A015448 -id:A015448 - cp's OEIS frontend

A287819 Number of nonary sequences of length n such that no two consecutive terms have distance 4.

1, 9, 71, 561, 4433, 35031, 276827, 2187585, 17287073, 136608591, 1079529611, 8530826457, 67413620993, 532726379847, 4209793089371, 33267280400913, 262889866978817, 2077449112980255, 16416740845208075, 129730917736941417, 1025179795159015841

Offset: 0

David Nacin, Jun 02 2017

			For n=2 the a(2) = 81 - 10 = 71 sequences contain every combination except these ten: 04,40,15,51,26,62,37,73,48,84.

Index entries for linear recurrences with constant coefficients, signature (8,1,-14).

Cf. A040000, A003945, A083318, A078057, A003946, A126358, A003946, A055099, A003947, A015448, A126473. A287804-A287819.

LinearRecurrence[{8, 1, -14}, {1, 9, 71, 561}, 40]

def a(n):
    if n in [0, 1, 2, 3]:
        return [1, 9, 71, 561][n]
    return 8*a(n-1)+a(n-2)-14*a(n-3)

For n>2, a(n) = 8*a(n-1) + a(n-2) - 14*a(n-3), a(0)=1, a(1)=9, a(2)=71, a(3)=561.

G.f.: (1 + x - 2 x^2 - 2 x^3)/(1 - 8 x - x^2 + 14 x^3).

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

A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

Original entry on oeis.org

0, 2, 10, 44, 188, 798, 3382, 14328, 60696, 257114, 1089154, 4613732, 19544084, 82790070, 350704366, 1485607536, 6293134512, 26658145586, 112925716858, 478361013020, 2026369768940, 8583840088782, 36361730124070, 154030760585064, 652484772464328, 2763969850442378

Offset: 0

Ralf Stephan, Oct 30 2004

Partial sum of the even Fibonacci numbers. - Vladimir Joseph Stephan Orlovsky, Nov 28 2010

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 25.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Project Euler, Problem 2: Even Fibonacci Numbers.
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Partial sums of A014445.
Cf. A000045, A004794, A015448, A049652.
Cf. A087635.
Case k = 3 of partial sums of fibonacci(k*n): A000071, A027941, A058038, A138134, A053606.

[(Fibonacci(3*n+2) - 1)/2: n in [0..30]]; // G. C. Greubel, Jan 17 2018

CoefficientList[Series[2 x/((1 - x) (1 - 4 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 15 2014 *)
LinearRecurrence[{5, -3, -1}, {0, 2, 10}, 30] (* G. C. Greubel, Jan 17 2018 *)
Accumulate[Fibonacci[3Range[0, 19]]] (* Alonso del Arte, Dec 23 2018 *)

a(n) = sum(i=1, n, fibonacci(3*i)); \\ Michel Marcus, Mar 15 2014

a(n) = fibonacci(3*n+2)\2 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = (Fibonacci(3*n + 2) - 1)/2 = (A015448(n+1)-1)/2.
G.f.: 2*x/((1 - x)*(1 - 4*x - x^2)).
a(n) = (F(3n + 2) - 1)/2 = 2 * A049652(n).
a(n) = Sum_{0 <= j <= i <= n} binomial(i, j)*F(i + j). - Benoit Cloitre, May 21 2005
From Gary Detlefs, Dec 08 2010: (Start)
a(n) = 4*a(n - 1) + a(n - 2) + 2, n > 1.
a(n) = 5*a(n - 1) - 3*a(n - 2) - a(n - 3), n > 2.
a(n) = (Fibonacci(3*n + 3) + Fibonacci(3*n) - 2)/4. (End)
a(n) = (-10 + (5 - 3*sqrt(5))*(2 - sqrt(5))^n + (2 + sqrt(5))^n*(5 + 3*sqrt(5)))/20. - Colin Barker, Nov 26 2016
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + 3*sqrt(5)*sinh(sqrt(5)*x)) - 5)/10. - Stefano Spezia, Jun 03 2024

a(0) = 0 prepended by Joerg Arndt, Mar 13 2014

A287825 Number of sequences over the alphabet {0,1,...,9} such that no two consecutive terms have distance 1.

Original entry on oeis.org

1, 10, 82, 674, 5540, 45538, 374316, 3076828, 25291120, 207889674, 1708825732, 14046322404, 115458919774, 949057110644, 7801124426174, 64124215108032, 527092600834054, 4332631742719370, 35613662169258228, 292739611493034596, 2406281042646218328

Offset: 0

David Nacin, Jun 02 2017

Index entries for linear recurrences with constant coefficients, signature (9, -4, -21, 9, 5).

Cf. A040000, A003945, A083318, A078057, A003946, A126358, A003946, A055099, A003947, A015448, A126473. A287804-A287819. A287825-A287831.

LinearRecurrence[{9, -4, -21, 9, 5}, {1, 10, 82, 674, 5540, 45538}, 40]

def a(n):
    if n in [0, 1, 2, 3, 4, 5]:
        return [1, 10, 82, 674, 5540, 45538][n]
    return 9*a(n-1) - 4*a(n-2) - 21*a(n-3) + 9*a(n-4) + 5*a(n-5)

For n>5, a(n) = 9*a(n-1) - 4*a(n-2) - 21*a(n-3) + 9*a(n-4) + 5*a(n-5), a(0)=1, a(1)=10, a(2)=82, a(3)=674, a(4)=5540, a(5)=45538.

G.f.: (-1 - x + 4*x^2 + 3*x^3 - 3*x^4 - x^5)/(-1 + 9*x - 4*x^2 - 21*x^3 + 9*x^4 + 5*x^5).

A048876 a(n) = 4*a(n-1) + a(n-2); a(0)=1, a(1)=7.

Original entry on oeis.org

1, 7, 29, 123, 521, 2207, 9349, 39603, 167761, 710647, 3010349, 12752043, 54018521, 228826127, 969323029, 4106118243, 17393796001, 73681302247, 312119004989, 1322157322203, 5600748293801, 23725150497407, 100501350283429, 425730551631123, 1803423556807921

Offset: 0

Barry E. Williams

Generalized Pell equation with second term of 7.

T. D. Noe, Table of n, a(n) for n = 0..200
M. Bicknell, A Primer on the Pell Sequence and related sequences, Fibonacci Quarterly, Vol. 13, No. 4, 1975, pp. 345-349.
L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Pellian Representations, Fib. Quart. Vol. 10, No. 5, (1972), pp. 449-488.
Tanya Khovanova, Recursive Sequences
A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A001076, A001077, A015448, A033887.

with(combinat): a:=n->3*fibonacci(n-1,4)+fibonacci(n,4): seq(a(n), n=1..16); # Zerinvary Lajos, Apr 04 2008

f[n_] := Block[{s = Sqrt@ 5}, Simplify[((1 + s)(2 + s)^n + (1 - s)(2 - s)^n)/2]]; (* Or *)
f[n_] := Fibonacci[3 n + 3] - Fibonacci[3 n - 1]; (* Or *)
f[n_] := Mod[ Fibonacci[3n + 7], Fibonacci[3n + 3]]; Array[f, 22, 0]
a[n_] := 4a[n - 1] + a[n - 2]; a[0] = 1; a[1] = 7; Array[a, 22, 0] (* Or *)
CoefficientList[ Series[(1 + 3x)/(1 - 4x - x^2), {x, 0, 21}], x] (* Robert G. Wilson v *)
LinearRecurrence[{4,1},{1,7},30] (* Harvey P. Dale, Jun 13 2015 *)
Table[LucasL[3*n + 1], {n, 0, 20}] (* Rigoberto Florez, Apr 04 2019 *)

Vec((1+3*x)/(1-4*x-x^2) + O(x^30)) \\ Altug Alkan, Oct 07 2015

G.f.: (1+3*x)/(1-4*x-x^2). - Philippe Deléham, Nov 03 2008
a(n) = ((1+sqrt(5))*(2+sqrt(5))^n + (1-sqrt(5))*(2-sqrt(5))^n )/2.
a(n) = A000032(3*n+1). - Thomas Baruchel, Nov 26 2003
From Gary Detlefs, Mar 06 2011: (Start)
a(n) = Fibonacci(3*n+7) mod Fibonacci(3*n+3), n > 0.
a(n) = Fibonacci(3*n+3) - Fibonacci(3*n-1). (End)
a(n) = A001076(n+1)+3*A001076(n). - R. J. Mathar, Oct 22 2013
a(n) = 5*F(2*n)*F(n+1) - L(n-1)*(-1)^n. - J. M. Bergot, Mar 22 2016
a(n) = Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2)*2^(n-k). - Tony Foster III, Sep 03 2017

A048875 Generalized Pellian with second term of 6.

Original entry on oeis.org

1, 6, 25, 106, 449, 1902, 8057, 34130, 144577, 612438, 2594329, 10989754, 46553345, 197203134, 835365881, 3538666658, 14990032513, 63498796710, 268985219353, 1139439674122, 4826743915841, 20446415337486, 86612405265785, 366896036400626, 1554196550868289

Offset: 0

Barry E. Williams

			G.f. = 1 + 6*x + 25*x^2 + 106*x^3 + 449*x^4 + 1902*x^5 + 8057*x^6 + 34130*x^7 + ...

T. D. Noe, Table of n, a(n) for n = 0..200
M. Bicknell, A Primer on the Pell Sequence and related sequences, Fibonacci Quarterly, Vol. 13, No. 4, 1975, pp. 345-349.
L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Pellian Representations, Fib. Quart. Vol. 10, No. 5, (1972), pp. 449-488.
Tanya Khovanova, Recursive Sequences
A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A015448, A001076, A001077, A033887, A134418, A097924, A374439.

with(combinat): a:=n->2*fibonacci(n-1,4)+fibonacci(n,4): seq(a(n), n=1..17); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{4,1},{1,6},40] (* Harvey P. Dale, Nov 30 2011 *)
a[ n_] := (4 I ChebyshevT[ n + 1, -2 I] - 3 ChebyshevT[ n, -2 I]) I^n / 5; (* Michael Somos, Feb 23 2014 *)
a[ n_] := If[ n < 0, SeriesCoefficient[ (1 + 6 x) / (1 + 4 x - x^2), {x, 0, -n}], SeriesCoefficient[ (1 + 2 x) / (1 - 4 x - x^2), {x, 0, n}]]; (* Michael Somos, Feb 23 2014 *)

a[0]:1$ a[1]:6$ a[n]:=4*a[n-1]+a[n-2]$ makelist(a[n], n, 0, 30); /* Martin Ettl, Nov 03 2012 */

{a(n) = ( 4*I*polchebyshev( n+1, 1, -2*I) - 3*polchebyshev( n, 1, -2*I)) * I^n / 5}; /* Michael Somos, Feb 23 2014 */

{a(n) = if( n<0, polcoeff( (1 + 6*x) / (1 + 4*x - x^2) + x * O(x^-n), -n), polcoeff( (1 + 2*x) / (1 - 4*x - x^2) + x * O(x^n), n))}; \\ Michael Somos, Feb 23 2014

a(n) = ((4+sqrt(5))*(2+sqrt(5))^n - (4-sqrt(5))*(2-sqrt(5))^n)*sqrt(5)/2.
a(n) = 4a(n-1) + a(n-2); a(0)=1, a(1)=6.
Binomial transform of A134418: (1, 5, 14, 48, 152, ...). - Gary W. Adamson, Nov 23 2007
G.f.: (1+2*x)/(1-4*x-x^2). - Philippe Deléham, Nov 03 2008
a(-1 - n) = (-1)^n * A097924(n) for all n in Z. - Michael Somos, Feb 23 2014
a(n) = A001076(n+1) + 2*A001076(n). - R. J. Mathar, Sep 11 2019
a(n) = 4^n*Sum_{k=0..n} A374439(n, k)*(1/4)^k. - Peter Luschny, Jul 26 2024

Corrected by T. D. Noe, Nov 07 2006

A103134 a(n) = Fibonacci(6n+4).

Original entry on oeis.org

3, 55, 987, 17711, 317811, 5702887, 102334155, 1836311903, 32951280099, 591286729879, 10610209857723, 190392490709135, 3416454622906707, 61305790721611591, 1100087778366101931, 19740274219868223167, 354224848179261915075, 6356306993006846248183

Offset: 0

Creighton Dement, Jan 24 2005

Gives those numbers which are Fibonacci numbers in A103135.
Generally, for any sequence where a(0)= Fibonacci(p), a(1) = F(p+q) and Lucas(q)*a(1) +- a(0) = F(p+2q), then a(n) = L(q)*a(n-1) +- a(n-2) generates the following Fibonacci sequence: a(n) = F(q(n)+p). So for this sequence, a(n) = 18*a(n-1) - a(n-2) = F(6n+4): q=6, because 18 is the 6th Lucas number (L(0) = 2, L(1)=1); F(4)=3, F(10)=55 and F(16)=987 (F(0)=0 and F(1)=1). See Lucas sequence A000032. This is a special case where a(0) and a(1) are increasing Fibonacci numbers and Lucas(m)*a(1) +- a(0) is another Fibonacci. - Bob Selcoe, Jul 08 2013
a(n) = x + y where x and y are solutions to x^2 = 5*y^2 - 1. (See related sequences with formula below.) - Richard R. Forberg, Sep 05 2013

Colin Barker, Table of n, a(n) for n = 0..750
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Subsequence of A033887.
Cf. A000032, A000045, A001906, A001519, A015448, A014445, A033888, A033889, A033890, A033891, A049310, A049660, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.
Cf. A103135.

[Fibonacci(6*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[6n+4], {n, 0, 30}]
LinearRecurrence[{18,-1},{3,55},20] (* Harvey P. Dale, Mar 29 2023 *)
Table[ChebyshevU[3*n+1, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+4) \\ Charles R Greathouse IV, Feb 05 2013

G.f.: (x+3)/(x^2-18*x+1).
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=3, a(1)=55. - Philippe Deléham, Nov 17 2008
a(n) = A007805(n) + A075796(n), as follows from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((15-7*sqrt(5)+(9+4*sqrt(5))^(2*n)*(15+7*sqrt(5))))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n+1, 3) = 3*S(n,18) + S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Edited by N. J. A. Sloane, Aug 10 2010

A110679 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.

Original entry on oeis.org

1, 2, 11, 44, 189, 798, 3383, 14328, 60697, 257114, 1089155, 4613732, 19544085, 82790070, 350704367, 1485607536, 6293134513, 26658145586, 112925716859, 478361013020, 2026369768941, 8583840088782, 36361730124071, 154030760585064, 652484772464329

Offset: 0

Creighton Dement, Aug 02 2005

2tesseq[A*B*cyc(A)] (see program code) gives an alternative formula for A110528.

a(n) is the number of tilings of a 2 X n rectangle by using 1 X 1 squares, dominoes and right trominoes. - Roberto Tauraso, Mar 21 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3,5,1).

Cf. A110528, A110680, A001076, A110526, A110526, A033887.

[(Fibonacci(3*n+2) +(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Apr 19 2019

seriestolist(series((-1+x)/((x+1)*(x^2+4*x-1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: -1jesseq[A*B*cyc(A)] with A = - 'j + 'k - 'ii' - 'ij' - 'ik' and B = - .5'i - .5i' - .5'ii' + .5'jj' - .5'kk' + .5'jk' + .5'kj' - .5e

a[n_] := (Fibonacci[3*n+2] + (-1)^n)/2; a /@ Range[0, 22] (* Giovanni Resta, Mar 21 2017 *)

Vec((1 - x) / ((1 + x)*(1 - 4*x - x^2)) + O(x^30)) \\ Colin Barker, Mar 21 2017

{a(n) = -(-1)^n * (fibonacci(-2 - 3*n)\2)}; /* Michael Somos, Mar 26 2017 */

[(fibonacci(3*n+2) +(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Apr 19 2019

Program "FAMP" finds: 2*(-1^(n+1)) = A110528(n) - A001076(n+1) - 2*a(n). Program "Superseeker" finds: a(n) = A110526(n+1) - A110526(n); a(n) + a(n+1) = A033887(n+1).
a(n) = (-1)^n*Sum_{k=0..n} (-1)^k*Fibonacci(3*k+1). - Gary Detlefs, Jan 22 2013
a(n) = (Fibonacci(3*n+2)+(-1)^n)/2. - Roberto Tauraso, Mar 21 2017
From Colin Barker, Mar 21 2017: (Start)
G.f.: (1 - x) / ((1 + x)*(1 - 4*x - x^2)).
a(n) = 3*a(n-1) + 5*a(n-2) + a(n-3) for n>2.
(End)
a(n) = -(-1)^n * A049651(-1 - n) for all n in Z. - Michael Somos, Mar 26 2017
a(2*n) = A254627(2*n+1); a(2*n+1) = A077259(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
2*a(n) = A015448(n+1)+(-1)^n. - R. J. Mathar, Oct 03 2021

Typo in program code fixed by Creighton Dement, Dec 11 2009

A135597 Square array read by antidiagonals: row m (m >= 1) satisfies b(0) = b(1) = 1; b(n) = m*b(n-1) + b(n-2).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 5, 1, 1, 5, 13, 17, 8, 1, 1, 6, 21, 43, 41, 13, 1, 1, 7, 31, 89, 142, 99, 21, 1, 1, 8, 43, 161, 377, 469, 239, 34, 1, 1, 9, 57, 265, 836, 1597, 1549, 577, 55, 1, 1, 10, 73, 407, 1633, 4341, 6765, 5116, 1393, 89, 1, 1, 11, 91, 593, 2906

Offset: 1

N. J. A. Sloane, Mar 02 2008

For n > 1, the number of independent vertex sets in the graph K_m X P_{n-1}. For example, in K_3 X P_1 there are 4 independent vertex sets. - Andrew Howroyd, May 23 2017

			Array begins:
========================================================
m\n| 0 1 2  3   4    5     6      7       8        9
---|----------------------------------------------------
1  | 1 1 2  3   5    8    13     21      34       55 ...
2  | 1 1 3  7  17   41    99    239     577     1393 ...
3  | 1 1 4 13  43  142   469   1549    5116    16897 ...
4  | 1 1 5 21  89  377  1597   6765   28657   121393 ...
5  | 1 1 6 31 161  836  4341  22541  117046   607771 ...
6  | 1 1 7 43 265 1633 10063  62011  382129  2354785 ...
7  | 1 1 8 57 407 2906 20749 148149 1057792  7552693 ...
8  | 1 1 9 73 593 4817 39129 317849 2581921 20973217 ...
...

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Cf. A121875, A000045, A287376.

Rows 2-11 are A001333, A003688, A015448, A015449, A015451, A015453-A015457.

A135597 := proc(m,c) coeftayl( (m*x-x-1)/(x^2+m*x-1),x=0,c) ; end: for d from 1 to 15 do for c from 0 to d-1 do printf("%d,",A135597(d-c,c)) ; od: od: # R. J. Mathar, Apr 21 2008

a[, 0] = a[, 1] = 1; a[m_, n_] := m*a[m, n-1] + a[m, n-2]; Table[a[m-n+1, n], {m, 0, 11}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)

O.g.f. row m: (mx-x-1)/(x^2+mx-1). - R. J. Mathar, Apr 21 2008

More terms from R. J. Mathar, Apr 21 2008

A167808 Numerator of x(n), where x(n) = x(n-1) + x(n-2) with x(0)=0, x(1)=1/2.

Original entry on oeis.org

0, 1, 1, 1, 3, 5, 4, 13, 21, 17, 55, 89, 72, 233, 377, 305, 987, 1597, 1292, 4181, 6765, 5473, 17711, 28657, 23184, 75025, 121393, 98209, 317811, 514229, 416020, 1346269, 2178309, 1762289, 5702887, 9227465, 7465176, 24157817, 39088169, 31622993

Offset: 0

Reinhard Zumkeller, Nov 12 2009

Define a sequence c(n) by c(0)=0, c(1)=1; thereafter c(n) = (c(n-2)*c(n-1)-1)/(c(n-2)+c(n-1)+2). Then it appears that (apart from signs), a(n) is the denominator of c(n). - Jonas Holmvall, Jun 21 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,1).

Cf. A000045, A130196 (denominator).

The a(2*n) appear in A179135. - Johannes W. Meijer, Jul 01 2010

a:=[0,1,1,1,3,5];; for n in [7..40] do a[n]:=4*a[n-3]+a[n-6]; od; a; # Muniru A Asiru, Oct 16 2018

nmax:=39; x(0):=0: x(1):=1/2:for n from 2 to nmax do x(n):=x(n-1)+x(n-2) od: for n from 0 to nmax do a(n):= numer(x(n)) od: seq(a(n),n=0..nmax); # Johannes W. Meijer, Jul 01 2010
with(combinat):f:=n->fibonacci(n):L:=n->f(n)+2*f(n-1):seq(numer(f(n)/L(n)), n=0..39); # Gary Detlefs, Dec 11 2010

f[n_]:=Numerator[Fibonacci[n]/Fibonacci[n+3]];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011*)
Numerator[LinearRecurrence[{1,1},{0,1/2},40]] (* Harvey P. Dale, Aug 08 2014 *)
CoefficientList[Series[-x (1 + x + x^2 - x^3 + x^4)/((x^2 + x - 1) (x^4 - x^3 + 2 x^2 + x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 08 2014 *)
LinearRecurrence[{0, 0, 4, 0, 0, 1},{0, 1, 1, 1, 3, 5},40] (* Ray Chandler, Aug 03 2015 *)
a[n_]:=If[Mod[n,3]==0, Fibonacci[n]/2, Fibonacci[n]]; Array[a, 40, 0] (* Stefano Spezia, Oct 16 2018 *)

a(n) = (a(n-1)*A131534(n) + a(n-2)*A131534(n+2))/A131534(n+1) for n > 1.
a(3*n) = A001076(n) = (a(3*n-1) + a(3*n-2))/2;
a(3*n+1) = A033887(n) = 2*a(3*n-1) + a(3*n-2);
a(3*n+2) = A015448(n+1) = a(3*n-1) + 2*a(3*n-2).
From Johannes W. Meijer, Jul 01 2010: (Start)
a(2*n) = A001906(n)/A131534(n+1) for n >= 0 and a(2*n+1) = A179131(n)/5 for n >= 1.
a(6*n+2) - 2*a(6*n) = A134493(n);
2*a(6*n+1) - a(6*n+3) = A023039(n);
2*a(6*n+4) - a(6*n+2) = A134497(n);
a(6*n+5) - 2*a(6*n+3) = A103134(n);
2*a(6*n+4) - a(6*n+6) = A075796(n).
(End)
From Gary Detlefs, Dec 11 2010: (Start)
a(n) = numerator(A000045(n)/A000032(n)).
If n mod 3 = 0 then a(n) = Fibonacci(n)/2, else a(n)= Fibonacci(n). (End)
G.f.: -x*(1 + x + x^2 - x^3 + x^4) / ( (x^2 + x - 1)*(x^4 - x^3 + 2*x^2 + x + 1) ). - R. J. Mathar, Mar 08 2011
a(n) = 4*a(n-3) + a(n-6). - Muniru A Asiru, Oct 16 2018

Typo in title corrected by Johannes W. Meijer, Jun 26 2010

cp's OEIS Frontend

A287819 Number of nonary sequences of length n such that no two consecutive terms have distance 4.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

Extensions

A287825 Number of sequences over the alphabet {0,1,...,9} such that no two consecutive terms have distance 1.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Formula

A048876 a(n) = 4*a(n-1) + a(n-2); a(0)=1, a(1)=7.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A048875 Generalized Pellian with second term of 6.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

PARI

Formula

Extensions

A103134 a(n) = Fibonacci(6n+4).

Views

Author

Keywords

Comments

A110679 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.