A006190 -id:A006190 - cp's OEIS frontend

A053028 Odd primes p with 4 zeros in any period of the Fibonacci numbers mod p.

5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149, 157, 173, 193, 197, 233, 257, 269, 277, 293, 313, 317, 337, 353, 373, 389, 397, 421, 433, 457, 557, 577, 593, 613, 617, 653, 661, 673, 677, 701, 733, 757, 761, 773, 797, 821, 829, 853, 857, 877, 937, 953

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Also, primes that do not divide any Lucas number. - T. D. Noe, Jul 25 2003
Although every prime divides some Fibonacci number, this is not true for the Lucas numbers. In fact, exactly 1/3 of all primes do not divide any Lucas number. See Lagarias and Moree for more details. The Lucas numbers separate the primes into three disjoint sets: (A053028) primes that do not divide any Lucas number, (A053027) primes that divide Lucas numbers of even index and (A053032) primes that divide Lucas numbers of odd index. - T. D. Noe, Jul 25 2003; revised by N. J. A. Sloane, Feb 21 2004
From Jianing Song, Jun 16 2024: (Start)
Primes p such that A001176(p) = 4.
For p > 2, p is in this sequence if and only if A001175(p) == 4 (mod 8), and if and only if A001177(p) is odd. For a proof of the equivalence between A001176(p) = 4 and A001177(p) being odd, see Section 2 of my link below.
This sequence contains all primes congruent to 13, 17 (mod 20). This corresponds to case (1) for k = 3 in the Conclusion of Section 1 of my link below. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

T. D. Noe, Table of n, a(n) for n=1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B2).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
J. C. Lagarias, The set of primes dividing the Lucas numbers has density 2/3, Pacific J. Math., 118. No. 2, (1985), 449-461.
J. C. Lagarias, Errata to: The set of primes dividing the Lucas numbers has density 2/3, Pacific J. Math., 162, No. 2, (1994), 393-396.
Diego Marques and Pavel Trojovsky, The order of appearance of the product of five consecutive Lucas numbers, Tatra Mountains Math. Publ. 59 (2014), 65-77.
Pieter Moree, Counting Divisors of Lucas Numbers, Pacific J. Math, Vol. 186, No. 2, 1998, pp. 267-284.
M. Renault, Fibonacci sequence modulo m
H. Sedaghat, Zero-Avoiding Solutions of the Fibonacci Recurrence Modulo A Prime, Fibonacci Quart. 52 (2014), no. 1, 39-45. See p. 45.
Jianing Song, Lucas sequences and entry point modulo p
Eric Weisstein's World of Mathematics, Lucas Number

Cf. A001175, A001177.
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+---------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | this seq | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

Lucas[n_] := Fibonacci[n+1] + Fibonacci[n-1]; badP={}; Do[p=Prime[n]; k=1; While[k0, k++ ]; If[k==p, AppendTo[badP, p]], {n, 200}]; badP

A prime p = prime(i) is in this sequence if p > 2 and A001602(i) is odd. - T. D. Noe, Jul 25 2003

Edited: Name clarified. Moree and Renault link updated. Ballot and Elia reference linked. - Wolfdieter Lang, Jan 20 2015

A098316 Decimal expansion of [3, 3, ...] = (3 + sqrt(13))/2.

Original entry on oeis.org

3, 3, 0, 2, 7, 7, 5, 6, 3, 7, 7, 3, 1, 9, 9, 4, 6, 4, 6, 5, 5, 9, 6, 1, 0, 6, 3, 3, 7, 3, 5, 2, 4, 7, 9, 7, 3, 1, 2, 5, 6, 4, 8, 2, 8, 6, 9, 2, 2, 6, 2, 3, 1, 0, 6, 3, 5, 5, 2, 2, 6, 5, 2, 8, 1, 1, 3, 5, 8, 3, 4, 7, 4, 1, 4, 6, 5, 0, 5, 2, 2, 2, 6, 0, 2, 3, 0, 9, 5, 4, 1, 0, 0, 9, 2, 4, 5, 3, 5, 8, 8, 3

Offset: 1

Eric W. Weisstein, Sep 02 2004

For reasons following from the formula section, this constant could be called "the bronze ratio". For this, compare with A001622 and A014176.
If c is this constant and n > 0, then for n even, c^n = [A100230(n), 1, A100230(n)-1, 1, A100230(n)-1, 1, A100230(n)-1, 1, ...], for n odd, c^n = [A100230(n)+1, A100230(n)+1, A100230(n)+1, ...]. - Gerald McGarvey, Dec 15 2007
This is the shape of a 3-extension rectangle; see A188640 for definitions. - Clark Kimberling, Apr 10 2011
From Vladimir Shevelev, Mar 02 2013: (Start)
An analog of Fermat theorem: for prime p, round(c^p) == 3 (mod p).
A generalization for "metallic" constants c_N = (N+sqrt(N^2+4))/2, N>=1: for prime p, round((c_N)^p) == N (mod p). (End)
This is the positive real algebraic number c of degree 2 with minimal polynomial x^3 - x - 1. The other negative root is 3 - c. - Wolfdieter Lang, Aug 29 2022
c^n = c*A006190(n) + A006190(n-1). - Gary W. Adamson, Apr 02 2024

			3.30277563...

G. C. Greubel, Table of n, a(n) for n = 1..10000
Wikipedia, Metallic mean
Index entries for algebraic numbers, degree 2

Cf. A001622, A014176, A098317, A098318, A000032, A006497, A080039, A049310.

RealDigits[(3 + Sqrt[13])/2, 10, 100][[1]] (* G. C. Greubel, Apr 16 2017 *)

(3 + sqrt(13))/2 \\ Charles R Greathouse IV, Jul 24 2013

3 plus the constant in A085550. - R. J. Mathar, Sep 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
Set c:=(3+sqrt(13))/2. Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).
c:=(3+sqrt(13))/2 satisfies c-c^(-1)=floor(c)=3, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.
1/c=(sqrt(13)-3)/2.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A014176 (the silver ratio: where floor(x)=2). (End)
c=3+sum{k>=1}(-1)^(k-1)/(A006190(k)*A006190(k+1)). - Vladimir Shevelev, Feb 23 2013
A generalization for "metallic" constants c_N = (N+sqrt(N^2+4))/2, N>=1. Let {A_N(n), n>=0} be the sequence 0, 1, N, N^2+1, N^3+2*N, N^4+3*N^2+1,..., a(N) = N*a(N-1) + a(N-2). Then c_N = N + sum_{n>=1} (-1)^(n-1)/(A_N(n)*A_N(n+1)) (cf. A001622, A014176, A098316, A098317, A098318). - Vladimir Shevelev, Feb 23 2013
Equals lim_{n->oo} S(n, sqrt(13))/S(n-1, sqrt(13)), with the S-Chebyshev polynomial (see A049310). - Wolfdieter Lang, Nov 15 2023

A168561 Riordan array (1/(1-x^2), x/(1-x^2)). Unsigned version of A049310.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 1, 0, 3, 0, 1, 0, 3, 0, 4, 0, 1, 1, 0, 6, 0, 5, 0, 1, 0, 4, 0, 10, 0, 6, 0, 1, 1, 0, 10, 0, 15, 0, 7, 0, 1, 0, 5, 0, 20, 0, 21, 0, 8, 0, 1, 1, 0, 15, 0, 35, 0, 28, 0, 9, 0, 1, 0, 6, 0, 35, 0, 56, 0, 36, 0, 10, 0, 1, 1, 0, 21, 0, 70, 0, 84, 0, 45, 0, 11, 0, 1

Offset: 0

Philippe Deléham, Nov 29 2009

Row sums: A000045(n+1), Fibonacci numbers.
A168561*A007318 = A037027, as lower triangular matrices. Diagonal sums : A077957. - Philippe Deléham, Dec 02 2009
T(n,k) is the number of compositions of n+1 into k+1 odd parts. Example: T(4,2)=3 because we have 5 = 1+1+3 = 1+3+1 = 3+1+1.
Coefficients of monic Fibonacci polynomials (rising powers of x). Ftilde(n, x) = x*Ftilde(n-1, x) + Ftilde(n-2, x), n >=0, Ftilde(-1,x) = 0, Ftilde(0, x) = 1. G.f.: 1/(1 - x*z - z^2). Compare with Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jul 29 2014

			The triangle T(n,k) begins:
n\k 0  1   2   3   4    5    6    7    8    9  10  11  12  13 14 15 ...
0:  1
1:  0  1
2:  1  0   1
3:  0  2   0   1
4:  1  0   3   0   1
5:  0  3   0   4   0    1
6:  1  0   6   0   5    0    1
7:  0  4   0  10   0    6    0    1
8:  1  0  10   0  15    0    7    0    1
9:  0  5   0  20   0   21    0    8    0    1
10: 1  0  15   0  35    0   28    0    9    0   1
11: 0  6   0  35   0   56    0   36    0   10   0   1
12: 1  0  21   0  70    0   84    0   45    0  11   0   1
13: 0  7   0  56   0  126    0  120    0   55   0  12   0   1
14: 1  0  28   0 126    0  210    0  165    0  66   0  13   0  1
15: 0  8   0  84   0  252    0  330    0  220   0  78   0  14  0  1
... reformatted by _Wolfdieter Lang_, Jul 29 2014.
------------------------------------------------------------------------

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - From _N. J. A. Sloane_, May 10 2012
Tom Copeland, Addendum to Elliptic Lie Triad
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A162515 (rows reversed), A112552, A102426 (deflated).

A168561:=proc(n,k) if n-k mod 2 = 0 then binomial((n+k)/2,k) else 0 fi end proc:
seq(seq(A168561(n,k),k=0..n),n=0..12) ; # yields sequence in triangular form

Table[If[EvenQ[n + k], Binomial[(n + k)/2, k], 0], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 16 2017 *)

T(n,k) = if ((n+k) % 2, 0, binomial((n+k)/2,k));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print();); \\ Michel Marcus, Oct 09 2016

Sum_{k=0..n} T(n,k)*x^k = A059841(n), A000045(n+1), A000129(n+1), A006190(n+1), A001076(n+1), A052918(n), A005668(n+1), A054413(n), A041025(n), A099371(n+1), A041041(n), A049666(n+1), A041061(n), A140455(n+1), A041085(n), A154597(n+1), A041113(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 respectively. - Philippe Deléham, Dec 02 2009
T(2n,2k) = A085478(n,k). T(2n+1,2k+1) = A078812(n,k). Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000045(n+1), A006131(n), A015445(n), A168579(n), A122999(n) for x = 0,1,2,3,4,5 respectively. - Philippe Deléham, Dec 02 2009
T(n,k) = binomial((n+k)/2,k) if (n+k) is even; otherwise T(n,k)=0.
G.f.: (1-z^2)/(1-t*z-z^2) if offset is 1.
T(n,k) = T(n-1,k-1) + T(n-2,k), T(0,0) = 1, T(0,1) = 0. - Philippe Deléham, Feb 09 2012
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 09 2012
From R. J. Mathar, Feb 04 2022: (Start)
Sum_{k=0..n} T(n,k)*k = A001629(n+1).
Sum_{k=0..n} T(n,k)*k^2 = 0,1,4,11,... = 2*A055243(n)-A099920(n+1).
Sum_{k=0..n} T(n,k)*k^3 = 0,1,8,29,88,236,... = 12*A055243(n) -6*A001629(n+2) +A001629(n+1)-6*(A001872(n)-2*A001872(n-1)). (End)

Typo in name corrected (1(1-x^2) changed to 1/(1-x^2)) by Wolfdieter Lang, Nov 20 2010

A057088 Scaled Chebyshev U-polynomials evaluated at i*sqrt(5)/2. Generalized Fibonacci sequence.

Original entry on oeis.org

1, 5, 30, 175, 1025, 6000, 35125, 205625, 1203750, 7046875, 41253125, 241500000, 1413765625, 8276328125, 48450468750, 283633984375, 1660422265625, 9720281250000, 56903517578125, 333118994140625, 1950112558593750, 11416157763671875, 66831351611328125, 391237546875000000

Offset: 0

Wolfdieter Lang, Aug 11 2000

a(n) gives the length of the word obtained after n steps with the substitution rule 0->11111, 1->111110, starting from 0. The number of 1's and 0's of this word is 5*a(n-1) and 5*a(n-2), resp.
a(n) / a(n-1) converges to (5 + (3 * sqrt(5))) / 2 as n approaches infinity. (5 + (3 * sqrt(5))) / 2 can also be written as phi^2 + (2 * phi), phi^3 + phi, phi + sqrt(5) + 2, (3 * phi) + 1, (3 * phi^2) - 2, phi^4 - 1 and (5 + (3 * (L(n) / F(n)))) / 2, where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity. - Ross La Haye, Aug 18 2003, on another version
Pisano period lengths: 1, 3, 3, 6, 1, 3, 24, 12, 9, 3, 10, 6, 56, 24, 3, 24,288, 9, 18, 6, ... - R. J. Mathar, Aug 10 2012

Indranil Ghosh, Table of n, a(n) for n = 0..1300
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=5, q=5.
Tanya Khovanova, Recursive Sequences
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs.(39) and (45),rhs, m=5.
Eric Weisstein's World of Mathematics, Horadam Sequence
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,5)

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015537, A015443, A015447, A030195, A053404, A057087, A083858, A085939, A090017, A091914, A099012, A180222, A180226.

I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1) + 5*Self(n-2): n in [0..30]]; // G. C. Greubel, Jan 16 2018

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=5*a[n-1]+5*a[n-2]od: seq(a[n], n=1..33); # Zerinvary Lajos, Dec 14 2008

LinearRecurrence[{5,5}, {1,5}, 30] (* G. C. Greubel, Jan 16 2018 *)

x='x+O('x^30); Vec(1/(1 - 5*x - 5*x^2)) \\ G. C. Greubel, Jan 16 2018

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

a(n) = 5*(a(n-1) + a(n-2)), a(-1)=0, a(0)=1.
a(n) = S(n, i*sqrt(5))*(-i*sqrt(5))^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.
G.f.: 1/(1 - 5*x - 5*x^2).
a(n) = (1/3)*Sum_{k=0..n} binomial(n, k)*Fibonacci(k)*3^k. - Benoit Cloitre, Oct 25 2003
a(n) = ((5 + 3*sqrt(5))/2)^n(1/2 + sqrt(5)/6) + (1/2 - sqrt(5)/6)((5 - 3*sqrt(5))/2)^n. - Paul Barry, Sep 22 2004
(a(n)) appears to be given by the floretion - 0.75'i - 0.5'j + 'k - 0.75i' + 0.5j' + 0.5k' + 1.75'ii' - 1.25'jj' + 1.75'kk' - 'ij' - 0.5'ji' - 0.75'jk' - 0.75'kj' - 1.25e ("jes"). - Creighton Dement, Nov 28 2004
a(n) = Sum_{k=0..n} 4^k*A063967(n,k). - Philippe Deléham, Nov 03 2006
G.f.: G(0)/(2-5*x), where G(k)= 1 + 1/(1 - x*(9*k-5)/(x*(9*k+4) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
From Ehren Metcalfe, Nov 18 2017: (Start)
With F(n) = A000045(n), L(n) = A000032(n), beta = (1-sqrt(5))/2:
a(2*n-1) = 5^n*F(4*n)/3 = (5^(n-1/2)*L(4*n) - 2*5^(n-1/2)*beta^(4*n))/3.
a(2*n) = 5^n*L(4*n+2)/3 = (5^(n+1/2)*F(4*n+2) + 2*5^n*beta^(4*n+2))/3.
a(n) = round 5^((n+1)/2)*F(2*(n+1))/3.
a(n) = round 5^(n/2)*L(2*(n+1))/3. (End)

A041025 Denominators of continued fraction convergents to sqrt(17).

Original entry on oeis.org

1, 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, 151693352, 1232221121, 10009462320, 81307919681, 660472819768, 5365090477825, 43581196642368, 354014663616769, 2875698505576520, 23359602708228929, 189752520171407952, 1541379764079492545

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A041024(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = +1, a(2*n) with b(2*n) := A041024(2*n), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = -1 (cf. Emerson reference).
Bisection: a(2*n) = T(2*n+1,sqrt(17))/sqrt(17) = A078988(n), n >= 0 and a(2*n+1) = 8*S(n-1,66), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003
Sqrt(17) = 8/2 + 8/65 + 8/(65*4289) + 8/(4289*283009) + ... . - Gary W. Adamson, Dec 26 2007
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 8's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
De Moivre's formula: a(n) = (r^n - s^n)/(r-s), for r > s, gives sequences with integers if r and s are conjugates. With r=4+sqrt(17) and s=4-sqrt(17), a(n+1)/a(n) converges to r=4+sqrt(17). - Sture Sjöstedt, Nov 11 2011
a(n) equals the number of words of length n on alphabet {0,1,...,8} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 8-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 8 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle, Chaos, Solitons & Fractals 2007; 33(1): 38-49.
S. Falcón and Á. Plaza, On k-Fibonacci sequences and polynomials and their derivatives, Chaos, Solitons & Fractals (2007).
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
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,1).

Cf. A041024, A040012.
Cf. A000045, A000129, A006190, A001076, A052918, A005668, A054413, A243399.
Row n=8 of A073133, A172236 and A352361.
Cf. A099369 (squares).

I:=[1, 8]; [n le 2 select I[n] else 8*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013

CoefficientList[Series[1/(-z^2 - 8 z + 1), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 23 2011 *)
Denominator[Convergents[Sqrt[17],30]] (* Harvey P. Dale, Aug 15 2011 *)
LinearRecurrence[{8,1}, {1,8}, 50] (* Sture Sjöstedt, Nov 11 2011 *)

Vec(1/(1-8*x-x^2)+O(x^99)) \\ Charles R Greathouse IV, Dec 09 2014

[lucas_number1(n,8,-1) for n in range(1, 20)] # Zerinvary Lajos, Apr 25 2009

G.f.: 1/(1 - 8*x - x^2).
a(n) = ((-i)^n)*S(n, 8*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind and i^2 = -1. See A049310.
a(n) = F(n, 8), the n-th Fibonacci polynomial evaluated at x=8. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = ((4 + sqrt(17))^n - (4 - sqrt(17))^n)/(2*sqrt(17));
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*8^(n-1-2i). (End)
Let T be the 2 X 2 matrix [0, 1; 1, 8]. Then T^n * [1, 0] = [a(n-2), a(n-1)]. - Gary W. Adamson, Dec 26 2007
a(n) = 8*a(n-1) + a(n-2), n > 1; a(0)=1, a(1)=8. - Philippe Deléham, Nov 20 2008
a(p-1) == 68^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009 [Corrected by Jason Yuen, Apr 05 2025. See A087475 for more info about this congruence.]
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = sqrt(17) - 4. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 8*x - x^2) = Sum_{n >= 0} x^n *( Product_{k = 1..n} (m*k + 8 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A053032 Odd primes p with one zero in Fibonacci numbers mod p.

Original entry on oeis.org

11, 19, 29, 31, 59, 71, 79, 101, 131, 139, 151, 179, 181, 191, 199, 211, 229, 239, 251, 271, 311, 331, 349, 359, 379, 419, 431, 439, 461, 479, 491, 499, 509, 521, 541, 571, 599, 619, 631, 659, 691, 709, 719, 739, 751, 809, 811, 839, 859, 911, 919, 941, 971

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Also, odd primes that divide Lucas numbers of odd index. - T. D. Noe, Jul 25 2003
From Charles R Greathouse IV, Dec 14 2016: (Start)
It seems that this sequence contains about 1/3 of the primes. In particular, members of this sequence constitute:
        35 of the first 10^2 primes
       330 of the first 10^3 primes
      3328 of the first 10^4 primes
     33371 of the first 10^5 primes
    333329 of the first 10^6 primes
   3333720 of the first 10^7 primes
  33333463 of the first 10^8 primes
etc. (End)
Of the Fibonacci-like sequences modulo a prime p that are not A000004, one of them has a period length less than A001175(p) if and only if p = 5 or p is in this sequence. - Isaac Saffold, Dec 18 2018
Odd primes in A053031. - Jianing Song, Jun 19 2019

			From _Michael B. Porter_, Jan 25 2019: (Start)
The Fibonacci numbers (mod 7) repeat the pattern 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1. Since there are two zeros, 7 is not in the sequence.
The Fibonacci numbers (mod 11) repeat the pattern 0, 1, 1, 2, 3, 5, 8, 2, 10, 1 which has only one zero, so 11 is in the sequence.
(End)

T. D. Noe, Table of n, a(n) for n = 1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B1).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m

Cf. A001175, A001177. See A112860 for another version.
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |    m=1     |   m=2   |   m=3
-----------------------------+------------+---------+---------
The sequence {x(n)}          | A000045    | A000129 | A006190
The sequence {w(k)}          | A001176    | A214027 | A322906
Primes p such that w(p) = 1  | A112860*   | A309580 | A309586
Primes p such that w(p) = 2  | A053027**  | A309581 | A309587
Primes p such that w(p) = 4  | A053028*** | A261580 | A309588
Numbers k such that w(k) = 1 | A053031    | A309583 | A309591
Numbers k such that w(k) = 2 | A053030    | A309584 | A309592
Numbers k such that w(k) = 4 | A053029    | A309585 | A309593
* and also this sequence U {2}
** also primes dividing Lucas numbers of even index
*** also primes dividing no Lucas number

Prime@ Rest@ Position[Table[Count[Drop[NestWhile[Append[#, Mod[Total@ Take[#, -2], n]] &, {1, 1}, If[Length@ # < 3, True, Take[#, -2] != {1, 1}] &], -2], 0], {n, Prime@ Range@ 168}], 1][[All, 1]] (* Michael De Vlieger, Aug 08 2018 *)

fibmod(n,m)=(Mod([1, 1; 1, 0], m)^n)[1, 2]
is(n)=my(k=n+[0, -1, 1, 1, -1][n%5+1]); k>>=valuation(k,2)-1; fibmod(k,n)==0 && fibmod(k/2,n) && isprime(n) \\ Charles R Greathouse IV, Dec 14 2016

A prime p = prime(i) is in this sequence if p > 2 and A001602(i)/2 is odd. - T. D. Noe, Jul 25 2003

A053027 Odd primes p with 2 zeros in Fibonacci numbers mod p.

Original entry on oeis.org

3, 7, 23, 41, 43, 47, 67, 83, 103, 107, 127, 163, 167, 223, 227, 241, 263, 281, 283, 307, 347, 367, 383, 401, 409, 443, 449, 463, 467, 487, 503, 523, 547, 563, 569, 587, 601, 607, 641, 643, 647, 683, 727, 743, 769, 787, 823, 827, 863, 881, 883, 887, 907, 929

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Also, odd primes that divide Lucas numbers of even index. - T. D. Noe, Jul 25 2003
Primes in A053030. - Jianing Song, Jun 19 2019
From Jianing Song, Jun 16 2024: (Start)
Primes p such that A001176(p) = 2.
For p > 2, p is in this sequence if and only if 8 divides of A001175(p), and if and only if 4 divides A001177(p). For a proof of the equivalence between A001176(p) = 2 and 4 dividing A001177(p), see Section 2 of my link below.
This sequence contains all primes congruent to 3, 7 (mod 20). This corresponds to case (2) for k = 3 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

T. D. Noe, Table of n, a(n) for n=1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B3).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m
Jianing Song, Lucas sequences and entry point modulo p

Cf. A001175, A001177.
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |    m=1    |   m=2   |   m=3
-----------------------------+-----------+---------+---------
The sequence {x(n)}          | A000045   | A000129 | A006190
The sequence {w(k)}          | A001176   | A214027 | A322906
Primes p such that w(p) = 1  | A112860*  | A309580 | A309586
Primes p such that w(p) = 2  | this seq  | A309581 | A309587
Primes p such that w(p) = 4  | A053028** | A261580 | A309588
Numbers k such that w(k) = 1 | A053031   | A309583 | A309591
Numbers k such that w(k) = 2 | A053030   | A309584 | A309592
Numbers k such that w(k) = 4 | A053029   | A309585 | A309593
* and also A053032 (primes dividing Lucas numbers of odd index) U {2}
** also primes dividing no Lucas number

A prime p = prime(i) is in this sequence if p > 2 and A001602(i)/2 is even. - T. D. Noe, Jul 25 2003

A003688 a(n) = 3*a(n-1) + a(n-2), with a(1)=1 and a(2)=4.

Original entry on oeis.org

1, 4, 13, 43, 142, 469, 1549, 5116, 16897, 55807, 184318, 608761, 2010601, 6640564, 21932293, 72437443, 239244622, 790171309, 2609758549, 8619446956, 28468099417, 94023745207, 310539335038, 1025641750321, 3387464586001, 11188035508324, 36951571110973

Offset: 1

Frans J. Faase

Number of 2-factors in K_3 X P_n.
Form the graph with matrix [1,1,1,1;1,1,1,0;1,1,0,1;1,0,1,1]. The sequence 1,1,4,13,... with g.f. (1-2*x)/(1-3*x-x^2) counts closed walks of length n at the vertex of degree 5. - Paul Barry, Oct 02 2004
a(n) is term (1,1) in M^n, where M is the 3x3 matrix [1,1,2; 1,1,1; 1,1,1]. - Gary W. Adamson, Mar 12 2009
Starting with 1, INVERT transform of A003945: (1, 3, 6, 12, 24, ...). - Gary W. Adamson, Aug 05 2010
Row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....3
.2..|..1.....3.....9
.3..|..1.....6.....9.....27
.4..|..1.....6....27.....27...81
.5..|..1.....9....27....108...81...243
.6..|..1.....9....54....108..405...243...729
.7..|..1....12....54....270..405..1458...729..2187
which is the triangle for numbers 3^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
Pisano period lengths: 1, 3, 1, 6, 12, 3, 16, 12, 6, 12, 8, 6, 52, 48, 12, 24, 16, 6, 40, 12, ... - R. J. Mathar, Aug 10 2012
a(n-1) is the number of length-n strings of 4 letters {0,1,2,3} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012

			G.f. = x + 4*x^2 + 13*x^3 + 43*x^4 + 142*x^5 + 469*x^6 + 1549*x^7 + 5116*x^8 + ...

F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Joerg Arndt, Matters Computational (The Fxtbook)
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
C. Bautista-Ramos and C. Guillen-Galvan, Fibonacci numbers of generalized Zykov sums, J. Integer Seq., 15 (2012), Article 12.7.8.
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Preliminary version of paper that appeared in Ars Combin. 49 (1998), 129-154.
F. Faase, Counting Hamiltonian cycles in product graphs
F. Faase, Results from the counting program
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 15.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 419
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,1).

Partial sums of A052906. Pairwise sums of A006190.
Cf. A003945, A006190, A049310, A055830, A136159, A290948.
Cf. A374439.

[n le 2 select 4^(n-1) else 3*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 19 2011

with(combinat): a:=n->fibonacci(n,3)-2*fibonacci(n-1,3): seq(a(n), n=2..25); # Zerinvary Lajos, Apr 04 2008

a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[1, 1]]; Table[ a[n], {n, 0, 23}] (* Robert G. Wilson v, Jan 13 2005 *)
LinearRecurrence[{3,1},{1,4},30] (* Harvey P. Dale, Mar 15 2015 *)

a(n)=([0,1; 1,3]^(n-1)*[1;4])[1,1] \\ Charles R Greathouse IV, Aug 14 2017

@CachedFunction
def a(n): # a = A003688
    if (n<3): return 4^(n-1)
    else: return 3*a(n-1) + a(n-2)
[a(n) for n in range(1,41)] # G. C. Greubel, Dec 26 2023

a(n) = ((13 - sqrt(13))/26)*((3 + sqrt(13))/2)^n + ((13 + sqrt(13))/26)*((3 - sqrt(13))/2)^n. - Paul Barry, Oct 02 2004
a(n) = Sum_{k=0..n} 2^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
Starting (1, 1, 4, 13, 43, 142, 469, ...), row sums (unsigned) of triangle A136159. - Gary W. Adamson, Dec 16 2007
G.f.: x*(1+x)/(1-3*x-x^2). - Philippe Deléham, Nov 03 2008
a(n) = A006190(n) + A006190(n-1). - Sergio Falcon, Nov 26 2009
For n>=2, a(n) = F_n(3) + F_(n+1)(3), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i) * x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
G.f.: G(0)*(1+x)/(2-3*x), where G(k) = 1 + 1/(1 - (x*(13*k-9))/( x*(13*k+4) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
a(n)^2 is the denominator of continued fraction [3,3,...,3, 5, 3,3,...,3], which has n-1 3's before, and n-1 3's after, the middle 5. - Greg Dresden, Sep 18 2019
a(n) = Sum_{k=0..n} A046854(n-1,k)*3^k. - R. J. Mathar, Feb 10 2024
a(n) = 3^n*Sum_{k=0..n} A374439(n, k)*(-1/3)^k. - Peter Luschny, Jul 26 2024

Formula added by Olivier Gérard, Aug 15 1997

Name clarified by Michel Marcus, Oct 16 2016

A015523 a(n) = 3a(n-1) + 5a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 14, 57, 241, 1008, 4229, 17727, 74326, 311613, 1306469, 5477472, 22964761, 96281643, 403668734, 1692414417, 7095586921, 29748832848, 124724433149, 522917463687, 2192374556806, 9191710988853, 38537005750589

Offset: 0

Olivier Gérard

From Johannes W. Meijer, Aug 01 2010: (Start)
a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner square (m = 1, 3, 7 and 9) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
For n >= 1, the sequence above corresponds to 24 red king vectors, i.e., A[5] vectors, with decimal values 27, 30, 51, 54, 57, 60, 90, 114, 120, 147, 150, 153, 156, 177, 180, 210, 216, 240, 282, 306, 312, 402, 408 and 432. These vectors lead for the side squares to A152187 and for the central square to A179606.
This sequence belongs to a family of sequences with g.f. 1/(1-3*x-k*x^2). Red king sequences that are members of this family are A007482 (k=2), A015521 (k=4), A015523 (k=5; this sequence), A083858 (k=6), A015524 (k=7) and A015525 (k=8). We observe that there is no red king sequence for k=3. Other members of this family are A049072 (k=-4), A057083 (k=-3), A000225 (k=-2), A001906 (k=-1), A000244 (k=0), A006190 (k=1), A030195 (k=3), A099012 (k=9), A015528 (k=10) and A015529 (k=11).
Inverse binomial transform of A052918 (with extra leading 0).
(End)
First differences in A197189. - Bruno Berselli, Oct 11 2011
Pisano period lengths:  1, 3, 4, 6, 4, 12, 3, 12, 12, 12, 120, 12, 12, 3, 4, 24, 288, 12, 72, 12, ... - R. J. Mathar, Aug 10 2012
This is the Lucas U(P=3, Q=-5) sequence, and hence for n >= 0, a(n+2)/a(n+1) equals the continued fraction 3 + 5/(3 + 5/(3 + 5/(3 + ... + 5/3))) with n 5's. - Greg Dresden, Oct 06 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. A072263, A072264, A152187, A179606, A197189.

[ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+5*Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 23 2011

Join[{a = 0, b = 1}, Table[c = 3 * b + 5 * a; a = b; b = c, {n, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
a[0] := 0; a[1] := 1; a[n_] := a[n] = 3a[n - 1] + 5a[n - 2]; Table[a[n], {n, 0, 49}] (* Alonso del Arte, Jan 16 2011 *)

x='x+O('x^30); concat([0], Vec(x/(1-3*x-5*x^2))) \\ G. C. Greubel, Jan 01 2018

[lucas_number1(n,3,-5) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009

a(n) = 3*a(n-1) + 5*a(n-2).
From Paul Barry, Jul 20 2004: (Start)
a(n) = ((3/2 + sqrt(29)/2)^n - (3/2 - sqrt(29)/2)^n)/sqrt(29).
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1,k)*5^k*3^(n-2*k-1). (End)
G.f.: x/(1 - 3*x - 5*x^2). - R. J. Mathar, Nov 16 2007
From Johannes W. Meijer, Aug 01 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A072263(n) + a(n)*sqrt(29))/2.
Limit_{n->oo} A072263(n)/a(n) = sqrt(29). (End)
G.f.: G(0)*x/(2-3*x), where G(k) = 1 + 1/(1 - x*(29*k-9)/(x*(29*k+20) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
E.g.f.: 2*exp(3*x/2)*sinh(sqrt(29)*x/2)/sqrt(29). - Stefano Spezia, Oct 06 2019

A041041 Denominators of continued fraction convergents to sqrt(26).

Original entry on oeis.org

1, 10, 101, 1020, 10301, 104030, 1050601, 10610040, 107151001, 1082120050, 10928351501, 110365635060, 1114584702101, 11256212656070, 113676711262801, 1148023325284080, 11593909964103601, 117087122966320090, 1182465139627304501, 11941738519239365100

Offset: 0

N. J. A. Sloane

Generalized Fibonacci sequence.
Sqrt(26) = 10/2 + 10/101 + 10/(101*10301) + 10/(10301*1050601) + ... - Gary W. Adamson, Jun 13 2008
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 10's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0, 1, ..., 10} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Bruno Berselli, May 03 2018: (Start)
Numbers k for which m*k^2 + (-1)^k is a perfect square:
m =  2: 0, 1, 2, 5, 12, 29, 70, 169, ...   (A000129);
m =  3: 0, 4, 56, 780, 10864, 151316, ...  (4*A007655);
m =  5: 0, 1, 4, 17, 72, 305, 1292, ...    (A001076);
m =  6: 0, 2, 20, 198, 1960, 19402, ...    (A001078);
m =  7: 0, 48, 12192, 3096720, ...         (2*A175672);
m =  8: 0, 6, 204, 6930, 235416, ...       (A082405);
m = 10: 0, 1, 6, 37, 228, 1405, 8658, ...  (A005668);
m = 11: 0, 60, 23880, 9504180, ...         [°];
m = 12: 0, 2, 28, 390, 5432, 75658, ...    (A011944);
m = 13: 0, 5, 180, 6485, 233640, ...       (5*A041613);
m = 14: 0, 4, 120, 3596, 107760, ...       (A068204);
m = 15: 0, 8, 496, 30744, 1905632, ...     [°];
m = 17: 0, 1, 8, 65, 528, 4289, 34840, ... (A041025);
m = 18: 0, 4, 136, 4620, 156944, ...       (A202299);
m = 19: 0, 13260, 1532829480, ...          [°];
m = 20: 0, 2, 36, 646, 11592, 208010, ...  (A207832);
m = 21: 0, 12, 1320, 145188, ...           (A174745);
m = 22: 0, 42, 16548, 6519870, ...         (A174766);
m = 23: 0, 240, 552480, 1271808720, ...    [°];
m = 24: 0, 10, 980, 96030, 9409960, ...    (A168520);
m = 26: 0, 1, 10, 101, 1020, 10301, ...    (this sequence);
m = 27: 0, 260, 702520, 1898208780, ...    [°];
m = 28: 0, 24, 6096, 1548360, ...          (A175672);
m = 29: 0, 13, 1820, 254813, 35675640, ... [°];
m = 30: 0, 2, 44, 966, 21208, 465610, ...  (2*A077421), etc.
[°] apparently without related sequences in the OEIS.
(End)
From Michael A. Allen, Mar 12 2023: (Start)
Also called the 10-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 10 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
Sergio Falcón and Ángel Plaza, On k-Fibonacci sequences and polynomials and their derivatives, Chaos, Solitons & Fractals (2007).
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
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,1).

Cf. A099374 (squares).
Cf. A041040.
Cf. A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A243399.
Row n=10 of A073133, A172236 and A352361.

I:=[1,10]; [n le 2 select I[n] else 10*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

seq(combinat:-fibonacci(n+1, 10), n=0..19); # Peter Luschny, May 04 2018

Denominator[Convergents[Sqrt[26], 30]] (* Vincenzo Librandi, Dec 10 2013 *)
LinearRecurrence[{10,1}, {1,10}, 30] (* G. C. Greubel, Jan 24 2018 *)

x='x+O('x^30); Vec(1/(1-10*x-x^2)) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,10,-1) for n in range(1, 19)] # Zerinvary Lajos, Apr 26 2009

G.f.: 1/(1 - 10*x - x^2).
a(n) = 10*a(n-1) + a(n-2), n>=1; a(-1):=0, a(0)=1.
a(n) = S(n, 10*i)*(-i)^n where i^2:=-1 and S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind. See A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = 5+sqrt(26), am = -1/ap = 5-sqrt(26).
a(n) = F(n+1, 10), the (n+1)-th Fibonacci polynomial evaluated at x=10. - T. D. Noe, Jan 19 2006
a(n) = Sum_{i=0..floor(n/2)} binomial(n-i,i)*10^(n-2*i). - Sergio Falcon, Sep 24 2007

Extended by T. D. Noe, May 23 2011

cp's OEIS Frontend

A053028 Odd primes p with 4 zeros in any period of the Fibonacci numbers mod p.

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A098316 Decimal expansion of [3, 3, ...] = (3 + sqrt(13))/2.

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A168561 Riordan array (1/(1-x^2), x/(1-x^2)). Unsigned version of A049310.

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A057088 Scaled Chebyshev U-polynomials evaluated at i*sqrt(5)/2. Generalized Fibonacci sequence.

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A041025 Denominators of continued fraction convergents to sqrt(17).

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A053032 Odd primes p with one zero in Fibonacci numbers mod p.

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A053027 Odd primes p with 2 zeros in Fibonacci numbers mod p.

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A015523 a(n) = 3a(n-1) + 5a(n-2), with a(0)=0, a(1)=1.