A005668 -id:A005668 - cp's OEIS frontend

A054413 a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.

1, 7, 50, 357, 2549, 18200, 129949, 927843, 6624850, 47301793, 337737401, 2411463600, 17217982601, 122937341807, 877779375250, 6267392968557, 44749530155149, 319514104054600, 2281348258537349, 16288951913816043, 116304011655249650, 830417033500563593

Offset: 0

Henry Bottomley, May 10 2000

In general, sequences with recurrence a(n) = k*a(n-1) + a(n-2) and a(0)=1 (and a(-1)=0) have the generating function 1/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n) = b(5n), a(3n+1) = b(5n+3), a(3n+2) = 2*b(5n+4) where b(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2/4+1).]
a(p-1) == 53^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
From Johannes W. Meijer, Jun 12 2010: (Start)
For the sequence given above k=7 which implies that it is associated with A041091.
For a similar statement about sequences with recurrence a(n) = k*a(n-1) + a(n-2) but with a(0) = 2, and a(-1) = 0, see A086902; a sequence that is associated with A041090.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 7'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,...,7} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 7-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 7 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.
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.
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 linear recurrences with constant coefficients, signature (7,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A000129, A001076, A005668, A006190, A052918, A243399.
Row n=7 of A073133, A172236 and A352361.
Cf. A099367 (squares).

I:=[1, 7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013

LinearRecurrence[{7, 1}, {1, 7}, 30] (* Vincenzo Librandi, Feb 23 2013 *)

a(n)=([0,1; 1,7]^n*[1;7])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,7,-1) for n in range(1, 19)] # Zerinvary Lajos, Apr 24 2009

a(3n) = A041091(5n), a(3n+1) = A041091(5n+3), a(3n+2) = 2*A041091(5n+4).
G.f.: 1/(1 - 7x - x^2).
a(n) = U(n, 7*i/2)*(-i)^n with i^2=-1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See A049310.
a(n) = F(n, 7), the n-th Fibonacci polynomial evaluated at x=7. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = (sigma^n - (-sigma)^(-n))/(sqrt(53)) with sigma = (7+sqrt(53))/2;
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*7^(n-1-2i). (End)
a(n) = ((7 + sqrt(53))^n - (7 - sqrt(53))^n)/(2^n*sqrt(53)). Offset 1. a(3)=50. - Al Hakanson (hawkuu(AT)gmail.com), Jan 17 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097836(n), a(2n) = A097838(n).
Lim_{k->oo} a(n+k)/a(k) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Lim_{n->oo} A086902(n)/A054413(n-1) = sqrt(53).
(End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(53)-7)/2. - Vladimir Shevelev, Feb 23 2013
From Kai Wang, Feb 24 2020: (Start)
Sum_{m>=0} 1/(a(m)*a(m+2)) = 1/49.
Sum_{m>=0} 1/(a(2*m)*a(2*m+2)) = (sqrt(53)-7)/14.
In general, for sequences with recurrence f(n)= k*f(n-1)+f(n-2) and f(0)=1,
Sum_{m>=0} 1/(f(m)*f(m+2)) = 1/(k^2).
Sum_{m>=0} 1/(f(2*m)*f(2*m+2)) = (sqrt(k^2+4) - k)/(2*k). (End)
E.g.f.: (1/53)*exp(7*x/2)*(53*cosh(sqrt(53)*x/2) + 7*sqrt(53)*sinh(sqrt(53)*x/2)). - Stefano Spezia, Feb 26 2020
G.f.: x/(1 - 7*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 7 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

Formula corrected by Johannes W. Meijer, May 30 2010, Jun 02 2010

Extended by T. D. Noe, May 23 2011

A157103 Array A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 3, 1, 1, 5, 12, 10, 4, 1, 1, 8, 29, 33, 17, 5, 1, 1, 13, 70, 109, 72, 26, 6, 1, 1, 21, 169, 360, 305, 135, 37, 7, 1, 1, 34, 408, 1189, 1292, 701, 228, 50, 8, 1, 1, 55, 985, 3927, 5473, 3640, 1405, 357, 65, 9, 1

Offset: 0

Gary W. Adamson, Feb 22 2009

From Michael A. Allen, Mar 30 2023: (Start)
Column k is the k-metallonacci sequence for k > 0.
T(n,k) is, for n > 0 and k > 0, 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 k kinds of squares available. (End)

			Array begins:
  1,  1,   1,    1,     1,     1,      1,      1, ... (A000012);
  1,  1,   2,    3,     4,     5,      6,      7, ... (A000027);
  1,  2,   5,   10,    17,    26,     37,     50, ... (A002522);
  1,  3,  12,   33,    72,   135,    228,    357, ...;
  1,  5,  29,  109,   305,   701,   1405,   2549, ...;
  1,  8,  70,  360,  1292,  3640,   8658,  18200, ...;
  1, 13, 169, 1189,  5473, 18901,  53353, 129949, ...;
  1, 21, 408, 3927, 23184, 98145, 328776, 927843, ...;
  ...
First few rows of the triangle:
  1;
  1,   1;
  1,   1,    1;
  1,   2,    2,     1;
  1,   3,    5,     3,     1;
  1,   5,   12,    10,     4,     1;
  1,   8,   29,    33,    17,     5,     1;
  1,  13,   70,   109,    72,    26,     6,     1;
  1,  21,  169,   360,   305,   135,    37,     7,    1;
  1,  34,  408,  1189,  1292,   701,   228,    50,    8,   1;
  1,  55,  985,  3927,  5473,  3640,  1405,   357,   65,   9,   1;
  1,  89, 2378, 12970, 23184, 18901,  8658,  2549,  528,  82,  10,  1;
  1, 144, 5741, 42837, 98209, 98145, 53353, 18200, 4289, 747, 101, 11, 1;
  ...
Example: Column 3 = (1, 3, 10, 33, 109, 360, ...) = A006190.

G. C. Greubel, Antidiagonals n = 0..50, flattened
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894 [math.NT], 2015. See Table 3.

Columns k=1 to 9: A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371.
Cf. A084844, A084845.
Essentially the transpose of A073133, A172236, A352361.

A157103:= func< n,k | k eq 0 or k eq n select 1 else Evaluate(DicksonSecond(n, -1), k) >;
[A157103(n-k, k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jan 11 2022

A157103 := proc(n,k)
    if k = 0 then
        1;
    else
        mul(k-2*I*cos(l*Pi/(n+1)),l=1..n) ;
        combine(%,trig) ;
        round(%) ;
    end if;
end proc:
seq( seq(A157103(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Feb 27 2023

(* First program *)
T[, 0]=1; T[n, n_]=1; T[, ]=0;
T[n_, k_] /; 0 <= k <= n := k T[n-1, k] + T[n-2, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Aug 07 2018 *)
(* Second program *)
T[n_, k_]:= If[k==0 || k==n, 1, Fibonacci[n-k+1, k]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 11 2022 *)

def A157103(n,k): return 1 if (k==0 or k==n) else lucas_number1(n+1, k, -1)
flatten([[A157103(n-k, k) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Jan 11 2022

A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1 (array).
A(n, 1) = A000045(n+1).
T(n, k) = k*T(n-1, k) + T(n-2, k) with T(n, 0) = T(n, n) = 1 (triangle).
From G. C. Greubel, Jan 11 2022: (Start)
T(n, k) = Fibonacci(n-k+1, k), with T(n, 0) = T(n, n) = 1.
T(2*n, n) = A084845(n) for n >= 1, with T(0, 0) = 1.
T(2*n+1, n+1) = A084844(n). (End)

Edited by G. C. Greubel, Jan 11 2022

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

A005667 Numerators of continued fraction convergents to sqrt(10).

Original entry on oeis.org

1, 3, 19, 117, 721, 4443, 27379, 168717, 1039681, 6406803, 39480499, 243289797, 1499219281, 9238605483, 56930852179, 350823718557, 2161873163521, 13322062699683, 82094249361619, 505887558869397, 3117419602578001, 19210405174337403, 118379850648602419

Offset: 0

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

a(2*n+1) with b(2*n+1) := A005668(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 10*b^2 = -1, a(2*n) with b(2*n) := A005668(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 10*b^2 = +1 (cf. Emerson reference).
Bisection: a(2*n) = T(n,19) = A078986(n), n >= 0 and a(2*n+1) = 3*S(2*n, 2*sqrt(10)), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
The initial 1 corresponds to a denominator 0 in A005668. But according to standard conventions, a continued fraction starts with b(0) = integer part of the number, and the sequence of convergents p(n)/q(n) start with (p(0),q(0)) = (b(0),1). A fraction 1/0 has no mathematical meaning, the only justification is that initial terms p(-1) = 1, q(-1) = 0 are consistent with the recurrent relations p(n) = b(n)*p(n-1) + b(n-2) and the same for q(n). - M. F. Hasler, Nov 02 2019

			G.f. = 1 + 3*x + 19*x^2 + 117*x^3 + 721*x^4 + 4443*x^5 + 27379*x^6 + ... - _Michael Somos_, Jul 14 2018

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
R. K. Guy, Letter to N. J. A. Sloane, 1987
Tian-Xiao He and Peter J.-S. Shiue, Identities for linear recursive sequences of order 2, Elect. Res. Archive (2021) Vol. 29, No. 5, 3489-3507.
Tanya Khovanova, Recursive Sequences
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.
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
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,1).

Cf. A010467, A040006, A084134, A005668 (denominators).

I:=[1, 3]; [n le 2 select I[n] else 6*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 09 2013

A005667:=(-1+3*z)/(-1+6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Join[{1},Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[10],n]]],{n,1,30}]] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2011 *)
CoefficientList[Series[(1-3x)/(1-6x-x^2), {x,0,30}], x] (* Vincenzo Librandi, Jun 09 2013 *)
Join[{1},Numerator[Convergents[Sqrt[10],30]]] (* or *) LinearRecurrence[ {6,1},{1,3},30] (* Harvey P. Dale, Aug 22 2016 *)
a[ n_] := (-I)^n ChebyshevT[ n, 3 I]; (* Michael Somos, Jul 14 2018 *)
LucasL[Range[0,30], 6]/2 (* G. C. Greubel, Jun 06 2019 *)

a(n)=([0,1;1,6]^n*[1;3])[1,1] \\ Charles R Greathouse IV, Jun 11 2015

((1-3*x)/(1-6*x-x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jun 06 2019

a(n) = 6*a(n-1) + a(n-2).
G.f.: (1-3*x)/(1-6*x-x^2).
a(n) = ((-i)^n)*T(n, 3*i) with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1.
From Paul Barry, Nov 15 2003: (Start)
Binomial transform of A084132.
E.g.f.: exp(3*x)*cosh(sqrt(10)*x).
a(n) = ((3+sqrt(10))^n + (3-sqrt(10))^n)/2.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k) * 10^k * 3^(n-2*k). (End)
a(n) = (-1)^n * a(-n) for all n in Z. - Michael Somos, Jul 14 2018 [This refers to the sequence extended to negative indices according to the recurrence relation, but not to the sequence as it is currently defined. - M. F. Hasler, Nov 02 2019]
a(n) = Lucas(n,6)/2, Lucas polynomial, L(n,x), evaluated at x = 6. - G. C. Greubel, Jun 06 2019

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

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

A097315 Pell equation solutions (3b(n))^2 - 10a(n)^2 = -1 with b(n) = A097314(n), n >= 0.

Original entry on oeis.org

1, 37, 1405, 53353, 2026009, 76934989, 2921503573, 110940200785, 4212806126257, 159975692596981, 6074863512559021, 230684837784645817, 8759948972303982025, 332647376109766671133, 12631840343198829521029, 479677285665445755127969, 18215105014943739865341793, 691694313282196669127860165

Offset: 0

Wolfdieter Lang, Aug 31 2004

Hypotenuses of primitive Pythagorean triples in A195616 and A195617. - Clark Kimberling, Sep 22 2011

			(x,y) = (3,1), (117,37), (4443,1405), ... give the positive integer solutions to x^2 - 10*y^2 = -1.
G.f. = 1 + 37*x + 1405*x^2 + 53353*x^3 + ... - _Michael Somos_, Feb 24 2023

Indranil Ghosh, Table of n, a(n) for n = 0..631
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A078987 (partial sums), A097314, A195616, A195617, A049310, A053120, A005668.
Row 3 of array A188647.
Cf. A221874.
Cf. similar sequences listed in A238379.

a:=[1,37];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1, 37]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 01 2019

CoefficientList[Series[(1-x)/(1-38x+x^2), {x,0,20}], x] (* Michael De Vlieger, Feb 04 2017 *)
LinearRecurrence[{38,-1}, {1,37}, 21] (* G. C. Greubel, Aug 01 2019 *)

Vec((1-x)/(1-38*x+x^2) + O(x^20)) \\ Michel Marcus, Jun 04 2015

from itertools import islice
def A097315_gen(): # generator of terms
    x, y = 30, 10
    while True:
        yield y//10
        x, y = x*19+y*60, x*6+y*19
A097315_list = list(islice(A097315_gen(),20)) # Chai Wah Wu, Apr 24 2025

((1-x)/(1-38*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = S(n, 38) - S(n-1, 38) = T(2*n+1, sqrt(10))/sqrt(10), with Chebyshev polynomials of the second and first kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 6*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-38*x+x^2).
a(n) = 38*a(n-1) - a(n-2) for n > 1. - Philippe Deléham, Nov 18 2008
a(n) = sqrt(2+(19-6*sqrt(10))^(1+2*n) + (19+6*sqrt(10))^(1+2*n))/(2*sqrt(10)). - Gerry Martens, Jun 04 2015
a(n) = A078987(n) - A078987(n-1). - R. J. Mathar, Dec 05 2015
a(n) = A005668(2*n+1). - Michael Somos, Feb 24 2023
E.g.f.: exp(19*x)*(10*cosh(6*sqrt(10)*x) + 3*sqrt(10)*sinh(6*sqrt(10)*x))/10. - Stefano Spezia, Apr 24 2025

Typo in recurrence formula corrected by Laurent Bonaventure (bonave(AT)free.fr), Oct 03 2010

More terms added by Indranil Ghosh, Feb 04 2017

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

A085447 a(n) = 6*a(n-1) + a(n-2), starting with a(0)=2 and a(1)=6.

Original entry on oeis.org

2, 6, 38, 234, 1442, 8886, 54758, 337434, 2079362, 12813606, 78960998, 486579594, 2998438562, 18477210966, 113861704358, 701647437114, 4323746327042, 26644125399366, 164188498723238, 1011775117738794

Offset: 0

Gary W. Adamson, Jul 01 2003

a(n+1)/a(n) converges to 3 + sqrt 10.

a(0)/a(1) = 1/3 = [3]; a(1)/a(2) = 6/38 = [6,3]; a(2)/a(3) = 38/234 = [6,6,3], a(3)/a(4) = 234/1442 = [6,6,6,3]; a(4)/a(5) = 1442/8886 = [6,6,6,6,3];...etc. Lim a(n)/a(n+1) as n approaches infinity = 0.1622776...= 1/(3 + sqrt10) = sqrt(10) - 3.

			a(4) = 6*a(3)+a(2) = 6*234+38 = 1442.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence L_{6,n}
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (6,1).

Cf. A005667, A005668.

I:=[2,6]; [n le 2 select I[n] else 6*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 6, a[n] == 6 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{6,1}, {2,6}, 30] (* G. C. Greubel, Nov 07 2018 *)

x='x+O('x^30); Vec(2*(1-3*x)/(1-6*x-x^2)) \\ G. C. Greubel, Nov 07 2018

a(n) = (3 + sqrt 10)^n + (3 - sqrt 10)^n = A005668(n+1) + A005668(n-1).
O.g.f.: 2*(-1+3*x)/(-1+6*x+x^2). - R. J. Mathar, Dec 02 2007
a(n) = 2*A005667(n). - R. J. Mathar, Nov 10 2013

Edited and extended by Henry Bottomley, Jul 13 2003

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

A040006 Continued fraction for sqrt(10).

Original entry on oeis.org

3, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 0

N. J. A. Sloane

Eventual period is (6). - Zak Seidov, Mar 05 2011
The convergents are given in A005667(n+1)/A005668(n+1), n >= 0. - Wolfdieter Lang, Nov 23 2017
Decimal expansion of 11/30. - Elmo R. Oliveira, Feb 16 2024

			3.162277660168379331998893544... = 3 + 1/(6 + 1/(6 + 1/(6 + 1/(6 + ...)))).

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 276.

Harry J. Smith, Table of n, a(n) for n = 0..20000
G. Xiao, Contfrac
Index entries for continued fractions for constants
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A010467 (decimal expansion), A005667(n+1)/A005668(n+1) (convergents), A248239 (Egyptian fraction).

Cf. A040000.

[6-3*(Binomial(2*n,n) mod 2): n in [0..100]]; // Vincenzo Librandi, Jan 03 2016

Digits := 100: convert(evalf(sqrt(N)),confrac,90,'cvgts'):

ContinuedFraction[Sqrt[10],300] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011 *)
PadRight[{3},120,{6}] (* Harvey P. Dale, Aug 25 2024 *)

contfrac(sqrt(10)) \\ For illustration.

A040006(n)=if(n,6,3) \\ M. F. Hasler, Nov 02 2019

a(n) = 3 + 3*sign(n). a(n) = 6, n > 0. - Wesley Ivan Hurt, Nov 01 2013
From Elmo R. Oliveira, Feb 16 2024: (Start)
G.f.: 3*(1+x)/(1-x).
E.g.f.: 6*exp(x) - 3.
a(n) = 3*A040000(n). (End)

cp's OEIS Frontend

A054413 a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.

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A157103 Array A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1, read by antidiagonals.

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A005667 Numerators of continued fraction convergents to sqrt(10).

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

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A097315 Pell equation solutions (3*b(n))^2 - 10*a(n)^2 = -1 with b(n) = A097314(n), n >= 0.

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A097315 Pell equation solutions (3b(n))^2 - 10a(n)^2 = -1 with b(n) = A097314(n), n >= 0.