A135597 -id:A135597 - cp's OEIS frontend

A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323, 318281039, 768398401, 1855077841, 4478554083, 10812186007, 26102926097, 63018038201, 152139002499, 367296043199

Offset: 0

N. J. A. Sloane and R. K. Guy

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].
Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011
Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012
Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman
a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.
In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017
Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002
a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.
a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
Binomial transform of A077957. - Paul Barry, Feb 25 2003
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004
For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004
Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004
Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]
Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008
The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008
Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008
Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008
From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then b(1,n)=a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
Cf. A000129, A142238-A142239, A153313-A153318.
(End)
This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010
Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(8,2) = (0 0 1 0)
              (0 1 0 1)
              (1 0 2 0)
              (0 2 0 1).
Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)
(End)
For n >= 1, row sums of triangle
  m/k.|..0.....1.....2.....3.....4.....5.....6.....7
  ==================================================
  .0..|..1
  .1..|..1.....2
  .2..|..1.....2.....4
  .3..|..1.....4.....4.....8
  .4..|..1.....4....12.....8....16
  .5..|..1.....6....12....32....16....32
  .6..|..1.....6....24....32....80....32....64
  .7..|..1.....8....24....80....80...192....64...128
which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012
The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012
Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012
This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012
a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016
a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018
For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020
For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021
a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022
From Greg Dresden, May 08 2023: (Start)
For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.
   _
  || _  
  |||_|||
  |_|
(End)
12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

			Convergents are 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
The 15 3 X 2 crossword grids, with white squares represented by an o:
  ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo
  ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo.
G.f. = 1 + x + 3*x^2 + 7*x^3 + 17*x^4 + 41*x^5 + 99*x^6 + 239*x^7 + 577*x^8 + ...

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Index entries for "core" sequences.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (2,1).

For denominators see A000129.
See A040000 for the continued fraction expansion of sqrt(2).
See also A078057 which is the same sequence without the initial 1.
Cf. also A002203, A152113.
Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle).
Row sums of A140750, A160756, A135837.
Equals A034182(n-1) + 2 and A084128(n)/2^n. First differences of A052937. Partial sums of A052542. Pairwise sums of A048624. Bisection of A002965.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Second row of the array in A135597.
Cf. A048211, A153588, A174283, A174284, A174285, A174286, A176499, A176500, A176501, A176502.
Cf. A055099.
Cf. A028859, A001906 / A088305, A033303, A000225, A095263, A003945, A006356, A002478, A214260, A001911 and A000217 for other restricted ternary words.
Cf. Triangle A106513 (alternating row sums).
Equals A293004 + 1.
Cf. A033539, A332602, A086395 (subseq. of primes).

a001333 n = a001333_list !! n
a001333_list = 1 : 1 : zipWith (+)
                       a001333_list (map (* 2) $ tail a001333_list)
-- Reinhard Zumkeller, Jul 08 2012

[n le 2 select 1 else 2*Self(n-1)+Self(n-2): n in [1..35]]; // Vincenzo Librandi, Nov 10 2018

A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*procname(n-1)+procname(n-2) fi end;
Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n);
with(numtheory): cf := cfrac (sqrt(2),1000): [seq(nthnumer(cf,i), i=0..50)];
a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^n):
seq(a(n), n=0..33);  # Alois P. Heinz, Aug 01 2008
A001333List := proc(m) local A, P, n; A := [1,1]; P := [1,1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-2]]);
A := [op(A), P[-1]] od; A end: A001333List(32); # Peter Luschny, Mar 26 2022

Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 08 2006 *)
Table[((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2, {n, 0, 29}] // Simplify (* Robert G. Wilson v, May 02 2006 *)
a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (* Robert G. Wilson v, May 02 2006 *)
Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, May 02 2006 *)
a=c=0;t={b=1}; Do[c=a+b+c; AppendTo[t,c]; a=b;b=c,{n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
LinearRecurrence[{2, 1}, {1, 1}, 40] (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
Join[{1}, Numerator[Convergents[Sqrt[2], 30]]] (* Harvey P. Dale, Aug 22 2011 *)
Table[(-I)^n ChebyshevT[n, I], {n, 10}] (* Eric W. Weisstein, Apr 04 2017 *)
CoefficientList[Series[(-1 + x)/(-1 + 2 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
Table[Sqrt[(ChebyshevT[n, 3] + (-1)^n)/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 17 2018 *)

{a(n) = if( n<0, (-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [1, 1]}; /* Michael Somos, Sep 02 2012 */

{a(n) = polchebyshev(n, 1, I) / I^n}; /* Michael Somos, Sep 02 2012 */

a(n) = real((1 + quadgen(8))^n); \\ Michel Marcus, Mar 16 2021

{ for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 12 2009

from functools import cache
@cache
def a(n): return 1 if n < 2 else 2*a(n-1) + a(n-2)
print([a(n) for n in range(32)]) # Michael S. Branicky, Nov 13 2022

from sage.combinat.sloane_functions import recur_gen2
it = recur_gen2(1,1,2,1)
[next(it) for i in range(30)] ## Zerinvary Lajos, Jun 24 2008

[lucas_number2(n,2,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

a(n) = A055642(A125058(n)). - Reinhard Zumkeller, Feb 02 2007
a(n) = 2a(n-1) + a(n-2);
a(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2.
a(n)+a(n+1) = 2 A000129(n+1). 2*a(n) = A002203(n).
G.f.: (1 - x) / (1 - 2*x - x^2) = 1 / (1 - x / (1 - 2*x / (1 + x))). - Simon Plouffe in his 1992 dissertation.
A000129(2n) = 2*A000129(n)*a(n). - John McNamara, Oct 30 2002
a(n) = (-i)^n * T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
a(n) = a(n-1) + A052542(n-1), n>1. a(n)/A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
E.g.f.: exp(x)cosh(x*sqrt(2)). - Paul Barry, May 08 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)2^k. - Paul Barry, May 13 2003
For n > 0, a(n)^2 - (1 + (-1)^(n))/2 = Sum_{k=0..n-1} ((2k+1)*A001653(n-1-k)); e.g., 17^2 - 1 = 288 = 1*169 + 3*29 + 5*5 + 7*1; 7^2 = 49 = 1*29 + 3*5 + 5*1. - Charlie Marion, Jul 18 2003
a(n+2) = A078343(n+1) + A048654(n). - Creighton Dement, Jan 19 2005
a(n) = A000129(n) + A000129(n-1) = A001109(n)/A000129(n) = sqrt(A001110(n)/A000129(n)^2) = ceiling(sqrt(A001108(n))). - Henry Bottomley, Apr 18 2000
Also the first differences of A000129 (the Pell numbers) because A052937(n) = A000129(n+1) + 1. - Graeme McRae, Aug 03 2006
a(n) = Sum_{k=0..n} A122542(n,k). - Philippe Deléham, Oct 08 2006
For another recurrence see A000129.
a(n) = Sum_{k=0..n} A098158(n,k)*2^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1,1; 2,1]^n. - Gary W. Adamson, Mar 12 2008
If p[1]=1, and p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
For n>=2, a(n)=F_n(2)+F_(n+1)(2), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(-n) = (-1)^n * a(n). - Michael Somos, Sep 02 2012
Dirichlet g.f.: (PolyLog(s,1-sqrt(2)) + PolyLog(s,1+sqrt(2)))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A000129(n) - A000129(n-1), where A000129(n) is the n-th Pell Number. Hence the continued fraction is of the form 1-(A000129(n-1)/A000129(n)). - Gregory L. Simay, Nov 09 2018
a(n) = (A000129(n+3) + A000129(n-3))/10, n>=3. - Paul Curtz, Jun 16 2021
a(n) = (A000129(n+6) - A000129(n-6))/140, n>=6. - Paul Curtz, Jun 20 2021
a(n) = round((1/2)*sqrt(Product_{k=1..n} 4*(1 + sin(k*Pi/n)^2))), for n>=1. - Greg Dresden, Dec 28 2021
a(n)^2 + a(n+1)^2 = A075870(n+1) = 2*(b(n)^2 + b(n+1)^2) for all n in Z where b(n) := A000129(n). - Michael Somos, Apr 02 2022
a(n) = 2*A048739(n-2)+1. - R. J. Mathar, Feb 01 2024
Sum_{n>=1} 1/a(n) = 1.5766479516393275911191017828913332473... - R. J. Mathar, Feb 05 2024
From Peter Bala, Jul 06 2025: (Start)
G.f.: Sum_{n >= 1} (-1)^(n+1) * x^(n-1) * Product_{k = 1..n} (1 - k*x)/(1 - 3*x + k*x^2).
The following series telescope:
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 1/a(2*n)) = 1/4, since 1/(a(2*n) + 1/a(2*n)) = 1/A077445(n) + 1/A077445(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) - 1/a(2*n+1)) = 1/8, since. 1/(a(2*n+1) - 1/a(2*n+1)) = 1/(4*Pell(2*n)) + 1/(4*Pell(2*n+2)), where Pell(n) = A000129(n).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) + 9/a(2*n+1)) = 1/10, since 1/(a(2*n+1) + 9/a(2*n+1)) = b(n) + b(n+1), where b(n) = A001109(n)/(2*Pell(2*n-1)*Pell(2*n+1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 1 - sqrt(2)/2 = A268682, since (-1)^(n+1)/(a(n)*a(n+1)) = Pell(n)/a(n) - Pell(n+1)/a(n+1). (End)

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A015449 Expansion of (1-4x)/(1-5x-x^2).

Original entry on oeis.org

1, 1, 6, 31, 161, 836, 4341, 22541, 117046, 607771, 3155901, 16387276, 85092281, 441848681, 2294335686, 11913527111, 61861971241, 321223383316, 1667978887821, 8661117822421, 44973567999926, 233528957822051

Offset: 0

Olivier Gérard

Row m=5 of A135597.
Binomial transform of A152187. - Johannes W. Meijer, Aug 01 2010
For n>=1, row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....5
.2..|..1.....5....25
.3..|..1....10....25.....125
.4..|..1....10....75.....125....625
.5..|..1....15....75.....500....625....3125
.6..|..1....15...150.....500...3125....3125...15625
.7..|..1....20...150....1250...3125...18750...15625...78125
which is triangle for numbers 5^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n+1) is (for n>=0) the number of length-n strings of 6 letters {0,1,2,3,4,5} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012
With offset 1, the sequence is the INVERT transform (1, 5, 5*4, 5*4^2, 5*4^3, ...); i.e., of A003947. The sequence can also be obtained by taking powers of the matrix [(1,5); (1,4)] and extracting the upper left terms. - Gary W. Adamson, Jul 31 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Joerg Arndt, Matters Computational (The Fxtbook)
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 16.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A084057, A108306, A164549.

a:=[1,1];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 23 2019

[n le 2 select 1 else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 06 2012

a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..21); # Zerinvary Lajos, Jul 26 2006

Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{1,5},#]}]&, {1,1},40]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
LinearRecurrence[{5,1}, {1,1}, 30] (* Vincenzo Librandi, Nov 06 2012 *)
CoefficientList[Series[(1-4*x)/(1-5*x-x^2), {x,0,30}], x] (* G. C. Greubel, Dec 19 2017 *)
Sum[Fibonacci[Range[30] +k-2, 5], {k,0,1}] (* G. C. Greubel, Oct 23 2019 *)

Vec((1-4*x)/(1-5*x-x^2) +O('x^30)) \\ _G. C. Greubel, Dec 19 2017

def A015449_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-4*x)/(1-5*x-x^2)).list()
A015449_list(30) # G. C. Greubel, Oct 23 2019

a(n) = 5*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 4^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-4*x)/(1-5*x-x^2). - Philippe Deléham, Nov 20 2008
For n >= 2, a(n) = F_n(5) + F_(n+1)(5), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = Sum_{k=0..n} A046854(n-1,k)*5^k. - R. J. Mathar, Feb 10 2024

A015451 a(n) = 6*a(n-1) + a(n-2) for n > 1, with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 7, 43, 265, 1633, 10063, 62011, 382129, 2354785, 14510839, 89419819, 551029753, 3395598337, 20924619775, 128943316987, 794584521697, 4896450447169, 30173287204711, 185936173675435, 1145790329257321

Offset: 0

Olivier Gérard

Row m=6 of A135597.
a(n) = term (1,1) in the 2 X 2 matrix [1,2; 3,5]^n. - Gary W. Adamson, May 30 2008
a(n)/a(n-1) tends to sqrt(10) + 3 = 6.16227766... - Gary W. Adamson, May 30 2008
For n >= 1, row sums of triangle for numbers 6^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
Z[sqrt(10)] is not a unique factorization domain, since, for example, 6 = 2 * 3 = (-1)(2 - sqrt(10))(2 + sqrt(10)) = (4 - sqrt(10))(4 + sqrt(10)). However, the latter two factorizations are not distinct, because 3 + sqrt(10) is a unit in Z[sqrt(10)], and (2 - sqrt(10))(-3 - sqrt(10)) = 4 + sqrt(10). In fact, (2 - sqrt(10))(-3 - sqrt(10))^n gives an algebraic integer b + a(n) * sqrt(10) which, when multiplied by its associate (and by -1 when n is even) is equal to 6. - Alonso del Arte, Mar 15 2014
For n >= 1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,5,6} containing no subwords 00, 11, 22, 33, 44, 55. - Milan Janjic, Jan 31 2015
a(n+1) equals the number of sequences over the alphabet {0,1,2,3,4,5,6} of length n such that no two consecutive terms differ by 4. - David Nacin, May 31 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,1).

Cf. A049310, A055830.

[n le 2 select 1 else 6*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=6*a[n-1]+a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-5*x)/(1-6*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec((1-5*x)/(1-6*x-x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) = Sum_{k=0..n} 5^k * A055830(n, k). - Philippe Deléham, Oct 18 2006
G.f.: (1-5*x)/(1-6*x-x^2). - Philippe Deléham, Nov 20 2008
For n >= 2, a(n) = F_n(6) + F_(n+1)(6), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i = 0..floor((n-1)/2)} C(n-i-1,i) * x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = Sum_{k=0..n} A046854(n-1,k)*6^k. - R. J. Mathar, Feb 14 2024

A015457 Generalized Fibonacci numbers.

Original entry on oeis.org

1, 1, 12, 133, 1475, 16358, 181413, 2011901, 22312324, 247447465, 2744234439, 30434026294, 337518523673, 3743137786697, 41512034177340, 460375513737437, 5105642685289147, 56622445051918054, 627952538256387741, 6964100365872183205, 77233056562850402996

Offset: 0

Olivier Gérard

For n>=1, row sums of triangle for numbers 11^k*binomial(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012

For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,11} containing no subwords ii, (i=0,1,...,10). - Milan Janjic, Jan 31 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (11,1).

Row m=11 of A135597.

Cf. A000045, A049310, A055830.

[n le 2 select 1 else 11*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

LinearRecurrence[{11, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-10*x)/(1-11*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec((1-10*x)/(1-11*x-x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) = 11*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 10^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-10*x)/(1-11*x-x^2). - Philippe Deléham, Nov 21 2008
For n>=2, a(n) = F_n(11)+F_(n+1)(11), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = (F(5*n-5) + F(5*n))/5 for F(n) the Fibonacci sequence A000045(n). - Greg Dresden, Aug 22 2021

A287376 Array read by antidiagonals: T(m,n) = number of independent vertex sets in the complete prism graph K_m X C_n.

Original entry on oeis.org

1, 3, 1, 4, 7, 1, 7, 13, 13, 1, 11, 35, 34, 21, 1, 18, 81, 121, 73, 31, 1, 29, 199, 391, 325, 136, 43, 1, 47, 477, 1300, 1361, 731, 229, 57, 1, 76, 1155, 4285, 5781, 3771, 1447, 358, 73, 1, 123, 2785, 14161, 24473, 19606, 8881, 2605, 529, 91, 1

Offset: 1

Andrew Howroyd, May 23 2017

Equivalently, the number of 0..m words of length n with cyclically adjacent letters unequal with the exception that 0's may be adjacent.

			Table starts:
====================================================
m\n| 1  2   3    4     5      6       7        8
---|------------------------------------------------
1  | 1  3   4    7    11     18      29       47 ...
2  | 1  7  13   35    81    199     477     1155 ...
3  | 1 13  34  121   391   1300    4285    14161 ...
4  | 1 21  73  325  1361   5781   24473   103685 ...
5  | 1 31 136  731  3771  19606  101781   528531 ...
6  | 1 43 229 1447  8881  54763  337429  2079367 ...
7  | 1 57 358 2605 18551 132504  946037  6754805 ...
8  | 1 73 529 4361 35361 287305 2333745 18957321 ...
...

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Rows 2-7 are A051927, A051928, A051929, A051930, A051931, A051932.

Cf. A135597 (K_m X P_n), A106512, A175243.

max = 10; row[m_] := ((m+1) - (m^2 - 2)*x - (2*m - 1)*x^2)/(1 - (m-1)*x - (m+1)*x^2 - x^3) + O[x]^(max+1) // CoefficientList[#, x]& // Rest;
T = Table[row[m], {m, 1, max}];
Table[T[[m-n+1, n]], {m, 1, max}, {n, m, 1, -1}] // Flatten (* Jean-François Alcover, Jun 06 2017 *)

RowGf(m,x)=((m+1)-(m^2-2)*x-(2*m-1)*x^2)/(1-(m-1)*x-(m+1)*x^2-x^3);
for (m=1,8,for(n=1,8,print1(Vec(RowGf(m,x)+O(x^(n+1)))[n+1], " "));print);

Row g.f.: ((m+1)-(m^2-2)*x-(2*m-1)*x^2)/(1-(m-1)*x-(m+1)*x^2-x^3).

A015453 Generalized Fibonacci numbers.

Original entry on oeis.org

1, 1, 8, 57, 407, 2906, 20749, 148149, 1057792, 7552693, 53926643, 385039194, 2749201001, 19629446201, 140155324408, 1000716717057, 7145172343807, 51016923123706, 364263634209749, 2600862362591949, 18570300172353392

Offset: 0

Olivier Gérard

Row m=7 of A135597.
For n >= 1, row sums of triangle for numbers 7^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
For n >= 1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,5,6,7} containing no subwords ii, (i=0,1,...,6). - Milan Janjic, Jan 31 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 18.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,1)

Cf. A054413, A055830.

Row m=7 of A135597.

[n le 2 select 1 else 7*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

LinearRecurrence[{7, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-6*x)/(1-7*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

my(x='x+O('x^30)); Vec((1-6*x)/(1-7*x-x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number1(n+1, 7, -1) - 6*lucas_number1(n, 7, -1) for n in (0..30)] # G. C. Greubel, Dec 24 2021

a(n) = 7*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 6^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-6*x)/(1-7*x-x^2). - Philippe Deléham, Nov 20 2008
For n >= 2, a(n) = F_(n)(7) + F_(n+1)(7), where F_(n)(x) is Fibonacci polynomial (cf. A049310): F_(n)(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = A054413(n) - 6*A054413(n-1). - R. J. Mathar, Jul 06 2012
a(n) = Sum_{k=0..n} A046854(n-1,k)*7^k. - R. J. Mathar, Feb 14 2024

A015454 Generalized Fibonacci numbers.

Original entry on oeis.org

1, 1, 9, 73, 593, 4817, 39129, 317849, 2581921, 20973217, 170367657, 1383914473, 11241683441, 91317382001, 741780739449, 6025563297593, 48946287120193, 397595860259137, 3229713169193289, 26235301213805449, 213112122879636881

Offset: 0

Olivier Gérard

a(n)/a(n-1) tends to (8 + 2*sqrt(17))/2 = exp ArcSinh 4 = A176458. - Gary W. Adamson, Dec 26 2007
For n>=1, row sums of triangle for numbers 8^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,8} containing no subwords ii, (i=0,1,...,7). - Milan Janjic, Jan 31 2015
a(n+1) is the number of nonary sequences of length n such that no two consecutive terms have distance 5. - David Nacin, May 31 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (8,1)

Row m=8 of A135597.

[n le 2 select 1 else 8*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

LinearRecurrence[{8, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-7*x)/(1-8*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec((1-7*x)/(1-8*x-x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) = 8*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 7^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-7*x)/(1-8*x-x^2). - Philippe Deléham, Nov 20 2008
For n>=2, a(n) = F_n(8)+F_(n+1)(8), where F_n(x) is Fibonacci polynomial (cf.A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = A041025(n) -7*A041025(n-1). - R. J. Mathar, Jul 06 2012

A015456 Generalized Fibonacci numbers.

Original entry on oeis.org

1, 1, 11, 111, 1121, 11321, 114331, 1154631, 11660641, 117761041, 1189271051, 12010471551, 121293986561, 1224950337161, 12370797358171, 124932923918871, 1261700036546881, 12741933289387681, 128681032930423691, 1299552262593624591, 13124203658866669601

Offset: 0

Olivier Gérard

For n>=1, row sums of triangle for numbers 10^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,10} containing no subwords ii, (i=0,1,...,9). - Milan Janjic, Jan 31 2015
a(n) equals the number of sequences over the alphabet {0,1,...,9,10} such that no two consecutive terms have distance 6. - David Nacin, Jun 02 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (10,1).

Row m=10 of A135597.

[n le 2 select 1 else 10*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

LinearRecurrence[{10, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-9*x)/(1-10*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec((1-9*x)/(1-10*x-x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) = 10*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 9^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-9*x)/(1-10*x-x^2). - Philippe Deléham, Nov 20 2008
For n>=2, a(n) = F_(n)(10) + F_(n+1)(10), where F_n(x) is Fibonacci polynomial (cf.A049310): F_n(x) = Sum_{i=0,...,floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012

A015455 a(n) = 9*a(n-1) + a(n-2) for n>1; a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 10, 91, 829, 7552, 68797, 626725, 5709322, 52010623, 473804929, 4316254984, 39320099785, 358197153049, 3263094477226, 29726047448083, 270797521509973, 2466903741037840, 22472931190850533

Offset: 0

Olivier Gérard

Generalized Fibonacci numbers.
As R. K. Guy suggested on the SeqFan list, the sequence could be extended "to the left side" by ..., 10, 1, 1, -8, 73, -665, 6058, -55187, 502741, -4579856, 41721445, ... by using the recurrence equation to get a(n-2) = a(n) - 9 a(n-1). The sequence 1,-8,73,... would have g.f. (1+x)/(1+9x-x^2).
For n>=1, row sums of triangle for numbers 9^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,9} containing no subwords ii, (i=0,1,...,8). - Milan Janjic, Jan 31 2015

R. K. Guy, "A further family of sequences", SeqFan mailing list (www.seqfan.eu), Jun 13 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (9,1).

Row m=9 of A135597.

[n le 2 select 1 else 9*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 01 2015

CoefficientList[Series[(1 - 8*x)/(1 - 9*x - x^2), {x, 0, 50}], x] (* or *) LinearRecurrence[{9,1}, {1,1}, 50] (* G. C. Greubel, Dec 19 2017 *)

a(n) = polcoeff((1-(O(x^n)+8)*x)/(1-9*x-x^2),n) \\ M. F. Hasler, Jun 14 2008

G.f.: (1 - 8*x)/(1 - 9*x - x^2). - M. F. Hasler, Jun 14 2008
a(n) = Sum_{k, 0<=k<=n} 8^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
a(n) = round(1/2*(9/2 - 1/2*sqrt(85))^n + 7/170*sqrt(85)*(9/2 - 1/2*sqrt(85))^n - 7/170*sqrt(85)*(9/2 + 1/2*sqrt(85))^n + 1/2*(9/2 + 1/2*sqrt(85))^n). - Alexander R. Povolotsky, Jun 13 2008
For n>=2, a(n)=F_n(9)+F_(n+1)(9), where F_n(x) is Fibonacci polynomial (cf.A049310): F_n(x) = Sum_{i=0,...,floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012

Edited by M. F. Hasler, Jun 14 2008

A121875 Triangular array read by rows: see Comments for definition.

Original entry on oeis.org

1, 2, 3, 3, 7, 13, 5, 17, 43, 89, 8, 41, 142, 377, 836, 13, 99, 469, 1597, 4341, 10063, 21, 239, 1549, 6765, 22541, 62011, 148149, 34, 577, 5116, 28657, 117046, 382129, 1057792, 2581921, 55, 1393, 16897, 121393, 607771, 2354785, 7552693, 20973217

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 09 2006

Form the square array in which row m satisfies r(0) = r(1) = 1; r(n) = m*r(n-1) + r(n-2):
1 1 2 3 5 8 13 21 ...
1 1 3 7 17 41 99 ...
1 1 4 13 43 142 ...
1 1 5 21 89 377 ...
...
Now form a triangle by taking the first k terms of column k:
1
2, 3
3, 7, 13
5, 17, 43, 89
8, 41, 142, 377, 836
...

Cf. A135597.

f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[n] == m*a[n - 1] + a[n - 2], a[0] == 1, a[1] == 1}, a[n], n][[1]] // FullSimplify] a = Table[Table[Rationalize[N[f[n], 100], 0], {m, 1, n}], {n, 1, 10}] Flatten[a]

Edited by N. J. A. Sloane, Mar 02 2008

cp's OEIS Frontend

A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

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A015449 Expansion of (1-4*x)/(1-5*x-x^2).

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A015451 a(n) = 6*a(n-1) + a(n-2) for n > 1, with a(0) = a(1) = 1.

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A015457 Generalized Fibonacci numbers.

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A287376 Array read by antidiagonals: T(m,n) = number of independent vertex sets in the complete prism graph K_m X C_n.

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A015453 Generalized Fibonacci numbers.

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A015449 Expansion of (1-4x)/(1-5x-x^2).