A001911 -id:A001911 - cp's OEIS frontend

A180673 a(n) = Fibonacci(n+8) - Fibonacci(8).

0, 13, 34, 68, 123, 212, 356, 589, 966, 1576, 2563, 4160, 6744, 10925, 17690, 28636, 46347, 75004, 121372, 196397, 317790, 514208, 832019, 1346248, 2178288, 3524557, 5702866, 9227444, 14930331, 24157796, 39088148, 63245965, 102334134

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) (terms doubled) are the Kn17 sums of the Fibonacci(n) triangle A104763. See A180662 for information about these knight and other chess sums.

Vincenzo Librandi, Table of n, a(n) for n = 0..275
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A000071.

Cf. A131524 (Kn11), A001911 (Kn12), A006327 (Kn13), A167616 (Kn14), A180671 (Kn15), A180672 (Kn16), A180673 (Kn17), A180674 (Kn18).

List([0..40], n-> Fibonacci(n+8)-21); # G. C. Greubel, Jul 13 2019

[Fibonacci(n+8) - Fibonacci(8): n in [0..40]]; // Vincenzo Librandi, Apr 24 2011

nmax:=40: with(combinat): for n from 0 to nmax do a(n):=fibonacci(n+8)-fibonacci(8) od: seq(a(n),n=0..nmax);

Fibonacci[8 +Range[0, 40]] -21 (* G. C. Greubel, Jul 13 2019 *)

concat(0, Vec(x*(13+8*x)/((1-x)*(1-x-x^2)) + O(x^40))) \\ Colin Barker, Feb 24 2017

a(n)=fibonacci(n+8)-21 \\ Charles R Greathouse IV, Feb 24 2017

[fibonacci(n+8)-21 for n in (0..40)] # G. C. Greubel, Jul 13 2019

a(n) = F(n+8) - F(8) with F(n) the Fibonacci numbers A000045.
a(n) = a(n-1) + a(n-2) + 21 for n>1, a(0)=0, a(1)=13, and where 21 = F(8).
G.f.: x*(13 + 8*x)/((1 - x)*(1 - x - x^2)). - Ilya Gutkovskiy, Feb 24 2017
a(n) = 13*A000071(n+2) + 8*A000071(n+1). - Bruno Berselli, Feb 24 2017
From Colin Barker, Feb 24 2017: (Start)
a(n) = (-21 + (2^(-1-n)*((1-sqrt(5))^n*(-47+21*sqrt(5)) + (1+sqrt(5))^n*(47+21*sqrt(5)))) / sqrt(5)).
a(n) = 2*a(n-1) - a(n-3) for n>2. (End)

A180674 a(n) = Fibonacci(n+9) - Fibonacci(9).

Original entry on oeis.org

0, 21, 55, 110, 199, 343, 576, 953, 1563, 2550, 4147, 6731, 10912, 17677, 28623, 46334, 74991, 121359, 196384, 317777, 514195, 832006, 1346235, 2178275, 3524544, 5702853, 9227431, 14930318, 24157783, 39088135, 63245952, 102334121

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) (terms doubled) are the Kn18 sums of the Fibonacci(n) triangle A104763. See A180662 for information about these knight and other chess sums.

Vincenzo Librandi, Table of n, a(n) for n = 0..275
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A000071.

Cf. A131524 (Kn11), A001911 (Kn12), A006327 (Kn13), A167616 (Kn14), A180671 (Kn15), A180672 (Kn16), A180673 (Kn17), A180674 (Kn18).

List([0..40], n-> Fibonacci(n+9)-34); # G. C. Greubel, Jul 13 2019

[Fibonacci(n+9) - Fibonacci(9): n in [0..40]]; // Vincenzo Librandi, Apr 24 2011

nmax:=31: with(combinat): for n from 0 to nmax do a(n):=fibonacci(n+9)-fibonacci(9) od: seq(a(n),n=0..nmax);

Fibonacci[9 +Range[0, 40]] -34 (* G. C. Greubel, Jul 13 2019 *)
LinearRecurrence[{2,0,-1},{0,21,55},40] (* Harvey P. Dale, Aug 24 2024 *)

concat(0, Vec(x*(21+13*x)/((1-x)*(1-x-x^2)) + O(x^40))) \\ Colin Barker, Feb 24 2017

a(n) = fibonacci(n+9) - fibonacci(9) \\ Charles R Greathouse IV, Feb 24 2017

[fibonacci(n+9)-34 for n in (0..40)] # G. C. Greubel, Jul 13 2019

a(n) = F(n+9) - F(9) with F = A000045.
a(n) = a(n-1) + a(n-2) + 34 for n>1, a(0)=0, a(1)=21, and where 34 = F(9).
G.f.: x*(21 + 13*x)/((1 - x)*(1 - x - x^2)). - Ilya Gutkovskiy, Feb 24 2017
a(n) = 21*A000071(n+2) + 13*A000071(n+1). - Bruno Berselli, Feb 24 2017
From Colin Barker, Feb 24 2017: (Start)
a(n) = (-34 + (2^(-n)*((1-sqrt(5))^n*(-38+17*sqrt(5)) + (1+sqrt(5))^n*(38+17*sqrt(5)))) / sqrt(5)).
a(n) = 2*a(n-1) - a(n-3) for n>2. (End)

A232580 Number of binary sequences of length n that contain at least one contiguous subsequence 011.

Original entry on oeis.org

0, 0, 0, 1, 4, 12, 31, 74, 168, 369, 792, 1672, 3487, 7206, 14788, 30185, 61356, 124308, 251199, 506578, 1019920, 2050785, 4119280, 8267216, 16580799, 33236622, 66594636, 133385689, 267089188, 534692604, 1070217247, 2141780762, 4285739832, 8575004241

Offset: 0

Geoffrey Critzer, Nov 26 2013

From Gus Wiseman, Jun 26 2022: (Start)
Also the number of integer compositions of n + 1 with an even part other than the first or last. For example, the a(3) = 1 through a(5) = 12 compositions are:
  (121)  (122)   (123)
         (221)   (141)
         (1121)  (222)
         (1211)  (321)
                 (1122)
                 (1212)
                 (1221)
                 (2121)
                 (2211)
                 (11121)
                 (11211)
                 (12111)
The odd version is A274230.
(End)

			a(4) = 4 because we have: 0011, 0110, 0111, 1011.

Colin Barker, Table of n, a(n) for n = 0..1000
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, page 34.
Index entries for linear recurrences with constant coefficients, signature (4,-4,-1,2).

The complement is counted by A000071(n) = A001911(n) + 1.
For the contiguous pattern (1,1) or (0,0) we have A000225.
For the contiguous pattern (1,0,1) or (0,1,0) we have A000253.
For the contiguous pattern (1,0) or (0,1) we have A000295.
Numbers whose binary expansion is of this type are A004750.
For the contiguous pattern (1,1,1) or (0,0,0) we have A050231.
The not necessarily contiguous version is A324172.
Cf. A034691, A261983, A274230, A335455, A335457, A335458, A335516.

nn=40;a=x/(1-x);CoefficientList[Series[a^2 x/(1-a x)/(1-2x),{x,0,nn}],x]
(* second program *)
Table[Length[Select[Tuples[{0,1},n],MatchQ[#,{_,0,1,1,_}]&]],{n,0,10}] (* Gus Wiseman, Jun 26 2022 *)

concat(vector(3), Vec(x^3/(-2*x^4+x^3+4*x^2-4*x+1) + O(x^40))) \\ Colin Barker, Nov 03 2016

O.g.f.: x^3/( (1-x)^2*(1-x^2/(1-x))*(1-2x) ).
a(n) ~ 2^n.
From Colin Barker, Nov 03 2016: (Start)
a(n) = (1 + 2^n - (2^(-n)*((1-sqrt(5))^n*(-2+sqrt(5)) + (1+sqrt(5))^n*(2+sqrt(5))))/sqrt(5)).
a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4) for n > 3. (End)
a(n) = 2^n - Fibonacci(n+3) + 1. - Ehren Metcalfe, Dec 27 2018
E.g.f.: 2*exp(x/2)*(5*exp(x)*cosh(x/2) - 5*cosh(sqrt(5)*x/2) - 2*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Apr 06 2022

A104004 Expansion of (1-x) * (1+x) / ((1-2x)(1-x-x^2)).

Original entry on oeis.org

1, 3, 7, 16, 35, 75, 158, 329, 679, 1392, 2839, 5767, 11678, 23589, 47555, 95720, 192427, 386451, 775486, 1555153, 3117071, 6245088, 12507887, 25044431, 50135230, 100345485, 200812363, 401821144, 803960099, 1608434427, 3217700894, 6436748057

Offset: 0

Creighton Dement, Feb 24 2005

A floretion-generated sequence relating to Fibonacci numbers and powers of 2. The sequence results from a particular transform of the sequence A000079*(-1)^n (powers of 2).

Floretion Algebra Multiplication Program, FAMP Code: 1jesforseq[ ( 5'i + .5i' + .5'ii' + .5e)*( + .5j' + .5'kk' + .5'ki' + .5e ) ], 1vesforseq = A000079(n+1)*(-1)^(n+1), ForType: 1A. Identity used: jesfor = jesrightfor + jesleftfor

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2).

Cf. A000045, A000079, A001911, A008466, A016777, A022958, A027934, A042950, A078024, A099036, A221719.

[3*2^n-Fibonacci(n+3): n in [0..40]]; // Vincenzo Librandi, Aug 18 2017

with (combinat):a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=fibonacci(n-1)+2*a[n-1] od: seq(a[n], n=1..26); # Zerinvary Lajos, Mar 17 2008

LinearRecurrence[{3, -1, -2}, {1, 3, 7}, 80] (* Vincenzo Librandi, Aug 18 2017 *)
CoefficientList[Series[(1-x)(1+x)/((2x-1)(x^2+x-1)),{x,0,40}],x] (* Harvey P. Dale, Oct 12 2024 *)
A104004[n_]:= 3*2^n -Fibonacci[n+3]; (* G. C. Greubel, Jun 05 2025 *)

def A104004(n): return 3*2**n - fibonacci(n+3) # G. C. Greubel, Jun 05 2025

4*a(n) = A008466(n+3) + A027973(n) (FAMP result).
Suggestions made by Superseeker: (Start)
a(n+2) - a(n+1) - a(n) = A042950(n+1).
Coefficients of g.f.*(1-x)/(1+x) match A099036.
Coefficients of g.f./(1+x) match A027934.
Coefficients of g.f./(1-x^2) match A008466. (End)
a(n) = A101220(3, 2, n+1) - A101220(3, 2, n). - Ross La Haye, Aug 05 2005
a(n) = 3*2^n - Fibonacci(n+3) = A221719(n) + 1. - Ralf Stephan, May 20 2007, Hugo Pfoertner, Mar 06 2024
a(n) = (3*2^n - (2^(-n)*((1-sqrt(5))^n*(-2+sqrt(5)) + (1+sqrt(5))^n*(2+sqrt(5)))) / sqrt(5)). - Colin Barker, Aug 18 2017
From G. C. Greubel, Jun 05 2025: (Start)
Sum_{k=0..n} A022958(k+1)*a(n-k) = A001911(n+1).
Sum_{k=0..n} (-1)^k*A016777(k)*a(n-k) = A078024(n).
E.g.f.: 3*exp(2*x) - (2/sqrt(5))*exp(x/2)*( 2*sinh(sqrt(5)*x/2) + sqrt(5)*cosh(sqrt(5)*x/2) ). (End)

A105438 Triangle, row sums = (Fibonacci numbers - 2).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 4, 2, 1, 5, 6, 5, 2, 1, 6, 9, 8, 6, 2, 1, 7, 12, 14, 10, 7, 2, 1, 8, 16, 20, 20, 12, 8, 2, 1, 9, 20, 30, 30, 27, 14, 9, 2, 1, 10, 25, 40, 50, 42, 35, 16, 10, 2, 1, 11, 30, 55, 70, 77, 56, 44, 18, 11, 2, 1

Offset: 0

Gary W. Adamson, Apr 09 2005

Row sums = 1, 3, 6, 11, 19, 32, 53...(Fibonacci numbers - 2; starting with F(4)) The first few rows of the triangle are:
Row sums = (Fibonacci numbers - 2; starting 1, 3, 6...).
Column 1 = A002620; Column 2 = A006918; Column 3 = A096338.
Inverse array is A105522. - Paul Barry, Apr 11 2005
Diagonal sums are A027383(n). - Philippe Deléham, Jan 16 2014

			Column 2: 1, 2, 5, 8, 14, 20, 30...is generated by using the partial sum operator on 1, 1, 3, 3, 6, 6, 10, 10...
The first few rows of the triangle are:
  1;
  2, 1;
  3, 2, 1;
  4, 4, 2, 1;
  5, 6, 5, 2, 1;
  6, 9, 8, 6, 2, 1;
  7, 12, 14, 10, 7, 2, 1;
  8, 16, 20, 20, 12, 8, 2, 1;
  9, 20, 30, 30, 27, 14, 9, 2, 1;
  10, 25, 40, 50, 42, 35, 16, 10, 2, 1;
  ...

T. Mansour and A. O. Munagi, Alternating subsets modulo m, Rocky Mt. J. Math. 42, No. 4, 1313-1325 (2012), eq. (2)

Cf. A001911, A002620, A006918, A096338, A105108.

By columns (k = 0, 1, 2...); use partial sum operator on (bin(n, k) numbers repeated).
T(n,k) = Sum_{j=0..n-k} C((j+2k)/2, k)*(1+(-1)^j)+C((j-1+2k)/2, k)*(1-(-1)^j)/2; Riordan array (1/(1-x)^2, x/(1-x^2)). - Paul Barry, Apr 11 2005
T(n,k) = T(n-1,k)+T(n-1,k-1)+T(n-2,k)-T(n-2,k-1)-T(n-3,k), T(0,0)=1, T(1,0)=2, T(1,1)= 1, T(n,k)= 0 if k<0 or if k>n. - Philippe Deléham, Jan 16 2014

A181631 Triangle by rows, number of leading 1's in the maximal Fibonacci representation (A104326).

Original entry on oeis.org

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

Offset: 1

Gary W. Adamson, Nov 02 2010

Row sums = A001911: (1, 3, 6, 11, 19, 32, ...).
A112310 = number of 1's in the maximal Fibonacci representation, which has headings of (..., 8, 5, 3, 2, 1) filling entries from the right to left; as opposed to the minimal Fibonacci representation (A014417) which starts from the left. For example, 8 in maximal = 1011 = (5 + 2 + 1) whereas in minimal = (10000) = (8).
Rows have (1, 2, 3, 5, 8, ...) terms.

			First few rows of the triangle:
  1;
  1, 2;
  1, 2, 3;
  1, 1, 2, 3, 4;
  1, 1, 1, 2, 2, 3, 4, 5;
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 6;
  ...
Example: a(14) = 1 since 14 in the maximal Fibonacci representation is 10111.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000045 (row lengths), A003754, A001911 (row sums), A014417, A090996, A104326, A112310.

f[s_] := Module[{i = FirstPosition[s, 0]}, If[MissingQ[i], Length[s], i[[1]] - 1]]; f /@ Select[IntegerDigits[#, 2] & /@ Range[300], SequencePosition[#, {0, 0}] == {} &] (* Amiram Eldar, May 31 2025 *)

a(n) = A090996(A003754(n+1)). - Amiram Eldar, May 31 2025

More terms from Amiram Eldar, May 31 2025

A054469 a(n) = a(n-1) + a(n-2) + (n+2)*binomial(n+3, 3)/2, with a(0) = 1, a(1) = 7.

Original entry on oeis.org

1, 7, 28, 85, 218, 499, 1053, 2092, 3970, 7272, 12958, 22596, 38739, 65535, 109714, 182185, 300620, 493635, 807555, 1317360, 2144396, 3485032, 5657028, 9174560, 14869613, 24088399, 39009168, 63156437, 102233030, 165466347, 267786673

Offset: 0

Barry E. Williams, Mar 31 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
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 (6,-14,15,-5,-4,4,-1).

Cf. A000045, A001891, A001911.

Right-hand column 11 of triangle A011794.

A054469:= func< n | Fibonacci(n+12) -(1/12)*(1716 +802*n +173*n^2 +20*n^3 +n^4) >;
[A054469(n): n in [0..40]]; // G. C. Greubel, Oct 21 2024

RecurrenceTable[{a[0]==1,a[1]==7,a[n]==a[n-1]+a[n-2]+(n+2) Binomial[ n+3,3]/2},a,{n,30}] (* Harvey P. Dale, Sep 22 2013 *)
CoefficientList[Series[(1+x)/((1-x)^5*(1-x-x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 23 2013 *)

a(n) = sum(i=1,(n+2)\2,binomial(n+5-i,n+2-2*i))+2*sum(i=1,(n+1)\2,binomial(n+5-i,n+1-2*i)) \\ Jason Yuen, Aug 13 2024

def A054469(n): return fibonacci(n+12) - (1716 + 802*n + 173*n^2 + 20*n^3 + n^4)//12
[A054469(n) for n in range(41)] # G. C. Greubel, Oct 21 2024

a(n) = a(n-1) + a(n-2) + (n+1)*(n+2)^2*(n+3)/12.
a(-n) = 0.
a(n) = (Sum_{i=1..floor((n+2)/2)} binomial(n+5-i, n+2-2*i)) + 2*(Sum_{i=1..floor((n+1)/2)} binomial(n+5-i, n+1-2*i)).
G.f.: (1+x) / ((1-x)^5*(1-x-x^2)). - Colin Barker, Jun 11 2013
From G. C. Greubel, Oct 21 2024: (Start)
a(n) = Fibonacci(n+12) - Sum_{j=0..4} Fibonacci(11-2*j) * binomial(n+j, j).
a(n) = Fibonacci(n+12) - (1/12)*(1716 + 802*n + 173*n^2 + 20*n^3 + n^4). (End)

A129713 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and starting with exactly k 1's (0<=k<=n). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 5, 3, 2, 1, 1, 1, 8, 5, 3, 2, 1, 1, 1, 13, 8, 5, 3, 2, 1, 1, 1, 21, 13, 8, 5, 3, 2, 1, 1, 1, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 233, 144

Offset: 0

Emeric Deutsch, May 12 2007

Row sums are the Fibonacci numbers (A000045). Sum(k*T(n,k), 0<=k<=n) = F(n+3)-2 = A001911(n).

			T(6,2) = 3 because we have 110110, 110111, 110101.
Triangle starts:
1;
1,1;
1,1,1;
2,1,1,1;
3,2,1,1,1;
5,3,2,1,1,1;
8,5,3,2,1,1,1;

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A000045, A001911.
Cf. A054123.
Cf. A007298. - Altug Alkan, May 03 2016

a129713 n k = a129713_tabl !! n !! k
a129713_row n = a129713_tabl !! n
a129713_tabl = [1] : [1, 1] : f [1] [1, 1] where
   f us vs = ws : f vs ws where
             ws = zipWith (+) (init us ++ [0, 0, 0]) (vs ++ [1])
-- Reinhard Zumkeller, May 26 2015

with(combinat): T:=proc(n,k) if k<=n-2 then fibonacci(n-k) elif k=n-1 or k=n then 1 else 0 fi end: for n from 0 to 15 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

nn=15;a=1/(1-y x);b=1/(1-x);Map[Select[#,#>0&]&,CoefficientList[Series[a (1+x)/(1-x^2b),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Dec 04 2013 *)

T(n,k) = F(n-k) if k<=n-2, T(n,n-1) = T(n,n) = 1, where F(j) are the Fibonacci numbers (F(0)=0, F(1)=1). G.f.: G(t,z) = (1-z^2)/[(1-z-z^2)(1-tz)].

a(n) = A007298(n+4) - A007298(n+3). - Altug Alkan, May 03 2016

A166472 a(n) = 2^F(n+2)*3^F(n+1)/12, where F(n) is the n-th Fibonacci number (A000045(n)).

Original entry on oeis.org

1, 6, 72, 5184, 4478976, 278628139008, 14975624970497949696, 50071566192138943522512952098816, 8998235963747242817865410245394871488270255869919232

Offset: 1

Matthew Vandermast, Nov 05 2009

nonn

G. C. Greubel, Table of n, a(n) for n = 1..15

Cf. A000045, A001911, A002110, A166469, A166470, A166471, A166473.

Subsequence of A003586, A025487.

[2^(Fibonacci(n+2)-2)*3^(Fibonacci(n+1)-1): n in [1..12]]; // G. C. Greubel, Jul 30 2024

Table[(2^Fibonacci[n+2]*3^Fibonacci[n+1])/12, {n,12}] (* G. C. Greubel, May 15 2016 *)
(3^#[[1]] 2^#[[2]])/12&/@Partition[Fibonacci[Range[2,15]],2,1] (* Harvey P. Dale, Jul 12 2021 *)

[2^(fibonacci(n+2)-2)*3^(fibonacci(n+1)-1) for n in range(1,13)] # G. C. Greubel, Jul 30 2024

a(n) = A166470(n+1)/12.
a(n) = 12*a(n-1)*a(n-2), for n > 1, with a(0) = 1/2, a(1) = 1.
A166469(A002110(m)*a(n)) = Fibonacci(m+n+1), for m > 1.
A166469(a(n)) = Fibonacci(n+3) - 2 = A001911(n).

A168193 a(n) = a(n-1) + a(n-2) + 4, with a(0)=0, a(1)=2.

Original entry on oeis.org

0, 2, 6, 12, 22, 38, 64, 106, 174, 284, 462, 750, 1216, 1970, 3190, 5164, 8358, 13526, 21888, 35418, 57310, 92732, 150046, 242782, 392832, 635618, 1028454, 1664076, 2692534, 4356614, 7049152, 11405770, 18454926, 29860700, 48315630, 78176334, 126491968

Offset: 0

Geoff Ahiakwo, Nov 19 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

I:=[0,2,6]; [n le 3 select I[n] else Self(n-1)+Self(n-2)+4: n in [1..40]]; // Vincenzo Librandi, Jul 16 2016

Fibonacci[Range[3,4! ]]*2-4 (* Vladimir Joseph Stephan Orlovsky, Mar 19 2010 *)
LinearRecurrence[{2, 0, -1}, {0, 2, 6}, 50] (* G. C. Greubel, Jul 15 2016 *)

From R. J. Mathar, Nov 22 2009: (Start)
a(n)= 2*a(n-1) - a(n-3) = 2*A001911(n).
G.f.: 2*x*(1+x)/((x-1)*(x^2+x-1)). (End)
a(n) = a(n-1) + 2*Fibonacci(n+1), with a(0)=0. - Taras Goy, Mar 24 2019
E.g.f.: 4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 2*sqrt(5)*sinh(sqrt(5)*x/2))/5 - 4*exp(x). - Stefano Spezia, Oct 14 2022
a(n) = A019274(n+1)+A019274(n+2). - R. J. Mathar, Jul 07 2023

Definition replaced by recurrence from R. J. Mathar, Nov 23 2009

cp's OEIS Frontend

A180673 a(n) = Fibonacci(n+8) - Fibonacci(8).

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A180674 a(n) = Fibonacci(n+9) - Fibonacci(9).

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A232580 Number of binary sequences of length n that contain at least one contiguous subsequence 011.

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A104004 Expansion of (1-x) * (1+x) / ((1-2*x)*(1-x-x^2)).

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A105438 Triangle, row sums = (Fibonacci numbers - 2).

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A181631 Triangle by rows, number of leading 1's in the maximal Fibonacci representation (A104326).

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A054469 a(n) = a(n-1) + a(n-2) + (n+2)*binomial(n+3, 3)/2, with a(0) = 1, a(1) = 7.

A104004 Expansion of (1-x) * (1+x) / ((1-2x)(1-x-x^2)).