A005013 -id:A005013 - 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

A026741 a(n) = n if n odd, n/2 if n even.

Original entry on oeis.org

0, 1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, 17, 9, 19, 10, 21, 11, 23, 12, 25, 13, 27, 14, 29, 15, 31, 16, 33, 17, 35, 18, 37, 19, 39, 20, 41, 21, 43, 22, 45, 23, 47, 24, 49, 25, 51, 26, 53, 27, 55, 28, 57, 29, 59, 30, 61, 31, 63, 32, 65, 33, 67, 34, 69, 35, 71, 36, 73, 37, 75, 38

Offset: 0

J. Carl Bellinger (carlb(AT)ctron.com)

a(n) is the size of largest conjugacy class in D_2n, the dihedral group with 2n elements. - Sharon Sela (sharonsela(AT)hotmail.com), May 14 2002
a(n+1) is the composition length of the n-th symmetric power of the natural representation of a finite subgroup of SL(2,C) of type D_4 (quaternion group). - Paul Boddington, Oct 23 2003
For n > 1, a(n) is the greatest common divisor of all permutations of {0, 1, ..., n} treated as base n + 1 integers. - David Scambler, Nov 08 2006 (see the Mathematics Stack Exchange link below).
From Dimitrios Choussos (choussos(AT)yahoo.de), May 11 2009: (Start)
Sequence A075888 and the above sequence are fitting together.
First 2 entries of this sequence have to be taken out.
In some cases two three or more sequenced entries of this sequence have to be added together to get the next entry of A075888.
Example: Sequences begin with 1, 3, 2, 5, 3, 7, 4, 9 (4 + 9 = 13, the next entry in A075888).
But it works out well up to primes around 50000 (haven't tested higher ones).
As A075888 gives a very regular graph. There seems to be a regularity in the primes. (End)
Starting with 1 = triangle A115359 * [1, 2, 3, ...]. - Gary W. Adamson, Nov 27 2009
From Gary W. Adamson, Dec 11 2009: (Start)
Let M be an infinite lower triangular matrix with (1, 1, 1, 0, 0, 0, ...) in every column, shifted down twice. This sequence starting with 1 = M * (1, 2, 3, ...)
M =
  1;
  1, 0;
  1, 1, 0;
  0, 1, 0, 0;
  0, 1, 1, 0, 0;
  0, 0, 1, 0, 0, 0;
  0, 0, 1, 1, 0, 0, 0;
  ...
A026741 = M * (1, 2, 3, ...); but A002487 = lim_{n->infinity} M^n, a left-shifted vector considered as a sequence. (End)
A particular case of sequence for which a(n+3) = (a(n+2) * a(n+1)+q)/a(n) for every n > n0. Here n0 = 1 and q = -1. - Richard Choulet, Mar 01 2010
For n >= 2, a(n+1) is the smallest m such that s_n(2*m*(n-1))/(n-1) is even, where s_b(c) is the sum of digits of c in base b. - Vladimir Shevelev, May 02 2011
A001477 and A005408 interleaved. - Omar E. Pol, Aug 22 2011
Numerator of n/((n-1)*(n-2)). - Michael B. Porter, Mar 18 2012
Number of odd terms of n-th row in the triangles A162610 and A209297. - Reinhard Zumkeller, Jan 19 2013
For n >= 3, a(n) is the periodic of integer of spiral length ratio of spiral that have (n-1) circle centers. See illustration in links. - Kival Ngaokrajang, Dec 28 2013
This is the sequence of Lehmer numbers u_n(sqrt(R), Q) with the parameters R = 4 and Q = 1. It is a strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all natural numbers n and m. Cf. A005013 and A108412. - Peter Bala, Apr 18 2014
The sequence of convergents of the 2-periodic continued fraction [0; 1, -4, 1, -4, ...] = 1/(1 - 1/(4 - 1/(1 - 1/(4 - ...)))) = 2 begins [0/1, 1/1, 4/3, 3/2, 8/5, 5/3, 12/7, ...]. The present sequence is the sequence of denominators; the sequence of numerators of the continued fraction convergents [0, 1, 4, 3, 8, 5, 12, ...] is A022998, also a strong divisibility sequence. - Peter Bala, May 19 2014
For n >= 3, (a(n-2)/a(n))*Pi = vertex angle of a regular n-gon. See illustration in links. - Kival Ngaokrajang, Jul 17 2014
For n > 1, the numerator of the harmonic mean of the first n triangular numbers. - Colin Barker, Nov 13 2014
The difference sequence is a permutation of the integers. - Clark Kimberling, Apr 19 2015
From Timothy Hopper, Feb 26 2017: (Start)
Given the function a(n, p) = n/p if n mod p = 0, else n, then a possible formula is: a(n, p) = n*(1 + (p-1)*((n^(p-1)) mod p))/p, p prime, (n^(p-1)) mod p = 1, n not divisible by p. (Fermat's Little Theorem). Examples: p = 2; a(n), p = 3; A051176(n), p = 5; A060791(n), p = 7; A106608(n).
Conjecture: lcm(n, p) = p*a(n, p), gcd(n, p) = n/a(n, p). (End)
Let r(n) = (a(n+1) + 1)/a(n+1) if n mod 2 = 1, a(n+1)/(a(n+1) + 2) otherwise; then lim_{k->oo} 2^(k+2) * Product_{n=0..k} r(n)^(k-n) = Pi. - Dimitris Valianatos, Mar 22 2021
Number of integers k from 1 to n such that gcd(n,k) is odd. - Amiram Eldar, May 18 2025

			G.f. = x + x^2 + 3*x^3 + 2*x^4 + 5*x^5 + 3*x^6 + 7*x^7 + 4*x^8 + ...

David Wells, Prime Numbers: The Most Mysterious Figures in Math. Hoboken, New Jersey: John Wiley & Sons (2005), p. 53.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, 2nd Ed. Penguin (1997), p. 79.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv:1105.3399 [math.GM], 2011.
Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, par. 2.
Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
Y. Ito and I. Nakamura, Hilbert schemes and simple singularities, New trends in algebraic geometry (Warwick, 1996), 151-233, Cambridge University Press, 1999.
Masanobu Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), Article 00.2.9.
Mathematics Stack Exchange, Permutations (with no duplicates) of decimal base digits 1,2,...,8,9,0.
Kival Ngaokrajang, Illustration of spiral with circle centers 2..5.
Kival Ngaokrajang, Illustration of vertex angle of regular n-gon for n = 3..7.
Eric Weisstein's World of Mathematics, Lehmer Number.
Eric Weisstein's World of Mathematics, Simplex Simplex Picking.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).
Index to divisibility sequences.

Signed version is in A030640. Partial sums give A001318.
Cf. A051176, A060819, A060791, A060789 for n / gcd(n, k) with k = 3..6. See also A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).
Cf. A045896, A022998, A060762, A126988, A109007, A130334, A109043, A115359, A002487, A220466.
Cf. A013942.
Cf. A227042 (first column). Cf. A005013 and A108412.
Cf. A000217, A256095.

import Data.List (transpose)
a026741 n = a026741_list !! n
a026741_list = concat $ transpose [[0..], [1,3..]]
-- Reinhard Zumkeller, Dec 12 2011

[2*n/(3+(-1)^n): n in [0..70]]; // Vincenzo Librandi, Aug 14 2011

A026741 := proc(n) if type(n,'odd') then n; else n/2; end if; end proc: seq(A026741(n), n=0..76); # R. J. Mathar, Jan 22 2011

Numerator[Abs[Table[Det[DiagonalMatrix[Table[1/i^2 - 1, {i, 1, n - 1}]] + 1], {n, 20}]]] (* Alexander Adamchuk, Jun 02 2006 *)
halfMax = 40; Riffle[Range[0, halfMax], Range[1, 2halfMax + 1, 2]] (* Harvey P. Dale, Mar 27 2011 *)
a[ n_] := Numerator[n / 2]; (* Michael Somos, Jan 20 2017 *)
Array[If[EvenQ[#],#/2,#]&,80,0] (* Harvey P. Dale, Jul 08 2023 *)

a(n) = numerator(n/2) \\ Rick L. Shepherd, Sep 12 2007

def A026741(n): return n if n % 2 else n//2 # Chai Wah Wu, Apr 02 2021

[lcm(n, 2) / 2 for n in range(77)] # Zerinvary Lajos, Jun 07 2009

G.f.: x*(1 + x + x^2)/(1-x^2)^2. - Len Smiley, Apr 30 2001
a(n) = 2*a(n-2) - a*(n-4) for n >= 4.
a(n) = n * 2^((n mod 2) - 1). - Reinhard Zumkeller, Oct 16 2001
a(n) = 2*n/(3 + (-1)^n). - Benoit Cloitre, Mar 24 2002
Multiplicative with a(2^e) = 2^(e-1) and a(p^e) = p^e, p > 2. - Vladeta Jovovic, Apr 05 2002
a(n) = n / gcd(n, 2). a(n)/A045896(n) = n/((n+1)*(n+2)).
For n > 0, a(n) = denominator of Sum_{i=1..n-1} 2/(i*(i+1)), numerator=A022998. - Reinhard Zumkeller, Apr 21 2012, Jul 25 2002 [thanks to Phil Carmody who noticed an error]
For n > 1, a(n) = GCD of the n-th and (n-1)-th triangular numbers (A000217). - Ross La Haye, Sep 13 2003
Euler transform of finite sequence [1, 2, -1]. - Michael Somos, Jun 15 2005
G.f.: x * (1 - x^3) / ((1 - x) * (1 - x^2)^2) = Sum_{k>0} k * (x^k - x^(2*k)). - Michael Somos, Jun 15 2005
a(n+3) + a(n+2) = 3 + a(n+1) + a(n). a(n+3) * a(n) = - 1 + a(n+2) * a(n+1). a(n) = -a(-n) for all n in Z. - Michael Somos, Jun 15 2005
For n > 1, a(n) is the numerator of the average of 1, 2, ..., n - 1; i.e., numerator of A000217(n-1)/(n-1), with corresponding denominators [1, 2, 1, 2, ...] (A000034). - Rick L. Shepherd, Jun 05 2006
Equals A126988 * (1, -1, 0, 0, 0, ...). - Gary W. Adamson, Apr 17 2007
For n >= 1, a(n) = gcd(n,A000217(n)). - Rick L. Shepherd, Sep 12 2007
a(n) = numerator(n/(2*n-2)) for n >= 2; A022998(n-1) = denominator(n/(2*n-2)) for n >= 2. - Johannes W. Meijer, Jun 18 2009
a(n) = A167192(n+2, 2). - Reinhard Zumkeller, Oct 30 2009
a(n) = A106619(n) * A109012(n). - Paul Curtz, Apr 04 2011
From R. J. Mathar, Apr 18 2011: (Start)
a(n) = A109043(n)/2.
Dirichlet g.f.: zeta(s-1)*(1 - 1/2^s). (End)
a(n) = A001318(n) - A001318(n-1) for n > 0. - Jonathan Sondow, Jan 28 2013
a((2*n+1)*2^p - 1) = 2^p - 1 + n*A151821(p+1), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 03 2013
a(n+1) = denominator(H(n, 1)), n >= 0, with H(n, 1) = 2*n/(n+1) the harmonic mean of n and 1. a(n+1) = A227042(n, 1). See the formula a(n) = n/gcd(n, 2) given above. - Wolfdieter Lang, Jul 04 2013
a(n) = numerator(n/2). - Wesley Ivan Hurt, Oct 02 2013
a(n) = numerator(1 - 2/(n+2)), n >= 0; a(n) = denominator(1 - 2/n), n >= 1. - Kival Ngaokrajang, Jul 17 2014
a(n) = Sum_{i = floor(n/2)..floor((n+1)/2)} i. - Wesley Ivan Hurt, Apr 27 2016
Euler transform of length 3 sequence [1, 2, -1]. - Michael Somos, Jan 20 2017
G.f.: x / (1 - x / (1 - 2*x / (1 + 7*x / (2 - 9*x / (7 - 4*x / (3 - 7*x / (2 + 3*x))))))). - Michael Somos, Jan 20 2017
From Peter Bala, Mar 24 2019: (Start)
a(n) = Sum_{d|n, n/d odd} phi(d), where phi(n) is the Euler totient function A000010.
O.g.f.: Sum_{n >= 1} phi(n)*x^n/(1 - x^(2*n)). (End)
a(n) = A256095(2*n,n). - Alois P. Heinz, Jan 21 2020
E.g.f.: x*(2*cosh(x) + sinh(x))/2. - Stefano Spezia, Apr 28 2023
From Ctibor O. Zizka, Oct 05 2023: (Start)
For k >= 0, a(k) = gcd(k + 1, k*(k + 1)/2).
If (k mod 4) = 0 or 2 then a(k) = (k + 1).
If (k mod 4) = 1 or 3 then a(k) = (k + 1)/2. (End)
Sum_{n=1..oo} 1/a(n)^2 = 7*Pi^2/24. - Stefano Spezia, Dec 02 2023
a(n)*a(n+1) = A000217(n). - Rémy Sigrist, Mar 19 2025

Better description from Jud McCranie

Edited by Ralf Stephan, Jun 04 2003

A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

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

Offset: 0

Len Smiley, Nov 29 2001

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020
Row sums give the Fibonacci sequence. So do the alternating row sums.
Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]
Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011
Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011
From Gary W. Adamson, Aug 25 2019: (Start)
Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.
For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.
Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:
   3...............1
   5...............1, 2
   7...............1, 2, 3
   9...............1, 2, 4
    ...............3 (singleton terms reduce to "1") (9 has two rows)
  11...............1, 2, 4, 3, 5
  13...............1, 2, 4, 5, 3, 6
  15...............1, 2, 4, 7
   ................3, 6  (dividing through by the gcd gives (1, 2))
   ................5. (singleton terms reduce to "1")
The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)
From Gary W. Adamson, Sep 30 2019: (Start)
The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)
From Gary W. Adamson, Aug 21 2019: (Start)
Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1,  0]; the f(x) trajectory is:
  ....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,
  ....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)
Received from Herb Conn, Jan 2004: (Start)
Let x = 2*cos(2A) (A = Angle); then
sin(A)/sin A  = 1
sin(3A)/sin A = x + 1
sin(5A)/sin A = x^2 + x - 1
sin(7A)/sin A = x^3 + x - 2x - 1
sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1
... (signed ++--++...). (End)
Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), 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 12 2012
The matrix inverse is given by
  1;
  1,  1;
  0, -1,  1;
  0,  1, -2,  1;
  0,  0,  1, -2,  1;
  0,  0, -1,  3, -3,  1;
  0,  0,  0, -1,  3, -3,  1;
  0,  0,  0,  1, -4,  6, -4,  1;
  0,  0,  0,  0,  1, -4,  6, -4,  1;
... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

			Triangle T(n, k) begins:
n\k 0  1  2  3   4   5  6   7  8  9 ...
---------------------------------------
[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2,  1,
[5] 1, 1, 4, 3,  3,  1,
[6] 1, 1, 5, 4,  6,  3,  1,
[7] 1, 1, 6, 5, 10,  6,  4,  1,
[8] 1, 1, 7, 6, 15, 10, 10,  4,  1,
[9] 1, 1, 8, 7, 21, 15, 20, 10,  5, 1,
---------------------------------------
From _Gary W. Adamson_, Oct 23 2019: (Start)
Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2 - 2x - 1) and the roots (a, b, c) are (-1.8019377..., 1.247697..., and -0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((a-b), (b-c), (c-a))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3 - 7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
  N =  5:  x^2 - 5;
  N =  7:  x^3 - 7x + 7;
  N = 11:  x^5 - 11x^3 + 11x^2 + 11x - 11;
  N = 13:  x^6 - 13x^4 + 13x^3 + 26x^2 - 39x + 13;
  N = 17:  x^8 - 17x^6 + 17x^5 + 68x^4 - 119x^3 + 17x^2 + 51x - 17;
  N = 19:  x^9 - 19x^7 + 19x^6 + 95x^5 - 171x^4 - 19x^3 + 190x^2 - 114x + 19. (End)

Nathaniel Johnston, Rows 0..100, flattened
Leonard E. Dickson, The Inscription of Regular Polygons, The American Mathematical Monthly, Vol. 1, No. 9 (Sep., 1894), pp. 299-301.
Henry W. Gould, A Variant of Pascal's Triangle, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271, with corrections.
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. See Table 4.
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. [Annotated scanned copy]
Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; p. 490.
Jay Kappraff, S. Jablan, G. W. Adamson, and R. Sazdanovich, Golden Fields, Generalized Fibonacci Sequences, and Chaotic Matrices, FORMA, Vol. 19, No. 4, 2005.
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Journal of Dynamical Systems and Geometric Theories, Vol 2 (2004), p 65.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, 2001 (Chapter 14)
E. Munarini and N. Z. Salvi, Binary strings without zigzags, Séminaire Lotharingien de Combinatoire, B49h (2004), 15 pp.
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017.
P.C. Parks, A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov , Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) p. 694-702.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A065942 (central stalk sequence), A000045 (row sums), A108299.
Reflected version of A046854.
Some triangle sums (see A180662): A000045 (Fi1), A016116 (Kn21), A000295 (Kn23), A094967 (Fi2), A000931 (Ca2), A001519 (Gi3), A000930 (Ze3).
Cf. A003558, A182579, A059841, A333143.

a065941 n k = a065941_tabl !! n !! k
a065941_row n = a065941_tabl !! n
a065941_tabl = iterate (\row ->
   zipWith (+) ([0] ++ row) (zipWith (*) (row ++ [0]) a059841_list)) [1]
-- Reinhard Zumkeller, May 07 2012

[Binomial(n - Floor((k+1)/2), Floor(k/2)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 10 2019

A065941 := proc(n,k): binomial(n-floor((k+1)/2),floor(k/2)) end: seq(seq(A065941(n,k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
A065941 := proc(n,k) option remember: local j: if k=0 then 1 elif k=1 then 1: elif k>=2 then add(procname(j,k-2), j=k-2..n-2) fi: end: seq(seq(A065941(n,k), k=0..n), n=0..15);  # Johannes W. Meijer, Aug 11 2011
# The function qStirling2 is defined in A333143.
seq(print(seq(qStirling2(n, k, -1), k=0..n)), n=0..9);
# Peter Luschny, Mar 09 2020

Flatten[Table[Binomial[n-Floor[(k+1)/2],Floor[k/2]],{n,0,15},{k,0,n}]] (* Harvey P. Dale, Dec 11 2011 *)

T065941(n, k) = binomial(n-(k+1)\2, k\2); \\ Michel Marcus, Apr 28 2014

[[binomial(n - floor((k+1)/2), floor(k/2)) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 10 2019

T(n, k) = binomial(n-floor((k+1)/2), floor(k/2)).
As a square array read by antidiagonals, this is given by T1(n, k) = binomial(floor(n/2) + k, k). - Paul Barry, Mar 11 2003
Triangle is a reflection of that in A066170 (absolute values). - Gary W. Adamson, Feb 16 2004
Recurrences: T(k, 0) = 1, T(k, n) = T(k-1, n) + T(k-2, n-2), or T(k, n) = T(k-1, n) + T(k-1, n-1) if n even, T(k-1, n-1) if n odd. - Ralf Stephan, May 17 2004
G.f.: sum[n, sum[k, T(k, n)x^ky^n]] = (1+xy)/(1-y-x^2y^2). sum[n>=0, T(k, n)y^n] = y^k/(1-y)^[k/2]. - Ralf Stephan, May 17 2004
T(n, k) = A108299(n, k)*A087960(k) = abs(A108299(n, k)). - Reinhard Zumkeller, Jun 01 2005
From Johannes W. Meijer, Aug 11 2011: (Start)
  T(n,k) = A046854(n, n-k) = abs(A066170(n, n-k)).
  T(n+k, n-k) = A109223(n,k).
  T(n, k) = sum(T(j, k-2), j=k-2..n-2), 2 <= k <= n, n>=2;
  T(n, 0) =1, T(n+1, 1) = 1, n >= 0. (End)
For n > 1: T(n, k) = T(n-2, k) + T(n-1, k), 1 < k < n. - Reinhard Zumkeller, Apr 24 2013

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

Original entry on oeis.org

1, 19, 341, 6119, 109801, 1970299, 35355581, 634430159, 11384387281, 204284540899, 3665737348901, 65778987739319, 1180356041958841, 21180629767519819, 380070979773397901, 6820097006153642399, 122381675130992165281, 2196050055351705332659, 39406519321199703822581

Offset: 0

Clark Kimberling

x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.
The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.
a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

			Pell, n=1: (2*19)^2 - 5*17^2 = -1.

G. C. Greubel, Table of n, a(n) for n = 0..795
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Tanya Khovanova, Recursive Sequences.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Eric Weisstein's World of Mathematics, Lehmer Number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Bisection of A001077 divided by 2.
Cf. A000045, A007805, A049310, A049660, A100047, A188378.
Cf. A005013, A033890, A049685, A084608.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

[(Fibonacci(6*n+5) - Fibonacci(6*n+1))/4: n in [0..30]]; // G. C. Greubel, Dec 15 2017

with(numtheory): with(combinat):
seq((fibonacci(6*n+5)-fibonacci(6*n+1))/4,n=0..20); # Muniru A Asiru, Mar 25 2018

a[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[a, 16] (* Robert G. Wilson v, Oct 28 2010 *)

my(x='x+O('x^30)); Vec((1+x)/(1 - 18*x + x^2)) \\ G. C. Greubel, Dec 15 2017

a(n) ~ (1/4)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all members x of the sequence, 20*x^2 + 5 is a square. Lim_{n -> oo} a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the statement "20*x^2 + 5 is a square". - Gregory V. Richardson, Oct 12 2002
a(n) = (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)). - Gregory V. Richardson, Oct 12 2002
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1+x)/(1 - 18x + x^2).
a(n) = A049660(n) + A049660(n+1). (End)
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - Philippe Deléham, Nov 17 2008
a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) + Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) - Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 1.
The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - Altug Alkan, Jan 24 2016
a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - Gerry Martens, Jul 25 2016
a(n) = Lucas(6*n + 3)/4. - Ehren Metcalfe, Feb 18 2017
From Peter Bala, Mar 23 2018: (Start)
a(n) = 1/4*( (sqrt(5) + 2)^(2*n+1) - (sqrt(5) - 2)^(2*n+1) ).
a(n) = 9*a(n-1) + 2*sqrt(5 + 20*a(n-1)^2).
a(n) = (1/2)*sinh((2*n + 1)*arcsinh(2)). (End)
From Peter Bala, May 09 2025: (Start)
a(n)^2 - 18*a(n)*a(n+1) + a(n+1)^2 = 20.
More generally, for real x, a(n+x)^2 - 18*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 20 with a(n) := (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)) as given above.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/20 (telescoping series).
Product_{n >= 1} ((a(n) + 1)/(a(n) - 1)) = sqrt(5)/2 (telescoping product). (End)

cp's OEIS Frontend

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

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A026741 a(n) = n if n odd, n/2 if n even.

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A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

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A049629 a(n) = (F(6*n+5) - F(6*n+1))/4 = (F(6*n+4) + F(6*n+2))/4, where F = A000045.

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A147600 Expansion of 1/(1 - 3*x^2 + x^4).

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A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.