A049685 -id:A049685 - cp's OEIS frontend

A004187 a(n) = 7*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

0, 1, 7, 48, 329, 2255, 15456, 105937, 726103, 4976784, 34111385, 233802911, 1602508992, 10983760033, 75283811239, 516002918640, 3536736619241, 24241153416047, 166151337293088, 1138818207635569, 7805576116155895, 53500214605455696, 366695926122033977

Offset: 0

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

Define the sequence T(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0 . A004187 (with initial 0 omitted) is T(1,7).
This is a divisibility sequence.
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 7's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) and b(n) := A056854(n) are the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 5*(3*a(n))^2 = +4. see the cross-reference to A056854 below. - Wolfdieter Lang, Jun 26 2013
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4,5,6}. - Milan Janjic, Jan 25 2015
The digital root is A253298, which shares its digital root with A253368. - Peter M. Chema, Jul 04 2016
Lim_{n->oo} a(n+1)/a(n) = 2 + 3*phi = 1+ A090550 = 6.854101...  - Wolfdieter Lang, Nov 16 2023

			a(2) = 7*a(1) - a(0) = 7*7 - 1 = 48. - _Michael B. Porter_, Jul 04 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
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.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3 , example 12
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993
Zvonko Cerin, Some alternating sums of Lucas numbers, Centr. Eur. J. Math. vol 3 no 1 (2005) 1-13.
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=7, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=9.
Index entries for sequences related to Chebyshev polynomials.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).
Index entries for Pisot sequences

Cf. A000027, A001906, A001353, A004254, A001109, A049685, A033888. a(n)=sqrt((A056854(n)^2 - 4)/45).

Second column of array A028412.

[Fibonacci(4*n)/3 : n in [0..30]]; // Vincenzo Librandi, Jun 07 2011

/* By definition: */ [n le 2 select n-1 else 7*Self(n-1)-Self(n-2): n in [1..23]]; // Bruno Berselli, Dec 24 2012

seq(combinat:-fibonacci(4*n)/3, n = 0 .. 30); # Robert Israel, Jan 26 2015

LinearRecurrence[{7,-1},{0,1},30] (* Harvey P. Dale, Jul 13 2011 *)
CoefficientList[Series[x/(1 - 7*x + x^2), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 23 2012 *)

a[0]:0$ a[1]:1$ a[n]:=7*a[n-1] - a[n-2]$ A004187(n):=a[n]$ makelist(A004187(n),n,0,30); /* Martin Ettl, Nov 11 2012 */

numlib::fibonacci(4*n)/3 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(4*n)/3 \\ Charles R Greathouse IV, Mar 09 2012

concat(0, Vec(x/(1-7*x+x^2) + O(x^99))) \\ Altug Alkan, Jul 03 2016

[lucas_number1(n,7,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(4*n)/3 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1-7*x+x^2).
a(n) = F(4*n)/3 = A033888(n)/3, where F=A000045 (the Fibonacci sequence).
a(n) = S(2*n-1, sqrt(9))/sqrt(9) = S(n-1, 7); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = Sum_{i = 0..n-1} C(2*n-1-i, i)*5^(n-i-1). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
[A049685(n-1), a(n)] = [1,5; 1,6]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167816(4*n). - Reinhard Zumkeller, Nov 13 2009
a(n) = (((7+sqrt(45))/2)^n-((7-sqrt(45))/2)^n)/sqrt(45). - Noureddine Chair, Aug 31 2011
a(n+1) = Sum_{k = 0..n} A101950(n,k)*6^k. - Philippe Deléham, Feb 10 2012
a(n) = (A081072(n)/3)-1. - Martin Ettl, Nov 11 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = (1/5)*(5 + 3*sqrt(5)).
Product {n >= 2} (1 - 1/a(n)) = (1/14)*(5 + 3*sqrt(5)). (End)
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -A(x)*A(-x), where A(x) = Sum_{n >= 1} Fibonacci(2*n)* x^n.
1 + 5*Sum_{n >= 1} a(n)*x^(2*n) = F(x)*F(-x) = G(x)*G(-x), where F(x) = 1 + A(x) and G(x) = 1 + 5*A(x).
1 + Sum_{n >= 1} a(n)*x^(2*n) = H(x)*H(-x) = I(x)*I(-x), where H(x) = 1 + Sum_{n >= 1} Fibonacci(2*n + 3)*x^n and I(x) = 1 + x + x*Sum_{n >= 1} Fibonacci(2*n - 1)*x^n. (End)
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/(3*sqrt(5)). - Ilya Gutkovskiy, Jul 03 2016
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*9^k*binomial(n+k, 2*k+1). - Peter Bala, Jul 17 2023
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*7^(n-2*k)*binomial(n-k, k). - Greg Dresden, Aug 03 2024
From Peter Bala, Jul 22 2025: (Start)
The following products telescope:
Product {n >= 2} (1 + (-1)^n/a(n)) = (3/14)*(3 + sqrt(5)).
Product {n >= 1} (1 - (-1)^n/a(n)) = (1/3)*(3 + sqrt(5)).
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = (3/5)*sqrt(5). (End)

Entry improved by comments from Michael Somos and Wolfdieter Lang, Aug 02 2000

A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

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

Offset: 0

Reinhard Zumkeller, Jun 01 2005

Matrix inverse of A124645.
Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:
L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));
Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);
Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);
abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);
T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;
T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;
T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;
T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;
T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;
T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;
T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;
T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;
T(n,n) = L(n,0) = (-1)^floor((n+1)/2);
L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);
L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;
L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;
L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;
L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;
L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;
L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;
L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;
L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;
L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;
L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;
L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;
L(n,13) = A085260(n);
L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;
L(n,n) = A108366(n); L(n,-n) = A108367(n).
Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005
From L. Edson Jeffery, Mar 12 2011: (Start)
Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form
  G_N=A_{N,1}=
  (0 1 0 ... 0)
  (1 0 1 0 ... 0)
  (0 1 0 1 0 ... 0)
  ...
  (0 ... 0 1 0 1)
  (0 ... 0 1 1),
with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,
  G_7=A_{7,1}=
  (0 1 0)
  (1 0 1)
  (0 1 1).
We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)
The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011
The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

			Triangle begins:
  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;
  ...

Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
Jay Kappraff, S. Jablan, G. Adamson, & R. Sazdonovich: "Golden Fields, Generalized Fibonacci Sequences, & Chaotic Matrices"; FORMA, Vol 19, No 4, (2005).

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Henry W. Gould, A Variant of Pascal's Triangle, Corrections, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271.
L. Edson Jeffery, Unit-primitive matrices.
Ju, Hyeong-Kwan On the sequence generated by a certain type of matrices. Honam Math. J. 39, No. 4, 665-675 (2017).
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894 [math.NT], 2015.
Frank Ruskey and Carla Savage, Gray codes for set partitions and restricted growth tails, Australasian Journal of Combinatorics, Volume 10(1994), pp. 85-96. See Table 1 p. 95.

Cf. A049310, A039961, A124645 (matrix inverse).
Triangle sums (see the comments): A193884 (Kn11), A154955 (Kn21), A087960 (Kn22), A000007 (Kn3), A010892 (Fi1), A134668 (Fi2), A078031 (Ca2), A193669 (Gi1), A001519 (Gi3), A193885 (Ze1), A050935 (Ze3). - Johannes W. Meijer, Aug 08 2011
Cf. A003558.
Cf. A033999, A059841.

a108299 n k = a108299_tabl !! n !! k
a108299_row n = a108299_tabl !! n
a108299_tabl = [1] : iterate (\row ->
   zipWith (+) (zipWith (*) ([0] ++ row) a033999_list)
               (zipWith (*) (row ++ [0]) a059841_list)) [1,-1]
-- Reinhard Zumkeller, May 06 2012

A108299 := proc(n,k): binomial(n-floor((k+1)/2), floor(k/2))*(-1)^floor((k+1)/2) end: seq(seq(A108299 (n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011

t[n_, k_?EvenQ] := I^k*Binomial[n-k/2, k/2]; t[n_, k_?OddQ] := -I^(k-1)*Binomial[n+(1-k)/2-1, (k-1)/2]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 16 2013 *)

{T(n,k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)),n,x)+y*O(y^k),k,y)} (Hanna)

T(n,k) = binomial(n-floor((k+1)/2),floor(k/2))*(-1)^floor((k+1)/2).
T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0 < k < n, T(n, 0) = 1, T(n, n) = (-1)^floor((n+1)/2).
G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - Paul D. Hanna, Jun 12 2005
The generating polynomial (in z) of row n >= 0 is (u^(2*n+1) + v^(2*n+1))/(u + v), where u and v are defined by u^2 + v^2 = 1 and u*v = z. - Emeric Deutsch, Jun 16 2011
From Johannes W. Meijer, Aug 08 2011: (Start)
abs(T(n,k)) = A065941(n,k) = abs(A187660(n,n-k));
T(n,n-k) = A130777(n,k); abs(T(n,n-k)) = A046854(n,k) = abs(A066170(n,k)). (End)

Corrected and edited by Philippe Deléham, Oct 20 2008

A033890 a(n) = Fibonacci(4*n + 2).

Original entry on oeis.org

1, 8, 55, 377, 2584, 17711, 121393, 832040, 5702887, 39088169, 267914296, 1836311903, 12586269025, 86267571272, 591286729879, 4052739537881, 27777890035288, 190392490709135, 1304969544928657, 8944394323791464, 61305790721611591, 420196140727489673

Offset: 0

N. J. A. Sloane

(x,y) = (a(n), a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033888. - Floor van Lamoen, Dec 10 2001
This sequence consists of the odd-indexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The even-indexed terms of A001906 are in A033888. Limit_{n->infinity} a(n)/a(n-1) = phi^4 = (7 + 3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Indices of square numbers which are also 12-gonal. - Sture Sjöstedt, Jun 01 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with 3's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
If we let b(0) = 0 and, for n >= 1, b(n) = A033890(n-1), then the sequence b(n) will be F(4n-2) and the first difference is L(4n) or A056854. F(4n-2) is also the ratio of golden spiral length (rounded to the nearest integer) after n rotations. L(4n) is also the pitch length ratio. See illustration in links. - Kival Ngaokrajang, Nov 03 2013
The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 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 = -5, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
Solutions y of Pell equation x^2 - 5*y^2 = 4; corresponding x values are in A342710 (see A342709). - Bernard Schott, Mar 19 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Nathan D. Cahill and Darren A. Narayan, Fibonacci and Lucas Numbers as Tridiagonal Matrix Determinants, Fib. Quart. 42, no. 3, Aug. 2004, pp. 216-221. See p. 219.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Kival Ngaokrajang, Illustration of golden spiral length and pitch ratio
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 linear recurrences with constant coefficients, signature (7,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001906, A100047.

Cf. A342709, A342710.

[Fibonacci(4*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

A033890 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,8]);
    else
        7*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

Table[Fibonacci[4n + 2], {n, 0, 14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{7, -1}, {1, 8}, 50] (* G. C. Greubel, Jul 13 2017 *)
a[n_] := (GoldenRatio^(2 (1 + 2 n)) - GoldenRatio^(-2 (1 + 2 n)))/Sqrt[5]
Table[a[n] // FullSimplify, {n, 0, 21}] (* Gerry Martens, Aug 20 2025 *)

PARI
```
a(n)=fibonacci(4*n+2);
```

G.f.: (1+x)/(1-7*x+x^2).
a(n) = 7*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=8.
a(n) = S(n,7) + S(n-1,7) = S(2*n,sqrt(9) = 3), where S(n,x) = U(n,x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n,7) = A004187(n+1), S(n,3) = A001906(n+1).
a(n) = ((7+3*sqrt(5))^n - (7-3*sqrt(5))^n + 2*((7+3*sqrt(5))^(n-1) - ((7-3*sqrt(5))^(n-1)))) / (3*(2^n)*sqrt(5)). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -9). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-7)*(-1)^n, where L is defined as in A108299; see also A049685 for L(n,+7). - Reinhard Zumkeller, Jun 01 2005
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),7/2) + f(a(n-2),7/2). - Marcos Carreira, Dec 27 2006
a(n+1) = 8*a(n) - 8*a(n-1) + a(n-2); a(1)=1, a(2)=8, a(3)=55. - Sture Sjöstedt, May 27 2009
a(n) = A167816(4*n+2). - Reinhard Zumkeller, Nov 13 2009
a(n)=b such that (-1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2-sin(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n) = A000045(A016825(n)). - Michel Marcus, Mar 22 2015
a(n) = A001906(2*n+1). - R. J. Mathar, Apr 30 2017
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + 3*sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Apr 14 2025
From Peter Bala, Jun 08 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/9 [telescoping series: 3/(a(n) - 1/a(n)) = 1/Fibonacci(4*n+4) + 1/Fibonacci(4*n)].
Product_{n >= 1} (a(n) + 3)/(a(n) - 3) = 5/2 [telescoping product:
(a(n) + 3)/(a(n) - 3) = b(n)/b(n-1), where b(n) = (Lucas(4*n+4) - 3)/(Lucas(4*n+4) + 3)].
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(9/5) [telescoping product:
(a(n) + 1)/(a(n) - 1) = c(n)/c(n-1) for n >= 1, where c(n) = Fibonacci(2*n+2)/Lucas(2*n+2)]. (End)
From Gerry Martens, Aug 20 2025: (Start)
a(n) = ((3 + sqrt(5))^(1 + 2*n) - (3 - sqrt(5))^(1 + 2*n)) / (2^(1 + 2*n)*sqrt(5)).
a(n) = Sum_{k=0..2*n} binomial(2*n + k + 1, 2*k + 1). (End)

A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 6, 16, 56, 176, 576, 1856, 6016, 19456, 62976, 203776, 659456, 2134016, 6905856, 22347776, 72318976, 234029056, 757334016, 2450784256, 7930904576, 25664946176, 83053510656, 268766806016, 869747654656, 2814562533376, 9108115685376, 29474481504256

Offset: 0

Paul Barry, May 10 2003

Inverse binomial transform of A001077. Binomial transform of expansion of cosh(sqrt(5)*x) (1,0,5,0,25,...).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - Cino Hilliard, Sep 25 2005
Numerators of fractions in the approximation of the square root of 5 satisfying: a(n) = (a(n-1)+c)/(a(n-1)+1), with c=5 and a(1)=1. For denominators see A063727. - Mark Dols, Jul 24 2009
Equals right border of triangle A143969. (1, 6, 16, 56, ...) = row sums of triangle A143969 and INVERT transform of (1, 5, 5, 5, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 5 types of other natural numbers. - Milan Janjic, Aug 13 2010
From Gary W. Adamson, Jul 30 2016: (Start)
The sequence is case N=1 in an infinite set obtained by taking powers of the 2 X 2 matrix M = [(1,5); (1,N)], then extracting the upper left terms. The infinite set begins:
N=1 (A084057):  1, 6, 16,  56,  176,   576,   1856, ...
N=2 (A108306):  1, 6, 21,  81,  306,  1161,   4401, ...
N=3 (A164549):  1, 6, 26, 116,  516,  2296,  10216, ...
N=4 (A015449):  1, 6, 31, 161,  836,  4341,  22541, ...
N=5 (A000400):  1, 6, 36, 216, 1296,  7776,  46656, ...
N=6 (A049685):  1, 6, 41, 281, 1926, 13201,  90481, ...
N=7 (.......):  1, 6, 46, 356, 2756, 21336, 222712, ...
...
Sequences in the above set can be obtained by taking INVERT transforms of the following:
N=1 INVERT transform of (1, 5,  5,  5,   5,    5, ...
N=2 ..."......"......". (1, 5, 10, 20,  40,   80, ...
N=3 ..."......"......". (1, 5, 15, 45, 135,  405, ...
N=4 ..."......"......". (1, 5, 20, 80, 320, 1280, ...
...
with the pattern (1, 5, N*5, (N^2)*5, (N^3)*5, ...
It appears that the sequence generated from powers (n>0) of the matrix P = [(1,a); (1,b)], (a,b > 0), then extracting the upper left terms, is equal to the INVERT transform of the sequence starting: (1, a, b*a, (b^2)*a, (b^3)*a, ...). (End)

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A046717, A002533, A098158, A143969, A269992.
a(n) = A087131(n)/2.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A108306, A164549, A015449, A000400, A049685

I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 31 2016

f[n_] := Simplify[((1 + Sqrt[5])^n + (1 - Sqrt[5])^n)/2]; Array[f, 28, 0] (* Or *)
LinearRecurrence[{2, 4}, {1, 1}, 28] (* Robert G. Wilson v, Sep 18 2013 *)
RecurrenceTable[{a[1] == 1, a[2] == 1, a[n] == 2 a[n-1] + 4 a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Jul 31 2016 *)
Table[2^(n-1) LucasL[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 19 2016 *)

lucas(n)=fibonacci(n-1)+fibonacci(n+1)
a(n)=lucas(n)/2*2^n \\ Charles R Greathouse IV, Sep 18 2013

from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,1,2,4, lambda n: 0); [next(it) for i in range(1,26)] # Zerinvary Lajos, Jul 09 2008

[lucas_number2(n,2,-4)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009

a(n) = ((1+sqrt(5))^n + (1-sqrt(5))^n)/2.
G.f.: (1-x) / (1-2*x-4*x^2).
E.g.f.: exp(x) * cosh(sqrt(5)*x).
a(2n+1) = 2*a(n)*a(n+1) - (-4)^n. - Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*5^k . - Paul Barry, Jul 25 2004
a(n) = Sum_{k=0..n} A098158(n,k)*5^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = 2^(n-1)*A000032(n). - Mark Dols, Jul 24 2009
If p(1)=1, and p(i)=5 for i>1, and if A is the Hessenberg matrix of order n defined by: A(i,j) = p(j-i+1) for 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
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = A063727(n) - A063272(n-1). - R. J. Mathar, Jun 06 2019
a(n) = 1 + 5*A014335(n). - R. J. Mathar, Jun 06 2019
Sum_{n>=1} 1/a(n) = A269992. - Amiram Eldar, Feb 01 2021

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681

Offset: 1

Ralf Stephan, May 31 2004

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.

Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...

Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.
Elizabeth Wilmer, A note on Stephan's conjecture 87
Elizabeth Wilmer, A note on Stephan's conjecture 87 [cached copy]

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)

PARI
```
T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)
```

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449

Offset: 0

Bruno Berselli, Feb 25 2014

First bisection of A041611.

Bruno Berselli, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (36,-1).

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

[n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020

CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]

a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016

m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
print(f.coefficients())

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

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)

A214992 Power ceiling-floor sequence of (golden ratio)^4.

Original entry on oeis.org

7, 47, 323, 2213, 15169, 103969, 712615, 4884335, 33477731, 229459781, 1572740737, 10779725377, 73885336903, 506417632943, 3471038093699, 23790849022949, 163064905066945, 1117663486445665, 7660579500052711

Offset: 0

Clark Kimberling, Nov 08 2012, Jan 24 2013

Let f = floor and c = ceiling.  For x > 1, define four sequences as functions of x, as follows:
p1(0) = f(x), p1(n) = f(x*p1(n-1));
p2(0) = f(x), p2(n) = c(x*p2(n-1)) if n is odd and p2(n) = f(x*p1(n-1)) if n is even;
p3(0) = c(x), p3(n) = f(x*p3(n-1)) if n is odd and p3(n) = c(x*p3(n-1)) if n is even;
p4(0) = c(x), p4(n) = c(x*p4(n-1)).
The present sequence is given by a(n) = p3(n).
Following the terminology at A214986, call the four sequences power floor, power floor-ceiling, power ceiling-floor, and power ceiling sequences.  In the table below, a sequence is identified with an A-numbered sequence if they appear to agree except possibly for initial terms.  Notation: S(t)=sqrt(t), r = (1+S(5))/2 = golden ratio, and Limit = limit of p3(n)/p2(n).
x ......p1..... p2..... p3..... p4.......Limit
r.......A000045 A000045 A000045 A000045..r
r^2.....A001519 A001654 A061646 A001906..-1+S(5)
r^3.....A024551 A001076 A015448 A049652..-1+S(5)
r^4.....A049685 A157335 A214992 A004187..-19+9*S(5)
r^5.....A214993 A049666 A015457 A214994...(-9+5*S(5))/2
r^6.....A007805 A156085 A214995 A049660..-151+68*S(5)
1+S(2)..A024537 A000129 A001333 A048739..S(2)
2+S(2)..A007052 A214996 A214997 A007070..(1+S(2))/2
1+S(3)..A057960 A002605 A028859 A077846..(1+S(3))/2
2+S(3)..A001835 A109437 A214998 A001353..-4+3*S(3)
S(5)....A214999 A215091 A218982 A218983..1.26879683...
2+S(5)..A024551 A001076 A015448 A049652..-1+S(5)
2+S(6)..A218984 A090017 A123347 A218985..S(3/2)
2+S(7)..A218986 A015530 A126473 A218987..(1+S(7))/3
2+S(8)..A218988 A057087 A086347 A218989..(1+S(2))/2
3+S(8)..A001653 A084158 A218990 A001109..-13+10*S(2)
3+S(10).A218991 A005668 A015451 A218992..-2+S(10)
...
Properties of p1, p2, p3, p4:
(1) If x > 2, the terms of p2 and p3 interlace: p2(0) < p3(0) < p2(1) < p3(1) < p2(2) < p3(2)... Also, p1(n) <= p2(n) <= p3(n) <= p4(n) <= p1(n+1) for all x>0 and n>=0.
(2) If x > 2, the limits L(x) = limit(p/x^n) exist for the four functions p(x), and L1(x) <= L2(x) <= L3(x) <= L4 (x). See the Mathematica programs for plots of the four functions; one of them also occurs in the Odlyzko and Wilf article, along with a discussion of the special case x = 3/2.
(3) Suppose that x = u + sqrt(v) where v is a nonsquare positive integer.  If u = f(x) or u = c(x), then p1, p2, p3, p4 are linear recurrence sequences.  Is this true for sequences p1, p2, p3, p4 obtained from x = (u + sqrt(v))^q for every positive integer q?
(4) Suppose that x is a Pisot-Vijayaraghavan number. Must p1, p2, p3, p4 then be linearly recurrent?  If x is also a quadratic irrational b + c*sqrt(d), must the four limits L(x) be in the field Q(sqrt(d))?
(5) The Odlyzko and Wilf article (page 239) raises three interesting questions about the power ceiling function; it appears that they remain open.

			a(0) = ceiling(r) = 7, where r = ((1+sqrt(5))/2)^4 = 6.8...; a(1) = floor(7*r) = 47; a(2) = ceiling(47) = 323.

Clark Kimberling, Table of n, a(n) for n = 0..250
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
Index entries for linear recurrences with constant coefficients, signature (6,6,-1).
Index entries for sequences related to the Josephus Problem.

Cf. A001622, A214986, A049685, A157335, A004187.

(* Program 1.  A214992 and related sequences *)
x = GoldenRatio^4; z = 30; (* z = # terms in sequences *)
z1 = 100; (* z1 = # digits in approximations *)
f[x_] := Floor[x]; c[x_] := Ceiling[x];
p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
p1[n_] := f[x*p1[n - 1]]
p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
p4[n_] := c[x*p4[n - 1]]
Table[p1[n], {n, 0, z}]  (* A049685 *)
Table[p2[n], {n, 0, z}]  (* A157335 *)
Table[p3[n], {n, 0, z}]  (* A214992 *)
Table[p4[n], {n, 0, z}]  (* A004187 *)
Table[p4[n] - p1[n], {n, 0, z}]  (* A004187 *)
Table[p3[n] - p2[n], {n, 0, z}]  (* A098305 *)
(* Program 2.  Plot of power floor and power ceiling functions, p1(x) and p4(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p1[x_, 0] := f[x]; p1[x_, n_] := f[x*p1[x, n - 1]];
p4[x_, 0] := c[x]; p4[x_, n_] := c[x*p4[x, n - 1]];
Plot[Evaluate[{p1[x, 10]/x^10, p4[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]
(* Program 3. Plot of power floor-ceiling and power ceiling-floor functions, p2(x) and p3(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p2[x_, 0] := f[x]; p3[x_, 0] := c[x];
p2[x_, n_] := If[Mod[n, 2] == 1, c[x*p2[x, n - 1]], f[x*p2[x, n - 1]]]
p3[x_, n_] := If[Mod[n, 2] == 1, f[x*p3[x, n - 1]], c[x*p3[x, n - 1]]]
Plot[Evaluate[{p2[x, 10]/x^10, p3[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]

a(n) = floor(r*a(n-1)) if n is odd and a(n) = ceiling(r*a(n-1)) if n is even, where a(0) = ceiling(r), r = (golden ratio)^4 = (7 + sqrt(5))/2.
a(n) = 6*a(n-1) + 6*a(n-2) - a(n-3).
G.f.: (7 + 5*x - x^2)/((1 + x)*(1 - 7*x + x^2)).
a(n) = (10*(-2)^n+(10+3*sqrt(5))*(7-3*sqrt(5))^(n+2)+(10-3*sqrt(5))*(7+3*sqrt(5))^(n+2))/(90*2^n). - Bruno Berselli, Nov 14 2012
a(n) = 7*A157335(n) + 5*A157335(n-1) - A157335(n-2). - R. J. Mathar, Feb 05 2020
E.g.f.: exp(-x)*(5 + 2*exp(9*x/2)*(155*cosh(3*sqrt(5)*x/2) + 69*sqrt(5)*sinh(3*sqrt(5)*x/2)))/45. - Stefano Spezia, Oct 28 2024

A033891 a(n) = Fibonacci(4*n+3).

Original entry on oeis.org

2, 13, 89, 610, 4181, 28657, 196418, 1346269, 9227465, 63245986, 433494437, 2971215073, 20365011074, 139583862445, 956722026041, 6557470319842, 44945570212853, 308061521170129, 2111485077978050

Offset: 0

N. J. A. Sloane

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000045, A004187, A049685, A101858, A167816.

List([0..30], n-> Fibonacci(4*n+3)); # G. C. Greubel, Jul 14 2019

[Fibonacci(4*n+3): n in [0..30]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[4*n+3],{n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{7,-1},{2,13},31] (* or *) CoefficientList[Series[ (2-x)/(1-7x+x^2),{x,0,30}],x] (* Harvey P. Dale, May 03 2011 *)

a(n)=fibonacci(4*n+3) \\ Charles R Greathouse IV, Sep 24 2015

[fibonacci(4*n+3) for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 7*a(n-1) - a(n-2). - Floor van Lamoen, Dec 10 2001
G.f.: (2-x)/(1-7*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = A167816(4*n+3). - Reinhard Zumkeller, Nov 13 2009
a(n) = Fibonacci(2*n+2)^2 + Fibonacci(2*n+1)^2. - Gary Detlefs, Oct 12 2011
a(n) = 2*A004187(n+1) - A004187(n). - R. J. Mathar, Nov 26 2011
a(n) = A004187(n+1) + A049685(n). - Yuriy Sibirmovsky, Sep 15 2016
From  Peter Bala, Aug 11 2022: (Start)
Let n ** m =  n*m + floor(phi*n)*floor(phi*m), where phi = (1 + sqrt(5))/2, denote the Porta-Stolarsky star product of the integers n and m (see A101858). Then a(n) = 2 ** 2 ** ... ** 2 (n+1 factors).
a(2*n+1) = a(n) ** a(n) = Fibonacci(8*n+7); a(3*n+2) = a(n) ** a(n) ** a(n) = Fibonacci(12*n+11) and so on. (End)

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A355980 Indices of primes in A049685.

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A004187 a(n) = 7*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

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A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

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A033890 a(n) = Fibonacci(4*n + 2).

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A084057 a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=1.

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A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

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A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

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