A238731 -id:A238731 - cp's OEIS frontend

A001519 a(n) = 3*a(n-1) - a(n-2) for n >= 2, with a(0) = a(1) = 1.

1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437, 1134903170, 2971215073, 7778742049, 20365011074, 53316291173, 139583862445, 365435296162, 956722026041

Offset: 0

N. J. A. Sloane

This is a bisection of the Fibonacci sequence A000045. a(n) = F(2*n-1), with F(n) = A000045(n) and F(-1) = 1.
Number of ordered trees with n+1 edges and height at most 3 (height=number of edges on a maximal path starting at the root). Number of directed column-convex polyominoes of area n+1. Number of nondecreasing Dyck paths of length 2n+2. - Emeric Deutsch, Jul 11 2001
Terms are the solutions x to: 5x^2-4 is a square, with 5x^2-4 in A081071 and sqrt(5x^2-4) in A002878. - Benoit Cloitre, Apr 07 2002
a(0) = a(1) = 1, a(n+1) is the smallest Fibonacci number greater than the n-th partial sum. - Amarnath Murthy, Oct 21 2002
The fractional part of tau*a(n) decreases monotonically to zero. - Benoit Cloitre, Feb 01 2003
Numbers k such that floor(phi^2*k^2) - floor(phi*k)^2 = 1 where phi=(1+sqrt(5))/2. - Benoit Cloitre, Mar 16 2003
Number of leftist horizontally convex polyominoes with area n+1.
Number of 31-avoiding words of length n on alphabet {1,2,3} which do not end in 3. (E.g., at n=3, we have 111, 112, 121, 122, 132, 211, 212, 221, 222, 232, 321, 322 and 332.) See A028859. - Jon Perry, Aug 04 2003
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=phi=(1+sqrt(5))/2. - Benoit Cloitre, Feb 24 2004
a(1) = 1, a(2) = 2, then the least number such that the square of any term is just less than the geometric mean of its neighbors. a(n+1)*a(n-1) > a(n)^2. - Amarnath Murthy, Apr 06 2004
All positive integer solutions of Pell equation b(n)^2 - 5*a(n+1)^2 = -4 together with b(n)=A002878(n), n >= 0. - Wolfdieter Lang, Aug 31 2004
Essentially same as Pisot sequence E(2,5).
Number of permutations of [n+1] avoiding 321 and 3412. E.g., a(3) = 13 because the permutations of [4] avoiding 321 and 3412 are 1234, 2134, 1324, 1243, 3124, 2314, 2143, 1423, 1342, 4123, 3142, 2413, 2341. - Bridget Tenner, Aug 15 2005
Number of 1324-avoiding circular permutations on [n+1].
A subset of the Markoff numbers (A002559). - Robert G. Wilson v, Oct 05 2005
(x,y) = (a(n), a(n+1)) are the solutions of x/(yz) + y/(xz) + z/(xy) = 3 with z=1. - Floor van Lamoen, Nov 29 2001
Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 1. - Herbert Kociemba, Jun 10 2004
With interpolated zeros, counts closed walks of length n at the start or end node of P_4. a(n) counts closed walks of length 2n at the start or end node of P_4. The sequence 0,1,0,2,0,5,... counts walks of length n between the start and second node of P_4. - Paul Barry, Jan 26 2005
a(n) is the number of ordered trees on n edges containing exactly one non-leaf vertex all of whose children are leaves (every ordered tree must contain at least one such vertex). For example, a(0) = 1 because the root of the tree with no edges is not considered to be a leaf and the condition "all children are leaves" is vacuously satisfied by the root and a(4) = 13 counts all 14 ordered trees on 4 edges (A000108) except (ignore dots)
  |..|
  .\/.
  which has two such vertices. - David Callan, Mar 02 2005
Number of directed column-convex polyominoes of area n. Example: a(2)=2 because we have the 1 X 2 and the 2 X 1 rectangles. - Emeric Deutsch, Jul 31 2006
Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons in extended (1,1)-nanotubes. See Table 1 on page 411 of I. Lukovits and D. Janezic. - Parthasarathy Nambi, Aug 22 2006
Number of free generators of degree n of symmetric polynomials in 3-noncommuting variables. - Mike Zabrocki, Oct 24 2006
Inverse: With phi = (sqrt(5) + 1)/2, log_phi((sqrt(5)*a(n) + sqrt(5*a(n)^2 - 4))/2) = n for n >= 1. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
Consider a teacher who teaches one student, then he finds he can teach two students while the original student learns to teach a student. And so on with every generation an individual can teach one more student then he could before. a(n) starting at a(2) gives the total number of new students/teachers (see program). - Ben Paul Thurston, Apr 11 2007
The Diophantine equation a(n)=m has a solution (for m >= 1) iff ceiling(arcsinh(sqrt(5)*m/2)/log(phi)) != ceiling(arccosh(sqrt(5)*m/2)/log(phi)) where phi is the golden ratio. An equivalent condition is A130255(m)=A130256(m). - Hieronymus Fischer, May 24 2007
a(n+1) = B^(n)(1), n >= 0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 2=`0`, 5=`00`, 13=`000`, ..., in Wythoff code.
Bisection of the Fibonacci sequence into odd-indexed nonzero terms (1, 2, 5, 13, ...) and even-indexed terms (1, 3, 8, 21, ...) may be represented as row sums of companion triangles A140068 and A140069. - Gary W. Adamson, May 04 2008
a(n) is the number of partitions pi of [n] (in standard increasing form) such that Flatten[pi] is a (2-1-3)-avoiding permutation. Example: a(4)=13 counts all 15 partitions of [4] except 13/24 and 13/2/4. Here "standard increasing form" means the entries are increasing in each block and the blocks are arranged in increasing order of their first entries. Also number that avoid 3-1-2. - David Callan, Jul 22 2008
Let P be the partial sum operator, A000012: (1; 1,1; 1,1,1; ...) and A153463 = M, the partial sum & shift operator. It appears that beginning with any randomly taken sequence S(n), iterates of the operations M * S(n), -> M * ANS, -> P * ANS, etc. (or starting with P) will rapidly converge upon a two-sequence limit cycle of (1, 2, 5, 13, 34, ...) and (1, 1, 3, 8, 21, ...). - Gary W. Adamson, Dec 27 2008
Number of musical compositions of Rhythm-music over a time period of n-1 units. Example: a(4)=13; indeed, denoting by R a rest over a time period of 1 unit and by N[j] a note over a period of j units, we have (writing N for N[1]): NNN, NNR, NRN, RNN, NRR, RNR, RRN, RRR, N[2]R, RN[2], NN[2], N[2]N, N[3] (see the J. Groh reference, pp. 43-48). - Juergen K. Groh (juergen.groh(AT)lhsystems.com), Jan 17 2010
Given an infinite lower triangular matrix M with (1, 2, 3, ...) in every column but the leftmost column shifted upwards one row. Then (1, 2, 5, ...) = lim_{n->infinity} M^n. (Cf. A144257.) - Gary W. Adamson, Feb 18 2010
As a fraction: 8/71 = 0.112676 or 98/9701 = 0.010102051334... (fraction 9/71 or 99/9701 for sequence without initial term). 19/71 or 199/9701 for sequence in reverse. - Mark Dols, May 18 2010
For n >= 1, a(n) is the number of compositions (ordered integer partitions) of 2n-1 into an odd number of odd parts. O.g.f.: (x-x^3)/(1-3x^2+x^4) = A(A(x)) where A(x) = 1/(1-x)-1/(1-x^2).
For n > 0, determinant of the n X n tridiagonal matrix with 1's in the super and subdiagonals, (1,3,3,3,...) in the main diagonal, and the rest zeros. - Gary W. Adamson, Jun 27 2011
The Gi3 sums, see A180662, of the triangles A108299 and A065941 equal the terms of this sequence without a(0). - Johannes W. Meijer, Aug 14 2011
The number of permutations for which length equals reflection length. - Bridget Tenner, Feb 22 2012
Number of nonisomorphic graded posets with 0 and 1 and uniform Hasse graph of rank n+1, with exactly 2 elements of each rank between 0 and 1. (Uniform used in the sense of Retakh, Serconek and Wilson. Graded used in R. Stanley's sense that all maximal chains have the same length.)
HANKEL transform of sequence and the sequence omitting a(0) is the sequence A019590(n). This is the unique sequence with that property. - Michael Somos, May 03 2012
The number of Dyck paths of length 2n and height at most 3. - Ira M. Gessel, Aug 06 2012
Pisano period lengths: 1, 3, 4, 3, 10, 12, 8, 6, 12, 30, 5, 12, 14, 24, 20, 12, 18, 12, 9, 30, ... - R. J. Mathar, Aug 10 2012
Primes in the sequence are 2, 5, 13, 89, 233, 1597, 28657, ... (apparently A005478 without the 3). - R. J. Mathar, May 09 2013
a(n+1) is the sum of rising diagonal of the Pascal triangle written as a square - cf. comments in A085812. E.g., 13 = 1+5+6+1. - John Molokach, Sep 26 2013
a(n) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 1, 1, 1; 0, 1, 1] or [1, 1, 1; 0, 1, 1; 1, 1, 1] or [1, 1, 0; 1, 1, 1; 1, 1, 1] or [1, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Except for the initial term, positive values of x (or y) satisfying x^2 - 3xy + y^2 + 1 = 0. - Colin Barker, Feb 04 2014
Except for the initial term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 64 = 0. - Colin Barker, Feb 16 2014
Positive values of x such that there is a y satisfying x^2 - xy - y^2 - 1 = 0. - Ralf Stephan, Jun 30 2014
a(n) is also the number of permutations simultaneously avoiding 231, 312 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
(1, a(n), a(n+1)), n >= 0, are Markoff triples (see A002559 and Robert G. Wilson v's Oct 05 2005 comment). In the Markoff tree they give one of the outer branches. Proof: a(n)*a(n+1) - 1 = A001906(2*n)^2 = (a(n+1) - a(n))^2 = a(n)^2 + a(n+1)^2 - 2*a(n)*a(n+1), thus 1^2 + a(n)^2 + a(n+1)^2 = 3*a(n)*a(n+1). - Wolfdieter Lang, Jan 30 2015
For n > 0, a(n) is the smallest positive integer not already in the sequence such that a(1) + a(2) + ... + a(n) is a Fibonacci number. - Derek Orr, Jun 01 2015
Number of vertices of degree n-2 (n >= 3) in all Fibonacci cubes, see Klavzar, Mollard, & Petkovsek. - Emeric Deutsch, Jun 22 2015
Except for the first term, this sequence can be generated by Corollary 1 (ii) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015
Precisely the numbers F(n)^k + F(n+1)^k that are also Fibonacci numbers with k > 1, see Luca & Oyono. - Charles R Greathouse IV, Aug 06 2015
a(n) = MA(n) - 2*(-1)^n where MA(n) is exactly the maximum area of a quadrilateral with lengths of sides in order L(n-2), L(n-2), F(n+1), F(n+1) for n > 1 and L(n)=A000032(n). - J. M. Bergot, Jan 28 2016
a(n) is the number of bargraphs of semiperimeter n+1 having no valleys (i.e., convex bargraphs). Equivalently, number of bargraphs of semiperimeter n+1 having exactly 1 peak. Example: a(5) = 34 because among the 35 (=A082582(6)) bargraphs of semiperimeter 6 only the one corresponding to the composition [2,1,2] has a valley. - Emeric Deutsch, Aug 12 2016
Integers k such that the fractional part of k*phi is less than 1/k. See Byszewski link p. 2. - Michel Marcus, Dec 10 2016
Number of words of length n-1 over {0,1,2,3} in which binary subwords appear in the form 10...0. - Milan Janjic, Jan 25 2017
With a(0) = 0 this is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the Fibonacci sequence A000045. See a Feb 17 2017 comment on A097805. - Wolfdieter Lang, Feb 17 2017
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) < e(k). [Martinez and Savage, 2.12] - Eric M. Schmidt, Jul 17 2017
Number of permutations of [n] that avoid the patterns 321 and 2341. - Colin Defant, May 11 2018
The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j^2 and j divides 1+i^2. In fact, the pairs (a(n), a(n+1)), n > 0, are all the solutions. - Tomohiro Yamada, Dec 23 2018
Number of permutations in S_n whose principal order ideals in the Bruhat order are lattices (equivalently, modular, distributive, Boolean lattices). - Bridget Tenner, Jan 16 2020
From Wolfdieter Lang, Mar 30 2020: (Start)
a(n) is the upper left entry of the n-th power of the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from A322602: a(n) = ((M_2)^n)[1,1].
Proof: (M_2)^2 = 3*M + 1_2 (with the 2 X 2 unit matrix 1_2) from the characteristic polynomial of M_2 (see a comment in A322602) and the Cayley-Hamilton theorem. The recurrence M^n = M*M^(n-1) leads to (M_n)^n = S(n, 3)*1_2 + S(n-a, 3)*(M - 3*1_2), for n >= 0, with S(n, 3) = F(2(n+1)) = A001906(n+1). Hence ((M_2)^n)[1,1] = S(n, 3) - 2*S(n-1, 3) = a(n) = F(2*n-1) = (1/(2*r+1))*r^(2*n-1)*(1 + (1/r^2)^(2*n-1)), with r = rho(5) = A001622 (golden ratio) (see the first Aug 31 2004 formula, using the recurrence of S(n, 3), and the Michael Somos Oct 28 2002 formula). This proves a conjecture of Gary W. Adamson in A322602.
The ratio a(n)/a(n-1) converges to r^2 = rho(5)^2 = A104457 for n -> infinity (see the a(n) formula in terms of r), which is one of the statements by Gary W. Adamson in A322602. (End)
a(n) is the number of ways to stack coins with a bottom row of n coins such that any coin not on the bottom row touches exactly two coins in the row below, and all the coins on any row are contiguous [Wilf, 2.12]. - Greg Dresden, Jun 29 2020
a(n) is the upper left entry of the (2*n)-th power of the 4 X 4 Jacobi matrix L with L(i,j)=1 if |i-j| = 1 and L(i,j)=0 otherwise. - Michael Shmoish, Aug 29 2020
All positive solutions of the indefinite binary quadratic F(1, -3, 1) := x^2 - 3*x*y + y^2, of discriminant 5, representing -1 (special Markov triples (1, y=x, z=y) if y <= z) are [x(n), y(n)] = [abs(F(2*n+1)), abs(F(2*n-1))], for n = -infinity..+infinity. (F(-n) = (-1)^(n+1)*F(n)). There is only this single family of proper solutions, and there are no improper solutions. [See also the Floor van Lamoen Nov 29 2001 comment, which uses this negative n, and my Jan 30 2015 comment.] - Wolfdieter Lang, Sep 23 2020
These are the denominators of the lower convergents to the golden ratio, tau; they are also the numerators of the upper convergents (viz. 1/1 < 3/2 < 8/5 < 21/13 < ... < tau < ... 13/8 < 5/3 < 2/1). - Clark Kimberling, Jan 02 2022
a(n+1) is the number of subgraphs of the path graph on n vertices. - Leen Droogendijk, Jun 17 2023
For n > 4, a(n+2) is the number of ways to tile this 3 x n "double-box" shape with squares and dominos (reflections or rotations are counted as distinct tilings). The double-box shape is made up of two horizontal strips of length n, connected by three vertical columns of length 3, and the center column can be located anywhere not touching the two outside columns.
     _   _   _   _  
  |||_|||_|||_|||_|||
  || _  |_|  _   _ ||
  |||_|||_|||_|||_|||. - Greg Dresden and Ruishan Wu, Aug 25 2024
a(n+1) is the number of integer sequences a_1, ..., a_n such that for any number 1 <= k <= n, (a_1 + ... + a_k)^2 = a_1^3 + ... + a_k^3. - Yifan Xie, Dec 07 2024

			a(3) = 13: there are 14 ordered trees with 4 edges; all of them, except for the path with 4 edges, have height at most 3.

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Milan Janjic, Determinants and Recurrence Sequences, Journal of Integer Sequences, 2012, Article 12.3.5. - From _N. J. A. Sloane_, Sep 16 2012
I. Jensen, Series expansions for self-avoiding polygons
O. Khadir, K. Liptai, and L. Szalay, On the Shifted Product of Binary Recurrences, J. Int. Seq. 13 (2010), 10.6.1.
Tanya Khovanova, Recursive Sequences
S. Kitaev, J. Remmel, and M. Tiefenbruck, Marked mesh patterns in 132-avoiding permutations I, arXiv preprint arXiv:1201.6243 [math.CO], 2012. - From _N. J. A. Sloane_, May 09 2012
Sergey Kitaev, Jeffrey Remmel, and Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132-Avoiding Permutations II, Electronic Journal of Combinatorial Number Theory, Volume 15 #A16. (arXiv:1302.2274)
S. Klavzar, M. Mollard, and M. Petkovsek, The degree sequence of Fibonacci and Lucas cubes, Discrete Math., 311, 2011, 1310-1322.
Ron Knott, Pi and the Fibonacci numbers
F. Luca and R. Oyono, An exponential Diophantine equation related to powers of two consecutive Fibonacci numbers, Proc. Japan Acad. Ser. A Math. Sci. 87 (2011), pp. 45-50.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
I. Lukovits and D. Janezic, Enumeration of conjugated circuits in nanotubes, J. Chem. Inf. Comput. Sci., vol. 44 (2004) pp. 410-414.
K. Manes, A. Sapounakis, I. Tasoulas, and P. Tsikouras, Equivalence classes of ballot paths modulo strings of length 2 and 3, arXiv:1510.01952 [math.CO], 2015.
Eric Marberg, Crossings and nestings in colored set partitions, arXiv preprint arXiv:1203.5738 [math.CO], 2012.
Megan A. Martinez and Carla D. Savage, Patterns in Inversion Sequences II: Inversion Sequences Avoiding Triples of Relations, arXiv:1609.08106 [math.CO], 2016.
M. D. McIlroy, The number of states of a dynamic storage system, Computer J., 25 (No. 3, 1982), 388-392. (Annotated scanned copy)
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
W. H. Mills, A system of quadratic Diophantine equations. Pacific J. Math. 3:1 (1953), 209-220, available from Project Euclid.
D. Nacin, The Minimal Non-Koszul A(Gamma), arXiv preprint arXiv:1204.1534 [math.QA], 2012. - From _N. J. A. Sloane_, Oct 05 2012
D. Necas and I. Ohlidal, Consolidated series for efficient calculation of the reflection and transmission in rough multilayers, Optics Express, Vol. 22, 2014, No. 4; DOI:10.1364/OE.22.004499. See Table 1.
László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
J. G. Penaud and O. Roques, Génération de chemins de Dyck à pics croissants, Discrete Mathematics, Vol. 246, no. 1-3 (2002), 255-267.
Permutation Pattern Avoidance Library (PermPAL), Av(123,1432)
T. K. Petersen and Bridget Eileen Tenner, The depth of a permutation, arXiv:1202.4765 [math.CO], 2012-2014.
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Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
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Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).
Index entries for "core" sequences
Index entries for two-way infinite sequences

Fibonacci A000045 = union of this sequence and A001906.
Cf. A001653, A055105, A055106, A055107, A074664, A101368, A124292, A124293, A124294, A124295, A140069, A153463, A153266, A153267, A144257, A211216, A002559, A082582.
a(n)= A060920(n, 0).
Row 3 of array A094954.
Equals A001654(n+1) - A001654(n-1), n > 0.
A122367 is another version. Inverse sequences A130255 and A130256. Row sums of A140068, A152251, A153342, A179806, A179745, A213948.
Cf. A001622, A001906, A104457, A153387, A322602.

a:=[1,1];; for n in [3..10^2] do a[n]:=3*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Sep 27 2017

a001519 n = a001519_list !! n
a001519_list = 1 : zipWith (-) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012
a001519_list = 1 : f a000045_list where f (_:x:xs) = x : f xs
-- Reinhard Zumkeller, Aug 09 2013

[1] cat [(Lucas(2*n) - Fibonacci(2*n))/2: n in [1..50]]; // Vincenzo Librandi, Jul 02 2014

A001519:=-(-1+z)/(1-3*z+z**2); # Simon Plouffe in his 1992 dissertation; gives sequence without an initial 1
A001519 := proc(n) option remember: if n=0 then 1 elif n=1 then 1 elif n>=2 then 3*procname(n-1)-procname(n-2) fi: end: seq(A001519(n), n=0..28); # Johannes W. Meijer, Aug 14 2011

Fibonacci /@ (2Range[29] - 1) (* Robert G. Wilson v, Oct 05 2005 *)
LinearRecurrence[{3, -1}, {1, 1}, 29] (* Robert G. Wilson v, Jun 28 2012 *)
a[ n_] := With[{c = Sqrt[5]/2}, ChebyshevT[2 n - 1, c]/c]; (* Michael Somos, Jul 08 2014 *)
CoefficientList[ Series[(1 - 2x)/(1 - 3x + x^2), {x, 0, 30}], x] (* Robert G. Wilson v, Feb 01 2015 *)

a[0]:1$ a[1]:1$ a[n]:=3*a[n-1]-a[n-2]$ makelist(a[n],n,0,30); /* Martin Ettl, Nov 15 2012 */

{a(n) = fibonacci(2*n - 1)}; /* Michael Somos, Jul 19 2003 */

{a(n) = real( quadgen(5) ^ (2*n))}; /* Michael Somos, Jul 19 2003 */

{a(n) = subst( poltchebi(n) + poltchebi(n - 1), x, 3/2) * 2/5}; /* Michael Somos, Jul 19 2003 */

[lucas_number1(n,3,1)-lucas_number1(n-1,3,1) for n in range(30)] # Zerinvary Lajos, Apr 29 2009

G.f.: (1-2*x)/(1-3*x+x^2).
G.f.: 1 / (1 - x / (1 - x / (1 - x))). - Michael Somos, May 03 2012
a(n) = A001906(n+1) - 2*A001906(n).
a(n) = a(1-n) for all n in Z.
a(n+2) = (a(n+1)^2+1)/a(n) with a(1)=1, a(2)=2. - Benoit Cloitre, Aug 29 2002
a(n) = (phi^(2*n-1) + phi^(1-2*n))/sqrt(5) where phi=(1+sqrt(5))/2. - Michael Somos, Oct 28 2002
a(n) = A007598(n-1) + A007598(n) = A000045(n-1)^2 + A000045(n)^2 = F(n)^2 + F(n+1)^2. - Henry Bottomley, Feb 09 2001
a(n) = Sum_{k=0..n} binomial(n+k, 2*k). - Len Smiley, Dec 09 2001
a(n) ~ (1/5)*sqrt(5)*phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = Sum_{k=0..n} C(n, k)*F(k+1). - Benoit Cloitre, Sep 03 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 1)=a(n) (this comment is essentially the same as that of L. Smiley). - Benoit Cloitre, Nov 10 2002
a(n) = (1/2)*(3*a(n-1) + sqrt(5*a(n-1)^2-4)). - Benoit Cloitre, Apr 12 2003
Main diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). - Benoit Cloitre, Aug 05 2003
Hankel transform of A002212. E.g., Det([1, 1, 3;1, 3, 10;3, 10, 36]) = 5. - Philippe Deléham, Jan 25 2004
Solutions x > 0 to equation floor(x*r*floor(x/r)) = floor(x/r*floor(x*r)) when r=phi. - Benoit Cloitre, Feb 15 2004
a(n) = Sum_{i=0..n} binomial(n+i, n-i). - Jon Perry, Mar 08 2004
a(n) = S(n-1, 3) - S(n-2, 3) = T(2*n-1, sqrt(5)/2)/(sqrt(5)/2) with S(n, x) = U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first kind. See triangle A049310, resp. A053120. - Wolfdieter Lang, Aug 31 2004
a(n) = ((-1)^(n-1))*S(2*(n-1), i), with the imaginary unit i and S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. - Wolfdieter Lang, Aug 31 2004
a(n) = Sum_{0<=i_1<=i_2<=n} binomial(i_2, i_1)*binomial(n, i_1+i_2). - Benoit Cloitre, Oct 14 2004
a(n) = L(n,3), where L is defined as in A108299; see also A002878 for L(n,-3). - Reinhard Zumkeller, Jun 01 2005
a(n) = a(n-1) + Sum_{i=0..n-1} a(i)*a(n) = F(2*n+1)*Sum_{i=0..n-1} a(i) = F(2*n). - Andras Erszegi (erszegi.andras(AT)chello.hu), Jun 28 2005
The i-th term of the sequence is the entry (1, 1) of the i-th power of the 2 X 2 matrix M = ((1, 1), (1, 2)). - Simone Severini, Oct 15 2005
a(n-1) = (1/n)*Sum_{k=0..n} B(2*k)*F(2*n-2*k)*binomial(2*n, 2*k) where B(2*k) is the (2*k)-th Bernoulli number. - Benoit Cloitre, Nov 02 2005
a(n) = A055105(n,1) + A055105(n,2) + A055105(n,3) = A055106(n,1) + A055106(n,2). - Mike Zabrocki, Oct 24 2006
a(n) = (2/sqrt(5))*cosh((2n-1)*psi), where psi=log(phi) and phi=(1+sqrt(5))/2. - Hieronymus Fischer, Apr 24 2007
a(n) = (phi+1)^n - phi*A001906(n) with phi=(1+sqrt(5))/2. - Reinhard Zumkeller, Nov 22 2007
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3); a(n) = ((sqrt(5) + 5)/10)*(3/2 + sqrt(5)/2)^(n-2) + ((-sqrt(5) + 5)/10)*(3/2 - sqrt(5)/2)^(n-2). - Antonio Alberto Olivares, Mar 21 2008
a(n) = A147703(n,0). - Philippe Deléham, Nov 29 2008
Sum_{n>=0} atan(1/a(n)) = (3/4)*Pi. - Jaume Oliver Lafont, Feb 27 2009
With X,Y defined as X = ( F(n) F(n+1) ), Y = ( F(n+2) F(n+3) ), where F(n) is the n-th Fibonacci number (A000045), it follows a(n+2) = X.Y', where Y' is the transpose of Y (n >= 0). - K.V.Iyer, Apr 24 2009
From Gary Detlefs, Nov 22 2010: (Start)
a(n) = Fibonacci(2*n+2) mod Fibonacci(2*n), n > 1.
a(n) = (Fibonacci(n-1)^2 + Fibonacci(n)^2 + Fibonacci(2*n-1))/2. (End)
INVERT transform is A166444. First difference is A001906. Partial sums is A055588. Binomial transform is A093129. Binomial transform of A000045(n-1). - Michael Somos, May 03 2012
a(n) = 2^n*f(n;1/2), where f(n;d), n=0,1,...,d, denote the so-called delta-Fibonacci numbers (see Witula et al. papers and comments in A000045). - Roman Witula, Jul 12 2012
a(n) = (Fibonacci(n+2)^2 + Fibonacci(n-3)^2)/5. - Gary Detlefs, Dec 14 2012
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 - x/(x*k + 1 )/Q(k+1); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 23 2013
G.f.: (1-2*x)*G(0)/(2-3*x), where G(k) = 1 + 1/( 1 - x*(5*k-9)/(x*(5*k-4) - 6/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 19 2013
G.f.: 1 + x*(1-x^2)*Q(0)/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + 2*x - x^2)/( x*(4*k+4 + 2*x - x^2 ) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 11 2013
G.f.: Q(0,u), where u=x/(1-x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
Sum_{n>=2} 1/(a(n) - 1/a(n)) = 1. Compare with A001906, A007805 and A097843. - Peter Bala, Nov 29 2013
Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n). Then for n > 0, a(n) = F(n)*L(n-1) + (-1)^n. - Charlie Marion, Jan 01 2014
a(n) = A238731(n,0). - Philippe Deléham, Mar 05 2014
1 = a(n)*a(n+2) - a(n+1)*a(n+1) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = (L(2*n+4) + L(2*n-6))/25 for L(n)=A000032(n). - J. M. Bergot, Dec 30 2014
a(n) = (L(n-1)^2 + L(n)^2)/5 with L(n)=A000032(n). - J. M. Bergot, Dec 31 2014
a(n) = (L(n-2)^2 + L(n+1)^2)/10 with L(n)=A000032(n). - J. M. Bergot, Oct 23 2015
a(n) = 3*F(n-1)^2 + F(n-3)*F(n) - 2*(-1)^n. - J. M. Bergot, Feb 17 2016
a(n) = (F(n-1)*L(n) + F(n)*L(n-1))/2 = (A081714(n-1) + A128534(n))/2. - J. M. Bergot, Mar 22 2016
E.g.f.: (2*exp(sqrt(5)*x) + 3 + sqrt(5))*exp(-x*(sqrt(5)-3)/2)/(5 + sqrt(5)). - Ilya Gutkovskiy, Jul 04 2016
a(n) = ((M_2)^n)[1,1] = S(n, 3) - 2*S(n-1, 3), with the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from A322602. For a proof see the Mar 30 2020 comment above. - Wolfdieter Lang, Mar 30 2020
Sum_{n>=1} 1/a(n) = A153387. - Amiram Eldar, Oct 05 2020
a(n+1) = Product_{k=1..n} (1 + 4*cos(2*Pi*k/(2*n + 1))^2). Special case of A099390. - Greg Dresden, Oct 16 2021
a(n+1) = 4^(n+1)*Sum_{k >= n} binomial(2*k,2*n)*(1/5)^(k+1). Cf. A102591. - Peter Bala, Nov 29 2021
a(n) = cosh((2*n-1)*arcsinh(1/2))/sqrt(5/4). - Peter Luschny, May 21 2022
From J. M. Bergot, May 27 2022: (Start)
a(n) = F(n-1)*L(n) - (-1)^n where L(n)=A000032(n) and F(n)=A000045(n).
a(n) = (L(n-1)^2 + L(n-1)*L(n+1))/5 + (-1)^n.
a(n) = 2*(area of a triangle with vertices at (L(n-2), L(n-1)), (F(n), F(n-1)), (L(n), L(n+1))) + 5*(-1)^n for n > 2. (End)
a(n) = A059929(n-1)+A059929(n-2), n>1. - R. J. Mathar, Jul 09 2024

Entry revised by N. J. A. Sloane, Aug 24 2006, May 13 2008

A016777 a(n) = 3*n + 1.

Original entry on oeis.org

1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97, 100, 103, 106, 109, 112, 115, 118, 121, 124, 127, 130, 133, 136, 139, 142, 145, 148, 151, 154, 157, 160, 163, 166, 169, 172, 175, 178, 181, 184, 187

Offset: 0

N. J. A. Sloane, Dec 11 1996

Numbers k such that the concatenation of the first k natural numbers is not divisible by 3. E.g., 16 is in the sequence because we have 123456789101111213141516 == 1 (mod 3).
Ignoring the first term, this sequence represents the number of bonds in a hydrocarbon: a(#of carbon atoms) = number of bonds. - Nathan Savir (thoobik(AT)yahoo.com), Jul 03 2003
n such that Sum_{k=0..n} (binomial(n+k,n-k) mod 2) is even (cf. A007306). - Benoit Cloitre, May 09 2004
Hilbert series for twisted cubic curve. - Paul Barry, Aug 11 2006
If Y is a 3-subset of an n-set X then, for n >= 3, a(n-3) is the number of 3-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
a(n) = A144390 (1, 9, 23, 43, 69, ...) - A045944 (0, 5, 16, 33, 56, ...). From successive spectra of hydrogen atom. - Paul Curtz, Oct 05 2008
Number of monomials in the n-th power of polynomial x^3+x^2+x+1. - Artur Jasinski, Oct 06 2008
A145389(a(n)) = 1. - Reinhard Zumkeller, Oct 10 2008
Union of A035504, A165333 and A165336. - Reinhard Zumkeller, Sep 17 2009
Hankel transform of A076025. - Paul Barry, Sep 23 2009
From Jaroslav Krizek, May 28 2010: (Start)
a(n) = numbers k such that the antiharmonic mean of the first k positive integers is an integer.
A169609(a(n-1)) = 1. See A146535 and A169609. Complement of A007494.
See A005408 (odd positive integers) for corresponding values A146535(a(n)). (End)
Apart from the initial term, A180080 is a subsequence; cf. A180076. - Reinhard Zumkeller, Aug 14 2010
Also the maximum number of triangles that n + 2 noncoplanar points can determine in 3D space. - Carmine Suriano, Oct 08 2010
A089911(4*a(n)) = 3. - Reinhard Zumkeller, Jul 05 2013
The number of partitions of 6*n into at most 2 parts. - Colin Barker, Mar 31 2015
For n >= 1, a(n)/2 is the proportion of oxygen for the stoichiometric combustion reaction of hydrocarbon CnH2n+2, e.g., one part propane (C3H8) requires 5 parts oxygen to complete its combustion. - Kival Ngaokrajang, Jul 21 2015
Exponents n > 0 for which 1 + x^2 + x^n is reducible. - Ron Knott, Oct 13 2016
Also the number of independent vertex sets in the n-cocktail party graph. - Eric W. Weisstein, Sep 21 2017
Also the number of (not necessarily maximal) cliques in the n-ladder rung graph. - Eric W. Weisstein, Nov 29 2017
Also the number of maximal and maximum cliques in the n-book graph. - Eric W. Weisstein, Dec 01 2017
For n>=1, a(n) is the size of any snake-polyomino with n cells. - Christian Barrientos and Sarah Minion, Feb 27 2018
The sum of two distinct terms of this sequence is never a square. See Lagarias et al. p. 167. - Michel Marcus, May 20 2018
It seems that, for any n >= 1, there exists no positive integer z such that digit_sum(a(n)*z) = digit_sum(a(n)+z). - Max Lacoma, Sep 18 2019
For n > 2, a(n-2) is the number of distinct values of the magic constant in a normal magic triangle of order n (see formula 5 in Trotter). - Stefano Spezia, Feb 18 2021
Number of 3-permutations of n elements avoiding the patterns 132, 231, 312. See Bonichon and Sun. - Michel Marcus, Aug 20 2022
Erdős & Sárközy conjecture that a set of n positive integers with property P must have some element at least a(n-1) = 3n - 2. Property P states that, for x, y, and z in the set and z < x, y, z does not divide x+y. An example of such a set is {2n-1, 2n, ..., 3n-2}. Bedert proves this for large enough n. (This is an upper bound, and is exact for all known n; I have verified it for n up to 12.) - Charles R Greathouse IV, Feb 06 2023
a(n-1) = 3*n-2 is the dimension of the vector space of all n X n tridiagonal matrices, equals the number of nonzero coefficients: n + 2*(n-1) (see Wikipedia link). - Bernard Schott, Mar 03 2023

			G.f. = 1 + 4*x + 7*x^2 + 10*x^3 + 13*x^4 + 16*x^5 + 19*x^6 + 22*x^7 + ... - _Michael Somos_, May 27 2019

W. Decker, C. Lossen, Computing in Algebraic Geometry, Springer, 2006, p. 22.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology, p. 264.
Konrad Knopp, Theory and Application of Infinite Series, Dover, p. 269.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Hacène Belbachir, Toufik Djellal, and Jean-Gabriel Luque, On the self-convolution of generalized Fibonacci numbers, arXiv:1703.00323 [math.CO], 2017.
Benjamin Bedert, On a problem of Erdős and Sárközy about sequences with no term dividing the sum of two larger terms, arXiv preprint, arXiv:2301.07065 [math.NT], 2023.
Nicolas Bonichon and Pierre-Jean Morel, Baxter d-permutations and other pattern avoiding classes, arXiv:2202.12677 [math.CO], 2022.
Paul Erdős and András Sárközy, On the divisibility properties of sequences of integers, Proc. London Math. Soc. (3), 21 (1970), pp. 97-101.
Leonhard Euler, Observatio de summis divisorum p. 9.
Leonhard Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009, see p. 9.
L. B. W. Jolley, Summation of Series, Dover, 1961, pp. 16, 38.
Tanya Khovanova, Recursive Sequences
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original German edition of "Theory and Application of Infinite Series")
J. C. Lagarias, A. M. Odlyzko, and J. B. Shearer, On the density of sequences of integers the sum of no two of which is a square. I. Arithmetic progressions, Journal of Combinatorial Theory. Series A, 33 (1982), pp. 167-185.
T. Mansour, Permutations avoiding a set of patterns from S_3 and a pattern from S_4, arXiv:math/9909019 [math.CO], 1999.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
Nathan Sun, On d-permutations and Pattern Avoidance Classes, arXiv:2208.08506 [math.CO], 2022.
Terrel Trotter, Normal Magic Triangles of Order n, Journal of Recreational Mathematics Vol. 5, No. 1, 1972, pp. 28-32.
Eric Weisstein's World of Mathematics, Book Graph
Eric Weisstein's World of Mathematics, Clique
Eric Weisstein's World of Mathematics, Cocktail Party Graph
Eric Weisstein's World of Mathematics, Independent Vertex Set
Eric Weisstein's World of Mathematics, Ladder Rung Graph
Eric Weisstein's World of Mathematics, Maximal Clique
Eric Weisstein's World of Mathematics, Maximum Clique
Wikipedia, Tridiagonal matrix.
Chengcheng Yang, A Problem of Erdös Concerning Lattice Cubes, arXiv:2011.15010 [math.CO], 2020. See Table p. 27.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000290, A001477, A005408, A007306, A007494, A016789, A016933.
Cf. A017569, A035504, A045944, A058183, A076025, A089911, A144390.
Cf. A145389, A146535, A165333, A165336, A169609, A180076, A180080.
Cf. A214550, A238731.
Cf. A007559 (partial products), A051536 (lcm).
First differences of A000326.
Row sums of A131033.
Complement of A007494. - Reinhard Zumkeller, Oct 10 2008
Some subsequences: A002476 (primes), A291745 (nonprimes), A135556 (squares), A016779 (cubes).

a016777 = (+ 1) . (* 3)
a016777_list = [1, 4 ..]  -- Reinhard Zumkeller, Feb 28 2013, Feb 10 2012

[3*n+1 : n in [1..70]]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006

Range[1, 199, 3] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
(* Start from Eric W. Weisstein, Sep 21 2017 *)
3 Range[0, 70] + 1
Table[3 n + 1, {n, 0, 70}]
LinearRecurrence[{2, -1}, {1, 4}, 70]
CoefficientList[Series[(1 + 2 x)/(-1 + x)^2, {x, 0, 70}], x]
(* End *)

A016777(n):=3*n+1$
makelist(A016777(n),n,0,30); /* Martin Ettl, Oct 31 2012 */

a(n)=3*n+1 \\ Charles R Greathouse IV, Jul 28 2015

[3*n+1 for n in range(1,71)] # G. C. Greubel, Mar 15 2024

G.f.: (1+2*x)/(1-x)^2.
a(n) = A016789(n) - 1.
a(n) = 3 + a(n-1).
Sum_{n>=1} (-1)^n/a(n) = (1/3)*(Pi/sqrt(3) + log(2)). [Jolley, p. 16, (79)] - Benoit Cloitre, Apr 05 2002
(1 + 4*x + 7*x^2 + 10*x^3 + ...) = (1 + 2*x + 3*x^2 + ...)/(1 - 2*x + 4*x^2 - 8*x^3 + ...). - Gary W. Adamson, Jul 03 2003
E.g.f.: exp(x)*(1+3*x). - Paul Barry, Jul 23 2003
a(n) = 2*a(n-1) - a(n-2); a(0)=1, a(1)=4. - Philippe Deléham, Nov 03 2008
a(n) = 6*n - a(n-1) - 1 (with a(0) = 1). - Vincenzo Librandi, Nov 20 2010
Sum_{n>=0} 1/a(n)^2 = A214550. - R. J. Mathar, Jul 21 2012
a(n) = A238731(n+1,n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-5)^k. - Philippe Deléham, Mar 05 2014
Sum_{i=0..n} (a(i)-i) = A000290(n+1). - Ivan N. Ianakiev, Sep 24 2014
From Wolfdieter Lang, Mar 09 2018: (Start)
a(n) = denominator(Sum_{k=0..n-1} 1/(a(k)*a(k+1))), with the numerator n = A001477(n), where the sum is set to 0 for n = 0. [Jolley, p. 38, (208)]
G.f. for {n/(1 + 3*n)}_{n >= 0} is (1/3)*(1-hypergeom([1, 1], [4/3], -x/(1-x)))/(1-x). (End)
a(n) = -A016789(-1-n) for all n in Z. - Michael Somos, May 27 2019

Better description from T. D. Noe, Aug 15 2002

Partially edited by Joerg Arndt, Mar 11 2010

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647

Offset: 0

N. J. A. Sloane

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths:  1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
From Gary W. Adamson, Jul 25 2016: (Start)
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
  1, 1, 0, 0, 0, 0, ...
  2, 0, 3, 0, 0, 0, ...
  2, 0, 0, 3, 0, 0, ...
  2, 0, 0, 0, 3, 0, ...
  2, 0, 0, 0, 0, 3, ...
  ...
The triangle generated from M is:
   1,
   1,  1,
   3,  1, 3,
  11,  3, 3, 9,
  41, 11, 9, 9, 27,
   ...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). (End)
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

Indranil Ghosh, Table of n, a(n) for n = 0..1745 (terms 0..200 from T. D. Noe)
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Krassimir T. Atanassov and Anthony G. Shannon, On intercalated Fibonacci sequences, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 218-223.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Pell, Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 43, 56.
Chris Caldwell, Primality Proving, Arndt's theorem.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., 2016.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Kiran S. Kedlaya, The 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Kiran S. Kedlaya, Solutions to the 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Tanya Khovanova, Recursive Sequences
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
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.
Eugene McDonnell, Heron's Rule and Integer-Area Triangles, At Play With J, 2010.
Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.
Yong Hao Ng, A partition in three classes of the set of all prime numbers?, Mathematics Stack Exchange.
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
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Bisections are A011943 and A094347.

Cf. A001353, A065918, A071954, A001571, A001834, A003500, A016064, A082840, A026150, A277434.

a001075 n = a001075_list !! n
a001075_list =
   1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
-- Reinhard Zumkeller, Aug 11 2011

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001075 := proc(n)
    orthopoly[T](n,2) ;
end proc:
seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018

Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)

{a(n) = subst(poltchebi(abs(n)), x, 2)};

{a(n) = real((2 + quadgen(12))^abs(n))};

{a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};

a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016

my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009

def a(n):
    Q = QuadraticField(3, 't')
    u = Q.units()[0]
    return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

More terms from James Sellers, Jul 10 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 23, 134, 781, 4552, 26531, 154634, 901273, 5253004, 30616751, 178447502, 1040068261, 6061962064, 35331704123, 205928262674, 1200237871921, 6995498968852, 40772755941191, 237641036678294, 1385073464128573, 8072799748093144, 47051725024430291, 274237550398488602

Offset: 0

Barry E. Williams, May 02 2000

This sequence gives one half of all positive solutions y = y1 = a(n) of the first class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding x=x1 terms see A054490(n). Therefore it also gives one fourth of all positive solutions x = x1 of the first class of the Pell equation x^2 - 2*y^2 = 14, with the y=y1 terms given by A054490. - Wolfdieter Lang, Feb 26 2015

			n = 2: A054490(2)^2 - 2*(2*a(2))^2 =
       65^2 - 2*(2*23)^2 = -7,
      (4*a(2))^2 - 2*A054490(2)^2 =
      (4*23)^2 - 2*65^2 = 14. - _Wolfdieter Lang_, Feb 26 2015
a(2) = (A253811(1) + A101386(1))/2 = (19 + 27)/2 = 23. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A001541, A001653, A054490.

A038725(n) = a(-n).

a[0]:=1: a[1]:=4: 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},{1,4},30] (* Harvey P. Dale, Aug 06 2020 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 + quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) + polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = ((4+sqrt(2))/8)*(3+2*sqrt(2))^(n-1) + ((4-sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
a(n) = A001653(n+1) - A001109(n). - Antonio Alberto Olivares, Mar 29 2008
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos, Sep 28 2008
G.f.: (1 - 2*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos, Sep 28 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*3^k. - Philippe Deléham, Mar 05 2014
a(n) = S(n,6) - 2*S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-1, x) = 0 evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell comment above. - Wolfdieter Lang, Feb 26 2015
a(0) = -(A038761(0) - A038762(0))/2, a(n) = (A253811(n-1) + A101386(n-1))/2, n >= 1. See the Mar 19 2015 comment on A054490. - Wolfdieter Lang, Mar 19 2015
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Apr 30 2020

More terms from James Sellers, May 03 2000

A033889 a(n) = Fibonacci(4*n + 1).

Original entry on oeis.org

1, 5, 34, 233, 1597, 10946, 75025, 514229, 3524578, 24157817, 165580141, 1134903170, 7778742049, 53316291173, 365435296162, 2504730781961, 17167680177565, 117669030460994, 806515533049393, 5527939700884757, 37889062373143906, 259695496911122585, 1779979416004714189

Offset: 0

N. J. A. Sloane

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

Bruno Berselli, Table of n, a(n) for n = 0..500
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Tanya Khovanova, Recursive Sequences.
Kai Wan, Problem H-90, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 3 (2022), p. 282; Solution to Problem H-90, by Ángel Plaza, ibid., Vol. 62, No. 1 (2024), pp. 95-96.
Roman Wituła, Binomials Transformation Formulae of Scaled Lucas Numbers, Demonstratio Mathematica, Vol. XLVI, No. 1 (2013), pp. 15-27.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A014445, A016813, A122070, A167816, A094567.

Cf. A003881, A049684, A081018, A081016, A081068, A172968.

[Fibonacci(4*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Table[Fibonacci[4*n+1], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)

a(n)=fibonacci(4*n+1) \\ Charles R Greathouse IV, Jul 15 2011

Vec((1-2*x)/(1-7*x+x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = 7*a(n-1) - a(n-2) for n >= 2. - Floor van Lamoen, Dec 10 2001
From R. J. Mathar, Jan 17 2008: (Start)
O.g.f.: (1 - 2*x)/(1 - 7*x + x^2).
a(n) = A004187(n+1) - 2*A004187(n). (End); corrected by Klaus Purath, Jul 29 2020
a(n) = A167816(4*n+1). - Reinhard Zumkeller, Nov 13 2009
a(n) = sqrt(1 + 2 * Fibonacci(2*n) * Fibonacci(2*n + 1) + 5 * (Fibonacci(2*n) * Fibonacci(2*n + 1))^2). - Artur Jasinski, Feb 06 2010
a(n) = Sum_{k=0..n} A122070(n,k)*2^k. - Philippe Deléham, Mar 13 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n)*Fibonacci(2*n+2) + 1. - Gary Detlefs, Apr 18 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n+1)^2. - Bruno Berselli, Apr 19 2012
a(n) = Sum_{k = 0..n} A238731(n,k)*4^k. - Philippe Deléham, Mar 05 2014
a(n) = A000045(A016813(n)). - Michel Marcus, Mar 05 2014
2*a(n) = Fibonacci(4*n) + Lucas(4*n). - Bruno Berselli, Oct 13 2017
a(n) = A094567(n-1) + A094567(n), assuming A094567(-1) = 0. - Klaus Purath, Jul 29 2020
Sum_{n>=0} (-1)^n * arctan(3/a(n)) = Pi/4 (A003881) (Wan, 2022). - Amiram Eldar, Mar 01 2024
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A054491 a(n) = 4*a(n-1) - a(n-2), a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 23, 86, 321, 1198, 4471, 16686, 62273, 232406, 867351, 3236998, 12080641, 45085566, 168261623, 627960926, 2343582081, 8746367398, 32641887511, 121821182646, 454642843073, 1696750189646, 6332357915511, 23632681472398

Offset: 0

Barry E. Williams, May 04 2000

Bisection (even part) of Chebyshev sequence with Diophantine property.

The odd part is A077234 with Diophantine companion A077235.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Andrej Dujella and László Szalay, Four squares from three numbers, arXiv:2506.14013 [math.NT], 2025. See p. 2.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001353, A001834, A077234, A077235.

a:=[1,6];; for n in [3..30] do a[n]:=4*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 15 2020

I:=[1,6]; [n le 2 select I[n] else 4*Self(n-1) -Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 15 2020

seq( simplify(ChebyshevU(n,2) +2*ChebyshevU(n-1,2)), n=0..30); # G. C. Greubel, Jan 15 2020

Table[ChebyshevU[n, 2] +2*ChebyshevU[n-1, 2], {n,0,30}] (* G. C. Greubel, Jan 15 2020 *)
LinearRecurrence[{4,-1},{1,6},30] (* Harvey P. Dale, Sep 04 2021 *)

a(n) = if (n==0, 1, if (n==1, 6, 4*a(n-1)-a(n-2))) \\ Michel Marcus, Jun 23 2013

a(n) = polchebyshev(n, 2, 2) + 2*polchebyshev(n-1, 2, 2); \\ Michel Marcus, Oct 13 2021

[chebyshev_U(n,2) +2*chebyshev_U(n-1,2) for n in (0..30)]; # G. C. Greubel, Jan 15 2020

-3*a(n)^2 + A077236(n)^2 = 13.
a(n) = ( 6*((2+sqrt(3))^n-(2-sqrt(3))^n) - ((2+sqrt(3))^(n-1)-(2-sqrt(3))^(n-1)) )/(2*sqrt(3)).
a(n) = 6*S(n-1, 4) - S(n-2, 4) = S(n, 4) + 2*S(n-1, 4), with S(n, x) := U(n, x/2) Chebyshev's polynomials of 2nd kind, A049310. S(-1, x) := 0, S(-2, x) := -1, S(n, 4)= A001353(n+1).
G.f.: (1+2*x)/(1-4*x+x^2).
a(n+1) = A001353(n+2) + 2*A001353(n+1). - Creighton Dement, Nov 28 2004. Comment from Vim Wenders, Mar 26 2008: This is easily verified using a(n) = (6*( (2+sqrt(3))^n - (2-sqrt(3))^n ) - ( (2+sqrt(3))^(n-1) - (2-sqrt(3))^(n-1) ))/(2*sqrt(3)) and A001353(n) = ( (2+sqrt(3))^n - (2-sqrt(3))^n )/(2*sqrt(3)).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-7)^k. - Philippe Deléham, Mar 05 2014
E.g.f.: (1/3)*exp(2*x)*(3*cosh(sqrt(3)*x) + 4*sqrt(3)*sinh(sqrt(3)*x)). - Stefano Spezia, Jan 27 2020

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A054488 Expansion of (1+2x)/(1-6x+x^2).

Original entry on oeis.org

1, 8, 47, 274, 1597, 9308, 54251, 316198, 1842937, 10741424, 62605607, 364892218, 2126747701, 12395593988, 72246816227, 421085303374, 2454265004017, 14304504720728, 83372763320351, 485932075201378, 2832219687887917

Offset: 0

Barry E. Williams, May 04 2000

Bisection (even part) of Chebyshev sequence with Diophantine property.
b(n)^2 - 8*a(n)^2 = 17, with the companion sequence b(n)= A077240(n).
The odd part is A077413(n) with Diophantine companion A077239(n).

			8 = a(1) = sqrt((A077240(1)^2 - 17)/8) = sqrt((23^2 - 17)/8)= sqrt(64) = 8.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A002203, A002315, A038761, A077239, A077240, A077413.

Cf. A077241 (even and odd parts).

a:=[1,8];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 19 2020

I:=[1,8]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 19 2020

R:=PowerSeriesRing(Integers(), 21); Coefficients(R!( (1+2*x)/(1-6*x+x^2))); // Marius A. Burtea, Jan 20 2020

a[0]:=1: a[1]:=8: 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},{1,8},30] (* Harvey P. Dale, Oct 09 2017 *)
Table[(LucasL[2*n+1, 2] + Fibonacci[2*n, 2])/2, {n,0,30}] (* G. C. Greubel, Jan 19 2020 *)

my(x='x+O('x^30)); Vec((1+2*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jan 19 2020

apply( {A054488(n)=[1,8]*([0,-1;1,6]^n)[,1]}, [0..30]) \\ M. F. Hasler, Feb 27 2020

[(lucas_number2(2*n+1,2,-1) + lucas_number1(2*n,2,-1))/2 for n in (0..30)] # G. C. Greubel, Jan 19 2020

a(n) = 6*a(n-1) - a(n-2), a(0)=1, a(1)=8.
a(n) = ((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1) + 2*((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n))/(4*sqrt(2)).
a(n) = S(n, 6) + 2*S(n-1, 6), with S(n, x) Chebyshev's polynomials of the second kind, A049310. S(n, 6) = A001109(n+1).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-9)^k. - Philippe Deléham, Mar 05 2014
a(n) = (Pell(2*n+2) + 2*Pell(2*n))/2 = (Pell-Lucas(2*n+1) + Pell(2*n))/2. - G. C. Greubel, Jan 19 2020
E.g.f.: (1/4)*exp(3*x)*(4*cosh(2*sqrt(2)*x) + 5*sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Jan 27 2020

More terms from James Sellers, May 05 2000

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A054486 Expansion of (1+2x)/(1-3x+x^2).

Original entry on oeis.org

1, 5, 14, 37, 97, 254, 665, 1741, 4558, 11933, 31241, 81790, 214129, 560597, 1467662, 3842389, 10059505, 26336126, 68948873, 180510493, 472582606, 1237237325, 3239129369, 8480150782, 22201322977, 58123818149, 152170131470, 398386576261, 1042989597313

Offset: 0

Barry E. Williams, May 06 2000

Binomial transform of A000285. - R. J. Mathar, Oct 26 2011

			G.f. = 1 + 5*x + 14*x^2 + 37*x^3 + 97*x^4 + 254*x^5 + 665*x^6 + 1741*x^7 + ...

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A000285, A002878, A100545, A104449, A228208.

F:=Fibonacci;; List([0..30], n-> F(2*n+2) +2*F(2*n) ); # G. C. Greubel, Nov 08 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+2*x)/(1-3*x+x^2)) ); // Marius A. Burtea, Nov 05 2019

a:=[1,5]; [n le 2 select a[n] else 3*Self(n-1)-Self(n-2): n in [1..30]]; // Marius A. Burtea, Nov 05 2019

with(combinat); f:=fibonacci; seq(f(2*n+2)+2*f(2*n), n=0..30); # G. C. Greubel, Nov 08 2019

CoefficientList[Series[(2*z+1)/(z^2-3*z+1), {z, 0, 30}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 15 2011 *)
a[ n_]:= 3 Fibonacci[2n] + Fibonacci[2n+1]; (* Michael Somos, Mar 17 2015 *)
LinearRecurrence[{3,-1},{1,5},40] (* Harvey P. Dale, Apr 24 2019 *)

Vec((1+2*x)/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jul 15 2011

{a(n) = 3*fibonacci(2*n) + fibonacci(2*n+1)}; /* Michael Somos, Mar 17 2015 */

f=fibonacci; [f(2*n+2) +2*f(2*n) for n in (0..30)] # G. C. Greubel, Nov 08 2019

a(n) = 3*a(n-1) - a(n-2), a(0)=1, a(1)=5.
a(n) = (5*(((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n) - (((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)))/sqrt(5).
a(n) + 7*A001519(n) = A005248(n). - Creighton Dement, Oct 30 2004
a(n) = Lucas(2*n+1) + Fibonacci(2*n) = A002878(n) + A001906(n) = A025169(n-1) + A001906(n+1).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-6)^k. - Philippe Deléham, Mar 05 2014
0 = -11 + a(n)^2 - 3*a(n)*a(n+1) + a(n+1)^2 for all n in Z. - Michael Somos, Mar 17 2015
a(n) = -2*F(n)^2 + 6*F(n)*F(n+1) + F(n+1)^2 for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = 3*F(2*n) + F(2*n+1) for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = -A100545(-2-n) for all n in Z. - Michael Somos, Mar 17 2015
a(n) = A000285(2*n) = A228208(2*n+1) = A104449(2*n+1) for all n in Z. - Michael Somos, Mar 17 2015
From Klaus Purath, Nov 05 2019: (Start)
a(n) = (a(n-m) + a(n+m))/Lucas(2*m), m <= n.
a(n) = sum of 2*m+1 consecutive terms starting with a(n-m) divided by Lucas(2*m+1), m <= n.
a(n) = alternating sum of 2*m+1 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+1), m <= n.
a(n) + a(n+1) = sum of 2*m+2 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+2), m <= n.
a(n) + a(n+1) = (a(n-m) + a(n+m+1))/Fibonacci(2*m+1), m <= n.
The following formulas are extended to negative indexes:
a(n) = 3*Fibonacci(2*n+1) - Fibonacci(2*n-3).
a(n) = (Fibonacci(2*n+5) - 3* Fibonacci(2*n-1))/2.
a(n) = (4*Lucas(2*n+2) - Lucas(2*n-4))/5.
a(n) = Fibonacci(2*n+5) - 4*Fibonacci(2*n+1).
a(n) = (5*Fibonacci(2*n+5) - Fibonacci(2*n-7))/12. (End)
E.g.f.: exp(-(1/2)*(-3+sqrt(5))*x)*(-7 + sqrt(5) + (7 + sqrt(5))*exp(sqrt(5)*x))/(2*sqrt(5)). - Stefano Spezia, Nov 19 2019
a(n) = 3*n + 1 + Sum_{k=1..n} k*a(n-k). - Yu Xiao, Jun 20 2020

"a(1)=5", not "a(0)=5" from Dan Nielsen (nielsed(AT)uah.edu), Sep 10 2009

A002320 a(n) = 5*a(n-1) - a(n-2).

Original entry on oeis.org

1, 3, 14, 67, 321, 1538, 7369, 35307, 169166, 810523, 3883449, 18606722, 89150161, 427144083, 2046570254, 9805707187, 46981965681, 225104121218, 1078538640409, 5167589080827, 24759406763726, 118629444737803

Offset: 0

Joe Keane (jgk(AT)jgk.org)

Together with A002310 these are the two sequences satisfying the requirement that (a(n)^2 + a(n-1)^2)/(1 - a(n)*a(n-1)) be an integer; in both cases this integer is -5. - Floor van Lamoen, Oct 26 2001

From a posting to Netnews group sci.math by ksbrown(AT)seanet.com (K. S. Brown) on Aug 15 1996.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See p. 44.
Tanya Khovanova, Recursive Sequences
MathPages, N = (x^2 + y^2)/(1+xy) is a Square
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A054477.

a002320 n = a002320_list !! n
a002320_list = 1 : 3 :
   (zipWith (-) (map (* 5) (tail a002320_list)) a002320_list)
-- Reinhard Zumkeller, Oct 16 2011

LinearRecurrence[{5,-1},{1,3},30] (* Harvey P. Dale, Nov 13 2014 *)

Sequences A002310, A002320 and A049685 have this in common: each one satisfies a(n+1) = (a(n)^2+5)/a(n-1) - Graeme McRae, Jan 30 2005
G.f.: (1-2x)/(1-5x+x^2). - Philippe Deléham, Nov 16 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*2^k. - _Philippe Deléham, Mar 05 2014
E.g.f.: exp(5*x/2)*(sqrt(21)*cosh(sqrt(21)*x/2) + sinh(sqrt(21)*x/2))/sqrt(21). - Stefano Spezia, Jul 07 2025
From Peter Bala, Jul 07 2025: (Start)
a(n) = ( (4 + sqrt(21))*(5 - sqrt(21))^(n+1) - (4 - sqrt(21))*(5 + sqrt(21))^(n+1) )/(2^(n+1)*sqrt(21)).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 5/a(2*n)) = 1/15, since 5/(a(2*n) + 5/a(2*n)) = 1/a(2*n-1) + 1/a(2*n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n-1) + 5/a(2*n-1)) = 1/5, since 5/(a(2*n-1) + 5/a(2*n-1)) = 1/a(2*n-2) + 1/a(2*n). (End)

A055271 a(n) = 5*a(n-1) - a(n-2) with a(0)=1, a(1)=7.

Original entry on oeis.org

1, 7, 34, 163, 781, 3742, 17929, 85903, 411586, 1972027, 9448549, 45270718, 216905041, 1039254487, 4979367394, 23857582483, 114308545021, 547685142622, 2624117168089, 12572900697823, 60240386321026, 288629030907307, 1382904768215509, 6625894810170238, 31746569282635681, 152106951603008167

Offset: 0

Barry E. Williams, May 10 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A030221.

I:=[1,7]; [n le 2 select I[n] else 5*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Mar 16 2020

A055271:= n-> simplify(ChebyshevU(n, 5/2) + 2*ChebyshevU(n-1, 5/2)); seq(A055271(n), n=0..30); # G. C. Greubel, Mar 16 2020

LinearRecurrence[{5,-1}, {1,7}, 30] (* G. C. Greubel, Mar 16 2020 *)

[chebyshev_U(n, 5/2) + 2*chebyshev_U(n-1, 5/2) for n in (0..30)] # G. C. Greubel, Mar 16 2020

a(n) = (7*(((5+sqrt(21))/2)^n - ((5-sqrt(21))/2)^n) - (((5+sqrt(21))/2)^(n-1) - ((5-sqrt(21))/2)^(n-1)))/sqrt(21).
G.f.: (1+2*x)/(1-5*x+x^2).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-8)^k. - Philippe Deléham, Mar 05 2014
a(n) = ChebyshevT(n, 5/2) + (9/2)*ChebyshevU(n-1,5/2) = ChebyshevU(n, 5/2) + 2*ChebyshevU(n-1, 5/2). - G. C. Greubel, Mar 16 2020

Terms a(22) onward added by G. C. Greubel, Mar 16 2020

cp's OEIS Frontend

A001519 a(n) = 3*a(n-1) - a(n-2) for n >= 2, with a(0) = a(1) = 1.

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A016777 a(n) = 3*n + 1.

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A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

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A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

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A054488 Expansion of (1+2x)/(1-6x+x^2).

A054486 Expansion of (1+2x)/(1-3x+x^2).