 474, 594, 511, 315, 140, 44, 9, 1, 89, 420, 942, 1324, 1295, 924, 490, 192, 54, 10, 1, 144, 744, 1836

Offset: 0

Floor van Lamoen, Jan 01 1999

T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), (1,0), (2,0). - Joerg Arndt, Jun 30 2011
T(n,k) is the number of lattice paths of length n, starting from the origin and ending at (n,k), using horizontal steps H=(1,0), up steps U=(1,1) and down steps D=(1,-1), never containing UUU, DD, HD. For instance, for n=4 and k=2, we have the paths; HHUU, HUHU, HUUH, UHHU, UHUH, UUHH, UUDU, UDUU, UUUD. - Emanuele Munarini, Mar 15 2011
Row sums form Pell numbers A000129, T(n,0) forms Fibonacci numbers A000045, T(n,1) forms A001629. T(n+k,n-k) is polynomial sequence of degree k.
T(n,k) gives a convolved Fibonacci sequence (A001629, A001872, etc.).
As a Riordan array, this is (1/(1-x-x^2),x/(1-x-x^2)). An interesting factorization is (1/(1-x^2),x/(1-x^2))*(1/(1-x),x/(1-x)) [abs(A049310) times A007318]. Diagonal sums are the Jacobsthal numbers A001045(n+1). - Paul Barry, Jul 28 2005
T(n,k) = T'(n+1,k+1), T' given by [0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 19 2005
Equals A049310 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Oct 28 2007
This triangle may also be obtained from the coefficients of the Morgan-Voyce polynomials defined by: Mv(x, n) = (x + 1)*Mv(x, n - 1) + Mv(x, n - 2). - Roger L. Bagula, Apr 09 2008
Row sums are A000129. - Roger L. Bagula, Apr 09 2008
Absolute value of coefficients of the characteristic polynomial of tridiagonal matrices with 1's along the main diagonal, and i's along the superdiagonal and the subdiagonal (where i=sqrt(-1), see Mathematica program). - John M. Campbell, Aug 23 2011
A037027 is jointly generated with A122075 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+(x+1)*v(n-1)x and v(n,x)=u(n-1,x)+x*v(n-1,x). See the Mathematica section at A122075. - Clark Kimberling, Mar 05 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
Row n, for n>=0, shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^n which is the denominator of the n-th convergent of the continued fraction [x+1, x+1, x+1, ...]; see A230000. - Clark Kimberling, Nov 13 2013
T(n,k) is the number of ternary words of length n having k letters 2 and avoiding a runs of odd length for the letter 0. - Milan Janjic, Jan 14 2017
Let T(m, n, k) be an m-bonacci Pascal's triangle, where T(m, n, 0) gives the values of F(m, n), the n-th m-bonacci number, and T(m, n, k) gives the values for the k-th convolution of F(m, n). Then the classic Pascal triangle is T(1, n, k) and this sequence is T(2, n, k). T(m, n, k) is the number of compositions of n using only the positive integers 1, 1' and 2 through m, with the part 1' used exactly k times. G.f. for k-th column of T(m, n, k): x/(1 - x - x^2 - ... - x^m)^k. The row sum for T(m, n, k) is the number of compositions of n using only the positive integers 1, 1' and 2 through m. G.f. for row sum of T(m, n, k): 1/(1 - 2x - x^2 - ... - x^m). - Gregory L. Simay, Jul 24 2021

			Ratio of row polynomials R(3)/R(2) = (3 + 5*x + 3*x^2 + x^3)/(2 + 2*x + x^2) = [1+x; 1+x, 1+x].
Triangle begins:
                                 1;
                              1,    1;
                           2,    2,    1;
                        3,    5,    3,    1;
                     5,   10,    9,    4,    1;
                  8,   20,   22,   14,    5,    1;
              13,   38,   51,   40,   20,    6,    1;
           21,   71,  111,  105,   65,   27,    7,    1;
        34,  130,  233,  256,  190,   98,   35,    8,    1;
     55,  235,  474,  594,  511,  315,  140,   44,    9,    1;
  89,  420,  942, 1324, 1295,  924,  490,  192,   54,   10,    1;

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Harlan J. Brothers, Pascal's Prism: Supplementary Material
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.
T. Mansour, Generalization of some identities involving the Fibonacci numbers, arXiv:math/0301157 [math.CO], 2003.
P. Moree, Convoluted convolved Fibonacci numbers, arXiv:math/0311205 [math.CO], 2003.
Yidong Sun, Numerical Triangles and Several Classical Sequences, Fib. Quart. 43, no. 4, Nov. 2005, pp. 359-370.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials.

A038112(n) = T(2n, n). A038137 is reflected version. Maximal row entries: A038149.
Diagonal differences are in A055830. Vertical sums are in A091186.
Cf. A007318, A049310, A000129, A155161, A122542, A059283, A228196, A228576.
Some other Fibonacci-Pascal triangles: A027926, A036355, A074829, A105809, A109906, A111006, A114197, A162741, A228074.

a037027 n k = a037027_tabl !! n !! k
a037027_row n = a037027_tabl !! n
a037027_tabl = [1] : [1,1] : f [1] [1,1] where
   f xs ys = ys' : f ys ys' where
     ys' = zipWith3 (\u v w -> u + v + w) (ys ++ [0]) (xs ++ [0,0]) ([0] ++ ys)
-- Reinhard Zumkeller, Jul 07 2012

T := (n,k) -> `if`(n=0,1,binomial(n,k)*hypergeom([(k-n)/2, (k-n+1)/2], [-n], -4)): seq(seq(simplify(T(n,k)), k=0..n), n=0..10); # Peter Luschny, Apr 25 2016
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> combinat:-fibonacci(n)); # Peter Luschny, Oct 07 2022

Mv[x, -1] = 0; Mv[x, 0] = 1; Mv[x, 1] = 1 + x; Mv[x_, n_] := Mv[x, n] = ExpandAll[(x + 1)*Mv[x, n - 1] + Mv[x, n - 2]]; Table[ CoefficientList[ Mv[x, n], x], {n, 0, 10}] // Flatten (* Roger L. Bagula, Apr 09 2008 *)
Abs[Flatten[Table[CoefficientList[CharacteristicPolynomial[Array[KroneckerDelta[#1,#2]+KroneckerDelta[#1,#2+1]*I+KroneckerDelta[#1,#2-1]*I&,{n,n}],x],x],{n,1,20}]]] (* John M. Campbell, Aug 23 2011 *)
T[n_, k_] := Binomial[n, k] Hypergeometric2F1[(k-n)/2, (k-n+1)/2, -n, -4];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 16 2019, after Peter Luschny *)

{T(n, k) = if( k<0 || k>n, 0, if( n==0 && k==0, 1, T(n-1, k) + T(n-1, k-1) + T(n-2, k)))}; /* Michael Somos, Sep 29 2003 */

PARI
```
T(n,k)=if(nPaul D. Hanna, Feb 27 2004
```

T(n, m) = T'(n-1, m) + T'(n-2, m) + T'(n-1, m-1), where T'(n, m) = T(n, m) for n >= 0 and 0< = m <= n and T'(n, m) = 0 otherwise.
G.f.: 1/(1 - y - y*z - y^2).
G.f. for k-th column: x/(1-x-x^2)^k.
T(n, m) = Sum_{k=0..n-m} binomial(m+k, m)*binomial(k, n-k-m), n >= m >= 0, otherwise 0. - Wolfdieter Lang, Jun 17 2002
T(n, m) = ((n-m+1)*T(n, m-1) + 2*(n+m)*T(n-1, m-1))/(5*m), n >= m >= 1; T(n, 0)= A000045(n+1); T(n, m)= 0 if n < m. - Wolfdieter Lang, Apr 12 2000
Chebyshev coefficient triangle (abs(A049310)) times Pascal's triangle (A007318) as product of lower triangular matrices. T(n, k) = Sum_{j=0..n} binomial((n+j)/2, j)*(1+(-1)^(n+j))*binomial(j, k)/2. - Paul Barry, Dec 22 2004
Let R(n) = n-th row polynomial in x, with R(0)=1, then R(n+1)/R(n) equals the continued fraction [1+x;1+x, ...(1+x) occurring (n+1) times ..., 1+x] for n >= 0. - Paul D. Hanna, Feb 27 2004
T(n,k) = Sum_{j=0..n} binomial(n-j,j)*binomial(n-2*j,k); in Egorychev notation, T(n,k) = res_w(1-w-w^2)^(-k-1)*w^(-n+k+1). - Paul Barry, Sep 13 2006
Sum_{k=0..n} T(n,k)*x^k = 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 respectively. - Philippe Deléham, Nov 29 2009
T((m+1)*n+r-1, m*n+r-1)*r/(m*n+r) = Sum_{k=1..n} k/n*T((m+1)*n-k-1, m*n-1)*(r+k,r), n >= m > 1.
T(n-1,m-1) = (m/n)*Sum_{k=1..n-m+1} k*A000045(k)*T(n-k-1,m-2), n >= m > 1. - Vladimir Kruchinin, Mar 17 2011
T(n,k) = binomial(n,k)*hypergeom([(k-n)/2, (k-n+1)/2], [-n], -4) for n >= 1. - Peter Luschny, Apr 25 2016

Examples from Paul D. Hanna, Feb 27 2004

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.

Original entry on oeis.org

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

A006497 a(n) = 3*a(n-1) + a(n-2) with a(0) = 2, a(1) = 3.

Original entry on oeis.org

2, 3, 11, 36, 119, 393, 1298, 4287, 14159, 46764, 154451, 510117, 1684802, 5564523, 18378371, 60699636, 200477279, 662131473, 2186871698, 7222746567, 23855111399, 78788080764, 260219353691, 859446141837, 2838557779202

Offset: 0

N. J. A. Sloane

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its [sic] Properties, Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92, Sequence L_{3,n}.
A. F. Horadam, Generating identities for generalized Fibonacci and Lucas triples, Fib. Quart., 15 (1977), 289-292.
Haruo Hosoya, What Can Mathematical Chemistry Contribute to the Development of Mathematics?, HYLE--International Journal for Philosophy of Chemistry, Vol. 19, No.1 (2013), pp. 87-105.
Tanya Khovanova, Recursive Sequences
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (3,1).

Cf. A006190, A100230, A001622, A014176, A080039, A098316.

a006497 n = a006497_list !! n
a006497_list = 2 : 3 : zipWith (+) (map (* 3) $ tail a006497_list) a006497_list
-- Reinhard Zumkeller, Feb 19 2011

[ n eq 1 select 2 else n eq 2 select 3 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 20 2011

a:= n-> (<<0|1>, <1|3>>^n. <<2, 3>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 26 2018

Table[LucasL[n, 3], {n, 0, 30}] (* Zerinvary Lajos, Jul 09 2009 *)
LucasL[Range[0, 30], 3] (* Eric W. Weisstein, Apr 17 2018 *)
LinearRecurrence[{3,1},{2,3},30] (* Harvey P. Dale, Feb 17 2020 *)

my(x='x+O('x^30)); Vec((2-3*x)/(1-3*x-x^2)) \\ G. C. Greubel, Jul 05 2017

apply( {A006497(n)=[2,3]*([0,1;1,3]^n)[,1]}, [0..30]) \\ M. F. Hasler, Mar 06 2020

[lucas_number2(n,3,-1) for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

G.f.: (2-3*x)/(1-3*x-x^2). - Simon Plouffe in his 1992 dissertation
From Gary W. Adamson, Jun 15 2003: (Start)
a(n) = ((3 + sqrt(13))/2)^n + ((3 - sqrt(13))/2)^n. See bronze mean (A098316).
A006190(n-2) + A006190(n) = a(n-1).
a(n)^2 - 13*A006190(n)^2 = 4(-1)^n. (End)
From Paul Barry, Nov 15 2003: (Start)
E.g.f.: 2*exp(3*x/2)*cosh(sqrt(13)*x/2).
a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2*k)* (13)^k * 3^(n-2*k).
a(n) = 2*T(n, 3i/2)*(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. (End)
From Hieronymus Fischer, Jan 02 2009: (Start)
fract(((3+sqrt(13))/2)^n) = (1/2)*(1+(-1)^n) - (-1)^n*((3+sqrt(13))/2)^(-n) = (1/2)*(1+(-1)^n) - ((3-sqrt(13))/2)^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
a(n) = round(((3+sqrt(13))/2)^n) for n > 0. (End)
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 3*A097783(n), a(2n) = A057076(n).
a(3n+1) = A041018(5n), a(3n+2) = A041018(5n+3) and a(3n+3) = 2*A041018(5n+4).
Limit_{k -> infinity} a(n+k)/a(k) = (a(n) + A006190(n)*sqrt(13))/2.
Limit_{n -> infinity} a(n)/A006190(n) = sqrt(13).
(End)
a(n) = sqrt(13*(A006190(n))^2 + 4*(-1)^n). - Vladimir Shevelev, Mar 13 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(13*k-9))/((x*(13*k+4)) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 13*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = Lucas(n,3), Lucas polynomials, L(n,x), evaluated at x=3. - G. C. Greubel, Jun 06 2019
a(n) = 2 * Sum_{k=0..n-2} A168561(n-2,k)*3^k + 3 * Sum_{k=0..n-1} A168561(n-1,k)*3^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 2*A006190(n+1) - 3*A006190(n). - R. J. Mathar, Feb 14 2024
a(2*n+1) = 3 + 3*Sum_{k=1..n} a(2*k). - Greg Dresden and Canran Wang, Jul 11 2024
From Peter Bala, Jul 14 2025: (Start)
The following series telescope (Cf. A000032):
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

Definition completed by M. F. Hasler, Mar 06 2020

A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

Original entry on oeis.org

1, 5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - Gary W. Adamson, Jul 01 2003, corrected Oct 07 2008, corrected by Jianing Song, Feb 01 2019
a(p-1) == 29^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A015523. - Johannes W. Meijer, Aug 01 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,5} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 5-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is 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 5 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
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 8.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 901
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Milan Janjić, 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
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Row 5 of A073133, A172236, and A352361.

Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).

a:=[1,5];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 16 2019

I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 23 2013

R:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // Marius A. Burtea, Oct 16 2019

spec := [S,{S=Sequence(Union(Z,Z,Z,Z,Z,Prod(Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006
with(combinat):a:=n->fibonacci(n,5):seq(a(n),n=1..30); # Zerinvary Lajos, Dec 07 2008

LinearRecurrence[{5, 1}, {1, 5}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
Table[Fibonacci[n+1, 5], {n,0,30}] (* Vladimir Reshetnikov, May 08 2016 *)

Vec(1/(1-5*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

[lucas_number1(n,5,-1) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009

G.f.: 1/(1 - 5*x - x^2).
a(3n) = A041047(5n), a(3n+1) = A041047(5n+3), a(3n+2) = 2*A041047(5n+4). - Henry Bottomley, May 10 2000
a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). - Gary W. Adamson, Jul 01 2003
a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle A049310.
Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. - Roger L. Bagula, May 29 2005
a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. - T. D. Noe, Jan 19 2006
a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... - Gary W. Adamson, Dec 21 2007
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
Limit_{k->oo} a(n+k)/a(k) = (A087130(n) + a(n-1)*sqrt(29))/2.
Limit_{n->oo} A087130(n)/a(n-1) = sqrt(29). (End)
From L. Edson Jeffery, Jan 07 2012: (Start)
Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
a(n) is the upper-right term of A^(n+1);
a(n) is the lower-left term of (1/5)*A^(n+1);
a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(29) - 5)/2. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 5*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 5 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

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

A086902 a(n) = 7*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7.

Original entry on oeis.org

2, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, 17555720002, 125348805407, 894997357851, 6390330310364, 45627309530399, 325781497023157, 2326097788692498, 16608466017870643, 118585359913786999, 846705985414379636

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 18 2003

a(n+1)/a(n) converges to (7+sqrt(53))/2 = 7.14005... = A176439.
Lim a(n)/a(n+1) as n approaches infinity = 0.1400549... = 2/(7+sqrt(53)) = (sqrt(53)-7)/2 = 1/A176439 = A176439 - 7.
From Johannes W. Meijer, Jun 12 2010: (Start)
In general sequences with recurrence a(n) = k*a(n-1)+a(n-2) with a(0)=2 and a(1)=k [and a(-1)=0] have generating function (2-k*x)/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n+1) = b(5n), a(3n+2)=b(5*n+3), a(3n+3)=2*b(5n+4) where b(n) is the sequence of numerators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n)/2, for n>=1, is the sequence of numerators of continued fraction convergents to sqrt(k^2/4+1).]
For the sequence given above k=7 which implies that it is associated with A041090.
For a similar statement about sequences with recurrence a(n) = k*a(n-1)+a(n-2) but with a(0)=1 [and a(-1)=0] see A054413; a sequence that is associated with A041091.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)

			a(4) = 7*a(3) + a(2) = 7*364 + 51 = 2599.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,1).

Cf. A058316, A176439.

Cf. A000032 (k=1), A006497 (k=3), A087130 (k=5), A086902 (k=7), A087798 (k=9), A001946 (k=11), A088316 (k=13), A090301 (k=15), A090306 (k=17). - Johannes W. Meijer, Jun 12 2010

I:=[2,7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 7, a[n] == 7 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{7,1},{2,7},30] (* Harvey P. Dale, May 25 2023 *)

a(n)=([0,1; 1,7]^n*[2;7])[1,1] \\ Charles R Greathouse IV, Apr 06 2016

a(n) = ((7+sqrt(53))/2)^n + ((7-sqrt(53))/2)^n.
E.g.f. : 2exp(7x/2)cosh(sqrt(53)x/2); a(n)=2^(1-n)sum{k=0..floor(n/2), C(n, 2k)53^k7^(n-2k)}. a(n)=2T(n, 7i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
G.f.: (2-7x)/(1-7x-x^2). - Philippe Deléham, Nov 16 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097837(n), a(2n) = A099368(n).
a(3n+1) = A041090(5n), a(3n+2) = A041090(5*n+3), a(3n+3) = 2*A041090(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Limit(A086902(n)/A054413(n-1), n=infinity) = sqrt(53). (End)

A049666 a(n) = Fibonacci(5*n)/5.

Original entry on oeis.org

0, 1, 11, 122, 1353, 15005, 166408, 1845493, 20466831, 226980634, 2517253805, 27916772489, 309601751184, 3433536035513, 38078498141827, 422297015595610, 4683345669693537, 51939099382224517, 576013438874163224

Offset: 0

Clark Kimberling

For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 11's along the main diagonal and 1's along the subdiagonal and the superdiagonal. - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,11} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
For n >= 1, a(n) equals the denominator of the continued fraction [11, 11, ..., 11] (with n copies of 11). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019
From Michael A. Allen, Mar 30 2023: (Start)
Also called the 11-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is 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 11 kinds of squares available. (End)

			G.f. = x + 11*x^2 + 122*x^3 + 1353*x^4 + 15005*x^5 + 166408*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..950
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (11,1).

A column of array A028412.
Cf. A000045, A102312, A243399.
Row n=11 of A073133, A172236 and A352361, and column k=11 of A157103.

[Fibonacci(5*n)/5: n in [0..30]]; // G. C. Greubel, Dec 02 2017

A049666 := proc(n)
    combinat[fibonacci](5*n)/5 ;
end proc: # R. J. Mathar, May 07 2024

Table[Fibonacci[5*n]/5, {n, 0, 100}] (* T. D. Noe, Oct 29 2009 *)
a[ n_] := Fibonacci[n, 11]; (* Michael Somos, May 28 2014 *)

numlib::fibonacci(5*n)/5 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(5*n)/5 \\ Charles R Greathouse IV, Feb 03 2014

from sage.combinat.sloane_functions import recur_gen3
it = recur_gen3(0,1,11,11,1,0)
[next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008

[lucas_number1(n,11,-1) for n in range(0, 19)] # Zerinvary Lajos, Apr 27 2009

[fibonacci(5*n)/5 for n in range(0, 19)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1 - 11*x - x^2).
a(n) = A102312(n)/5.
a(n) = 11*a(n-1) + a(n-2) for n > 1, a(0)=0, a(1)=1. With a=golden ratio and b=1-a, a(n) = (a^(5n)-b^(5n))/(5*sqrt(5)). - Mario Catalani (mario.catalani(AT)unito.it), Jul 24 2003
a(n) = F(n, 11), the n-th Fibonacci polynomial evaluated at x=11. - T. D. Noe, Jan 19 2006
a(n) = ((11+sqrt(125))^n-(11-sqrt(125))^n)/(2^n*sqrt(125)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n) = 11*A049670(n), a(2n+1) = A097843(n).
a(3n+1) = A041227(5n), a(3n+2) = A041227(5n+3), a(3n+3) = 2*A041227(5n+4).
Limit_{k->oo} a(n+k)/a(k) = (A001946(n) + A049666(n)*sqrt(125))/2.
Limit_{n->oo} A001946(n)/A049666(n) = sqrt(125).
(End)
a(n) = F(n) + (-1)^n*5*F(n)^3 + 5*F(n)^5, n >= 0. See the D. Jennings formula given in a comment on A111125, where also the reference is given. - Wolfdieter Lang, Aug 31 2012
a(-n) = -(-1)^n * a(n). - Michael Somos, May 28 2014
E.g.f.: (exp((1/2)*(11-5*sqrt(5))*x)*(-1 + exp(5*sqrt(5)*x)))/(5*sqrt(5)). - Stefano Spezia, Aug 02 2019

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

Original entry on oeis.org

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127

Offset: 0

Paul Barry, Aug 16 2003

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).
The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005
a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence L_{5,n}.
Tanya Khovanova, Recursive Sequences
Wikipedia, Metallic mean
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A006497, A014448, A085447.

Cf. A086902, A000032, A052918.

I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)

{a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */

[lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.
a(n) = A100236(n) + 1.
E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007
a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008
A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010
a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010
a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020
From Peter Bala, Jul 09 2025 : (Start)
The following series telescope (Cf. A000032):
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A099371 Expansion of g.f.: x/(1 - 9*x - x^2).

Original entry on oeis.org

0, 1, 9, 82, 747, 6805, 61992, 564733, 5144589, 46866034, 426938895, 3889316089, 35430783696, 322766369353, 2940328107873, 26785719340210, 244011802169763, 2222891938868077, 20250039251982456, 184473245206710181, 1680509246112374085, 15309056460218076946

Offset: 0

Wolfdieter Lang, Oct 18 2004

For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
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 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,9} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Mar 10 2023: (Start)
Also called the 9-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is 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 9 kinds of squares available. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
J. H. Han and M. D. Hirschhorn, Another Look at an Amazing Identity of Ramanujan, Mathematics Magazine, Vol. 79 (2006), pp. 302-304. See equation 6 on page 303.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,1).

Row n=9 of A073133, A172236 and A352361.
Cf. A099372 (squares).
Cf. A041151, A054413, A086902, A087798, A099371, A097839, A178765, A243399.

I:=[0,1]; [n le 2 select I[n] else 9*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

F:= gfun:-rectoproc({a(n)=9*a(n-1)+a(n-2),a(0)=0,a(1)=1},a(n),remember):
seq(F(n),n=0..30); # Robert Israel, Feb 01 2015

CoefficientList[Series[x/(1-9*x-x^2), {x,0,30}], x] (* G. C. Greubel, Apr 16 2017 *)
LinearRecurrence[{9,1}, {0,1}, 30] (* G. C. Greubel, Jan 24 2018 *)

my(x='x+O('x^30)); concat([0], Vec(1/(1-9*x-x^2)) ) \\ Charles R Greathouse IV, Feb 03 2014

from sage.combinat.sloane_functions import recur_gen3
it = recur_gen3(0,1,9,9,1,0)
[next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008

[lucas_number1(n,9,-1) for n in range(0, 20)] # Zerinvary Lajos, Apr 26 2009

G.f.: x/(1 - 9*x - x^2).
a(n) = 9*a(n-1) + a(n-2), n >= 2, a(0)=0, a(1)=1.
a(n) = (-i)^(n-1)*S(n-1, 9*i) with S(n, x) Chebyshev's polynomials of the second kind (see A049310) and i^2=-1.
a(n) = (ap^n - am^p)/(ap-am) with ap:= (9+sqrt(85))/2 and am:= (9-sqrt(85))/2 = -1/ap (Binet form).
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1-k, k)*9^(n-1-2*k) n >= 1.
a(n) = F(n, 9), the n-th Fibonacci polynomial evaluated at x=9. - T. D. Noe, Jan 19 2006
a(n) = ((9+sqrt(85))^n - (9-sqrt(85))^n)/(2^n*sqrt(85)). Offset 1. a(3)=82. - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 85^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+2) = 9*A097839(n), a(2n+1) = A097841(n).
a(3n+1) = A041151(5n), a(3n+2) = A041151(5n+3), a(3n+3) = 2*A041151(5n+4).
Limit_{k -> infinity} (a(n+k)/a(k)) = (A087798(n) + A099371(n)*sqrt(85))/2.
Lim_{n->infinity} A087798(n)/A099371(n) = sqrt(85). (End)
a(n) ~ 1/sqrt(85)*((9+sqrt(85))/2)^n. - Jean-François Alcover, Dec 04 2013
a(n) = [1,0] (M^n) [0,1]^T where M is the matrix [9,1; 1,0]. - Robert Israel, Feb 01 2015
E.g.f.: 2*exp(9*x/2)*sinh(sqrt(85)*x/2)/sqrt(85). - Stefano Spezia, Apr 06 2023

cp's OEIS Frontend

A037027 Skew Fibonacci-Pascal triangle read by rows.

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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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A006497 a(n) = 3*a(n-1) + a(n-2) with a(0) = 2, a(1) = 3.

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A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-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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