A001076 -id:A001076 - cp's OEIS frontend

A007805 a(n) = Fibonacci(6*n + 3)/2.

1, 17, 305, 5473, 98209, 1762289, 31622993, 567451585, 10182505537, 182717648081, 3278735159921, 58834515230497, 1055742538989025, 18944531186571953, 339945818819306129, 6100080207560938369

Offset: 0

James A. Raymond, Clark Kimberling

Hypotenuse (z) of Pythagorean triples (x,y,z) with |2x-y|=1.
x(n) := 2*A049629(n) and y(n) := a(n), n >= 0, give all positive solutions of the Pell equation x^2 - 5*y^2 = -1. See the Gregory V. Richardson formula, where his x is the y here and A075796(n+1) = x(n). - Wolfdieter Lang, Jun 20 2013
Positive numbers n such that 5*n^2 - 1 is a square (A075796(n+1)^2). - Gregory V. Richardson, Oct 13 2002

G. C. Greubel, Table of n, a(n) for n = 0..795 (terms 0..100 from T. D. Noe)
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.
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 (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045.
Row 18 of array A094954.
Row 2 of array A188647.
Cf. similar sequences listed in A238379.

a007805 = (`div` 2) . a000045 . (* 3) . (+ 1) . (* 2)
-- Reinhard Zumkeller, Mar 26 2013

I:=[1, 17]; [n le 2 select I[n] else 18*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

seq(combinat:-fibonacci(6*n+3)/2, n=0..30); # Robert Israel, Sep 10 2014

LinearRecurrence[{18, -1}, {1, 17}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
Table[Fibonacci[6n+3]/2, {n, 0, 20}] (* Harvey P. Dale, Dec 17 2011 *)
CoefficientList[Series[(1-x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

a(n)=fibonacci(6*n+3)/2 \\ Edward Jiang, Sep 09 2014

x='x+O('x^30); Vec((1-x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

G.f.: (1-x)/(1-18*x+x^2).
a(n) = 18*a(n-1) - a(n-2), n > 1, a(0)=1, a(1)=17.
a(n) = A134495(n)/2 = A001076(2n+1).
a(n+1) = 9*a(n) + 4*sqrt(5*a(n)^2-1). - Richard Choulet, Aug 30 2007, Dec 28 2007
a(n) = ((2+sqrt(5))^(2*n+1) - (2-sqrt(5))^(2*n+1))/(2*sqrt(5)). - Dean Hickerson, Dec 09 2002
a(n) ~ (1/10)*sqrt(5)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*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) = q(n, 16). - Benoit Cloitre, Dec 06 2002
a(n) = 19*a(n-1)- 19*a(n-2) + a(n-3); f(x) = (sqrt(5)/10)*((2+sqrt(5))*(9+4*sqrt(5))^(x-1) - (2-sqrt(5))*(9-4*sqrt(5))^(x-1)). - Antonio Alberto Olivares, May 15 2008
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3). - Antonio Alberto Olivares, Jun 19 2008
a(n) = b(n+1) - b(n), n >= 0, with b(n) := F(6*n)/F(6) = A049660(n). First differences. See the o.g.f.s. - Wolfdieter Lang, 2012
a(n) = S(n,18) - S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/4^2. Compare with A001519 and A097843. - Peter Bala, Nov 29 2013
a(n) = 9*a(n-1) + 8*A049629(n-1), n >= 1, a(0) = 1. This is just the rewritten Chebyshev S(n, 18) recurrence. - Wolfdieter Lang, Aug 26 2014
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.
The aerated sequence (b(n)) n>=1 = [1, 0, 17, 0, 305, 0, 5473, 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 = -20, 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) = sqrt(2 + (9-4*sqrt(5))^(1+2*n) + (9+4*sqrt(5))^(1+2*n))/(2*sqrt(5)). - Gerry Martens, Jun 04 2015

Better description and more terms from Michael Somos

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

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)

A040002 Continued fraction for sqrt(5).

Original entry on oeis.org

2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 0

N. J. A. Sloane

Decimal expansion of 11/45. - Natan Arie Consigli, Jan 19 2016

			2.236067977499789696409173668... = 2 + 1/(4 + 1/(4 + 1/(4 + 1/(4 + ...)))). - _Harry J. Smith_, Jun 01 2009

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 186.
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 of eventually constant sequences
Index of divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A002163 (decimal expansion), A001077/A001076 (convergents), A248235 (Egyptian fraction).
Cf. Continued fraction for sqrt(a^2+1) = (a, 2a, 2a, 2a....): A040000 (contfrac(sqrt(2)) = (1,2,2,...)), A040002, A040006, A040012, A040020, A040030, A040042, A040056, A040072, A040090, A040110 (contfrac(sqrt(122)) = (11,22,22,...)), A040132, A040156, A040182, A040210, A040240, A040272, A040306, A040342, A040380, A040420 (contfrac(sqrt(442)) = (21,42,42,...)), A040462, A040506, A040552, A040600, A040650, A040702, A040756, A040812, A040870, A040930 (contfrac(sqrt(962)) = (31,62,62,...)).
Essentially the same as A010709.

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

ContinuedFraction[Sqrt[5],300] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011 *)
PadRight[{2},120,{4}] (* Harvey P. Dale, Jul 06 2019 *)

{ allocatemem(932245000); default(realprecision, 26000); x=contfrac(sqrt(5)); for (n=0, 20000, write("b040002.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 01 2009

a(0) = 2, a(n) = 4 n>0. - Natan Arie Consigli, Jan 19 2016
From Elmo R. Oliveira, Feb 16 2024: (Start)
G.f.: 2*(1+x)/(1-x).
E.g.f.: 4*exp(x) - 2.
a(n) = 2*A040000(n). (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

A026671 Number of lattice paths from (0,0) to (n,n) with steps (0,1), (1,0) and, when on the diagonal, (1,1).

Original entry on oeis.org

1, 3, 11, 43, 173, 707, 2917, 12111, 50503, 211263, 885831, 3720995, 15652239, 65913927, 277822147, 1171853635, 4945846997, 20884526283, 88224662549, 372827899079, 1576001732485, 6663706588179, 28181895551161, 119208323665543, 504329070986033, 2133944799315027

Offset: 0

Clark Kimberling, Miklos Bona

1, 1, 3, 11, 43, 173, ... is the unique sequence for which both the Hankel transform of the sequence itself and the Hankel transform of its left shift are the powers of 2 (A000079). For example, det[{{1, 1, 3}, {1, 3, 11}, {3, 11, 43}}] = det[{{1, 3, 11}, {3, 11, 43}, {11, 43, 173}}] = 4. - David Callan, Mar 30 2007
From Paul Barry, Jan 25 2009: (Start)
a(n) is the image of F(2n+2) under the Catalan matrix (1,xc(x)) where c(x) is the g.f. of A000108.
The sequence 1,1,3,... is the image of A001519 under (1,xc(x)). This sequence has g.f. given by 1/(1-x-2x^2/(1-3x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction). (End)
Binomial transform of A111961. - Philippe Deléham, Feb 11 2009
From Paul Barry, Nov 03 2010: (Start)
The sequence 1,1,3,... has g.f. 1/(1-x/sqrt(1-4x)), INVERT transform of A000984.
It is an eigensequence of the sequence array for A000984. (End)

L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jean-Christophe Aval, Adrien Boussicault and Sandrine Dasse-Hartaut, The tree structure in staircase tableaux, arXiv:1109.4907 [math.CO], 2011-2013.
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.
Miklos Bona, The permutation classes equinumerous to the smooth class, Electron. J. Combin., 5 (1998), no. 1, Research Paper 31, 12 pp.
David Callan and Toufik Mansour, Five subsets of permutations enumerated as weak sorting permutations, arXiv:1602.05182 [math.CO], 2016.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
Huyile Liang, Jeffrey Remmel and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 16.

a(n) = T(2n-1, n-1), T given by A026736.
a(n) = T(2n, n), T given by A026670.
a(n) = T(2n+1, n+1), T given by A026725.
Row sums of triangle A054335.
Cf. A026781, A001076, A176280.

a:=[3,11,43];; for n in [4..30] do a[n]:=(2*(4*n-3)*a[n-1] - 3*(5*n-8)*a[n-2] - 2*(2*n-3)*a[n-3])/n; od; Concatenation([1], a); # G. C. Greubel, Jul 16 2019

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 1/(Sqrt(1-4*x)-x) )); // G. C. Greubel, Jul 16 2019

Table[SeriesCoefficient[1/(Sqrt[1-4*x]-x),{x,0,n}],{n,0,30}] (* Vaclav Kotesovec, Oct 08 2012 *)

{a(n)= if(n<0, 0, polcoeff( 1/(sqrt(1 -4*x +x*O(x^n)) -x), n))} /* Michael Somos, Apr 20 2007 */

my(x='x+O('x^66)); Vec( 1/(sqrt(1-4*x)-x) ) \\ Joerg Arndt, May 04 2013

(1/(sqrt(1-4*x)-x)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 16 2019

From Wolfdieter Lang, Mar 21 2000: (Start)
G.f.: 1/(sqrt(1-4*x)-x).
a(n) = Sum_{i=1..n} a(i-1)*binomial(2*(n-i), n-i) + binomial(2*n, n), n >= 1, a(0)=1. (End)
G.f.: 1/(1 -x -2*x*c(x)) where c(x) = g.f. for Catalan numbers A000108. - Michael Somos, Apr 20 2007
From Paul Barry, Jan 25 2009: (Start)
G.f.: 1/(1 - 3xc(x) + x^2*c(x)^2);
G.f.: 1/(1-3x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction).
a(0) = 1, a(n) = Sum_{k=0..n} (k/(2n-k))*C(2n-k,n-k)*F(2k+2). (End)
a(n) = Sum_{k=0..n} A039599(n,k) * A000045(k+2). - Philippe Deléham, Feb 11 2009
From Paul Barry, Feb 08 2009: (Start)
G.f.: 1/(1-x/(1-2x/(1-x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction);
G.f. of 1,1,3,... is 1/(1-x-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) = the upper left term in M^n, M = the infinite square production matrix:
  3, 2, 0, 0, 0, 0, ...
  1, 1, 1, 0, 0, 0, ...
  1, 1, 1, 1, 0, 0, ...
  1, 1, 1, 1, 1, 0, ...
  1, 1, 1, 1, 1, 1, ...
  ... (End)
From Vaclav Kotesovec, Oct 08 2012: (Start)
D-finite with recurrence: n*a(n) = 2*(4*n-3)*a(n-1) - 3*(5*n-8)*a(n-2) - 2*(2*n-3)*a(n-3).
a(n) ~ (2+sqrt(5))^n/sqrt(5). (End)
a(n) = Sum_{k=0..n+1} 4^(n+1-k) * binomial(n-k/2,n+1-k). - Seiichi Manyama, Mar 30 2025
From Peter Luschny, Mar 30 2025: (Start)
a(n) = 4^n*(binomial(n-1/2, n)*hypergeom([1, (1-n)/2, -n/2], [1/2, 1/2-n], -1/4) + hypergeom([(1-n)/2, 1-n/2], [1-n], -1/4)/4) for n > 0.
a(n) = A001076(n) + A176280(n). (End)

A028412 Rectangular array of numbers Fibonacci(m(n+1))/Fibonacci(m), m >= 1, n >= 0, read by downward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 8, 3, 1, 7, 17, 21, 5, 1, 11, 48, 72, 55, 8, 1, 18, 122, 329, 305, 144, 13, 1, 29, 323, 1353, 2255, 1292, 377, 21, 1, 47, 842, 5796, 15005, 15456, 5473, 987, 34, 1, 76, 2208, 24447, 104005, 166408, 105937, 23184, 2584, 55, 1, 123, 5777

Offset: 0

N. J. A. Sloane

Every integer-valued quotient of two Fibonacci numbers is in this array. - Clark Kimberling, Aug 28 2008

Not only does 5 divide row 5, but 50 divides (-5 + row 5), as in A214984. - Clark Kimberling, Nov 02 2012

			   1   1    1      1       1        1
   1   3    4      7      11       18
   2   8   17     48     122      323
   3  21   72    329    1353     5796
   5  55  305   2255   15005   104005
   8 144 1292  15456  166408  1866294
  13 377 5473 105937 1845493 33489287
  ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 142.

Clark Kimberling, Table of n, a(n) for n = 0..1829
I. Strazdins, Lucas factors and a Fibonomial generating function, in Applications of Fibonacci numbers, Vol. 7 (Graz, 1996), 401-404, Kluwer Acad. Publ., Dordrecht, 1998.

Columns include A000045, A001906, A001076, A004187, A049666, A049660, A049667, A049668, A049669, A049670.
Rows include (essentially) A000032, A047946, A083564, A103226.
Main diagonal is A051294.
Transpose is A214978.

max = 11; col[m_] := CoefficientList[ Series[ 1/(1 - LucasL[m]*x + (-1)^m*x^2), {x, 0, max}], x]; t = Transpose[ Table[ col[m], {m, 1, max}]] ; Flatten[ Table[ t[[n - m + 1, m]], {n, 1, max }, {m, n, 1, -1}]] (* Jean-François Alcover, Feb 21 2012, after Paul D. Hanna *)
f[n_] := Fibonacci[n]; t[m_, n_] := f[m*n]/f[n]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]] (* array *)
t = Flatten[Table[t[k, n + 1 - k], {n, 1, 120}, {k, 1, n}]] (* sequence *) (* Clark Kimberling, Nov 02 2012 *)

{T(n,m)=polcoeff(1/(1 - Lucas(m)*x + (-1)^m*x^2 +x*O(x^n)),n)}

T(n, m) = Sum_{i_1>=0} Sum_{i_2>=0} ... Sum_{i_m>=0} C(n-i_m, i_1)*C(n-i_1, i_2)*C(n-i_2, i_3)*...*C(n-i_{m-1}, i_m).

G.f. for column m >= 1: 1/(1 - Lucas(m)*x + (-1)^m*x^2), where Lucas(m) = A000204(m). - Paul D. Hanna, Jan 28 2012

More terms from Erich Friedman, Jun 03 2001
Edited by Ralf Stephan, Feb 03 2005
Better description from Clark Kimberling, Aug 28 2008

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

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

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