A052952 -id:A052952 - cp's OEIS frontend

A214656 Floor of the imaginary part of the zeros of the complex Fibonacci function on the left half-plane.

0, 0, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 10, 10, 11, 11, 12, 12, 13, 14, 14, 15, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 24, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 38, 38, 39

Offset: 0

Wolfdieter Lang, Jul 24 2012

nonn

See the comment on the Fibonacci Function F(z) and its zeros in A214315, where also the T. Koshy reference is given.

The imaginary part of the zeros, corresponding to the real part x_0(k) given in A214315, is y_0(k) = -b*k, with b = 4*Pi*log(phi)/(Pi^2 + (2*log(phi))^2) and phi = (1+sqrt(5))/2. Note that b is approximately 0.5601299084.

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A052952 (Fibonacci related formula), A214315 (real part).

R:= RealField(100); [Floor(4*n*Pi(R)*Log((1+Sqrt(5))/2)/(Pi(R)^2 + (2*Log((1+Sqrt(5))/2))^2)) : n in [0..100]]; // G. C. Greubel, Mar 09 2024

a[n_]:= Floor[4*n*Pi*Log[GoldenRatio]/(Pi^2 + 4*Log[GoldenRatio]^2)];
Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Jul 03 2013 *)

A214656(n,phi=(sqrt(5)+1)/2)=n*4*Pi*log(phi)\(Pi^2+(2*log(phi))^2)  \\ M. F. Hasler, Jul 24 2012

[floor(4*n*pi*log(golden_ratio)/(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024

a(n) = floor(b*n), n>=0, with b = -y_0(1) = 4*Pi*log(phi)/(Pi^2 + (2*log(phi))^2).

A099571 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+3, k).

Original entry on oeis.org

1, 1, 5, 6, 17, 23, 50, 73, 138, 211, 370, 581, 979, 1560, 2575, 4135, 6755, 10890, 17700, 28590, 46356, 74946, 121380, 196326, 317797, 514123, 832025, 1346148, 2178293, 3524441, 5702870, 9227311, 14930334, 24157645, 39088150, 63245795

Offset: 0

Paul Barry, Oct 23 2004

Fourth column of triangle A054450.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,4,-3,-6,3,4,-1,-1).

Cf. A000045, A052952, A054450, A054451, A099572.

[Fibonacci(n+4) +(-1)^n*(n^2+4*n+7)/16 -(n^2+12*n+39)/16: n in [0..40]]; // G. C. Greubel, Jul 25 2022

Table[Sum[Binomial[n-k+3,k],{k,0,Floor[n/2]}],{n,0,40}] (* or *) LinearRecurrence[{1,4,-3,-6,3,4,-1,-1},{1,1,5,6,17,23,50,73},40] (* Harvey P. Dale, Jun 04 2021 *)

[fibonacci(n+4) +(-1)^n*(n^2+4*n+7)/16 -(n^2+12*n+39)/16 for n in (0..40)] # G. C. Greubel, Jul 25 2022

G.f.: 1/((1-x^2)^3*(1-x-x^2)).
a(n) = a(n-1) + 4*a(n-2) - 3*a(n-3) - 6*a(n-4) + 3*a(n-5) + 4*a(n-6) - a(n-7) - a(n-8);
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2 +2, 2)*(1+(-1)^k)/2.
a(n) = Fibonacci(n+4) + (-1)^n*(n^2 + 4*n + 7)/16 - (n^2 + 12*n + 39)/16. - G. C. Greubel, Jul 25 2022

A124502 a(1)=a(2)=1; thereafter, a(n+1) = a(n) + a(n-1) + 1 if n is a multiple of 5, otherwise a(n+1) = a(n) + a(n-1).

Original entry on oeis.org

1, 1, 2, 3, 5, 9, 14, 23, 37, 60, 98, 158, 256, 414, 670, 1085, 1755, 2840, 4595, 7435, 12031, 19466, 31497, 50963, 82460, 133424, 215884, 349308, 565192, 914500, 1479693, 2394193, 3873886, 6268079, 10141965, 16410045, 26552010, 42962055, 69514065, 112476120

Offset: 1

N. J. A. Sloane, May 25 2008

nonn

If we split this sequence into 5 separate sequences of n mod 5, each individual sequence is of the form a(n) = 12*a(n-1) - 10*a(n-2) - a(n-3). For example, 12*98 - 10*9 - 1 = 1085. This is the same recurrence exhibited in A138134 and the n mod 5 =0 sequence...5, 60, 670, 7435 is A138134.

			a(6) = a(5) + a(4) + 1 = 5 + 3 + 1 = 9 because n=5 is a multiple of 5.
a(7) = a(6) + a(5) = 9 + 5 = 14 because n=6 is not a multiple of 5.

Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,1,-1,-1).

Cf. A052952, A004695, A080239, A131132.

A124502:=proc(n) option remember; local t1; if n <= 2 then return 1; fi: if n mod 5 = 1 then t1:=1 else t1:=0; fi: procname(n-1)+procname(n-2)+t1; end proc; [seq(A124502(n), n=1..100)]; # N. J. A. Sloane, May 25 2008

a=0; b=0; lst={a,b}; Do[z=a+b+1; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z; z=a+b; AppendTo[lst,z]; a=b; b=z,{n,4!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 16 2010 *)
nxt[{n_,a_,b_}]:={n+1,b,If[Divisible[n,5],a+b+1,a+b]}; NestList[nxt,{2,1,1},40][[All,2]] (* or *) LinearRecurrence[{1,1,0,0,1,-1,-1},{1,1,2,3,5,9,14},40] (* Harvey P. Dale, Jun 15 2017 *)

O.g.f.: x/((1-x)*(x^4 + x^3 + x^2 + x + 1)*(1 - x - x^2)). - R. J. Mathar, May 30 2008
a(n+5) = a(n) + Fibonacci(n+5), n>5.
a(n) = 12*a(n-5) - 10*a(n-10) - a(n-15). - Gary Detlefs, Dec 10 2010

Typo in Maple code corrected by R. J. Mathar, May 30 2008
More specific name from R. J. Mathar, Dec 09 2009
Indices in definition corrected by N. J. A. Sloane, Nov 25 2010

A214315 Floor of the real part of the zeros of the complex Fibonacci function on the right half-plane.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 10, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 42, 43, 45, 47, 49, 51, 53, 54, 56, 58, 60, 62, 63, 65, 67, 69, 71, 73, 74, 76, 78, 80, 82, 84, 85, 87, 89, 91, 93, 95, 96, 98, 100, 102, 104, 106, 107, 109, 111, 113, 115, 117, 118

Offset: 0

Wolfdieter Lang, Jul 24 2012

nonn

For the complex Fibonacci function and its complex zeros see the Koshy reference, pp. 523-524. See also the formula for F(z) given in the formula section of A052952. The real parts of the zeros of F are x_0(k) = alpha*k, with alpha = 2*(Pi^2)/(Pi^2 + (2*log(phi))^2), where phi = (1+sqrt(5))/2, and integer k. The corresponding imaginary parts are y_0(k) = - 4*Pi*log(phi)*k/(Pi^2 + (2*log(phi))^2). alpha is approximately 1.828404783. The zeros lie in the lower right and the upper left half-planes, and there is a zero at the origin.
a(n) = floor(alpha*n), n>=0, is a Beatty sequence with the complementary sequence b(n) = floor(beta*n), with beta = alpha/(alpha-1), approximately 2.207139336.
For the floor of the negative imaginary part see A214656.

			The complementary Beatty sequences start with:
n:    1   2 3  4  5  6   7   8   9  10  11  12  13  14  15  16
a(n): 0   1 3  5  7  9  10  12  14  16  18  20  21  23  25  27
b(n): (0) 2 4  6  8 11  13  15  17  19  22  24  26  28  30  33

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A052952 (Fibonacci related formula), A214656.

R:= RealField(100); [Floor(2*n*Pi(R)^2/(Pi(R)^2 + (2*Log((1+Sqrt(5))/2))^2)) : n in [0..100]]; // G. C. Greubel, Mar 09 2024

a[n_]:= Floor[2*n*Pi^2/(Pi^2 + 4*Log[GoldenRatio]^2)]; Table[a[n], {n, 0, 65}] (* Jean-François Alcover, Jul 03 2013 *)

[floor(2*n*pi^2/(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024

a(n) = floor(alpha*n), n>=0, with alpha = x_0(1) given in the comment section.

A074392 a(n) = Lucas(n+1) + (3*(-1)^n - 1)/2.

Original entry on oeis.org

2, 1, 5, 5, 12, 16, 30, 45, 77, 121, 200, 320, 522, 841, 1365, 2205, 3572, 5776, 9350, 15125, 24477, 39601, 64080, 103680, 167762, 271441, 439205, 710645, 1149852, 1860496, 3010350, 4870845, 7881197, 12752041, 20633240, 33385280, 54018522

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 22 2002

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

Cf. A000032, A074331.

Cf. A004146.

CoefficientList[Series[(2-x)/(1-x-2*x^2+x^3+x^4), {x, 0, 40}], x]

Vec((2-x) / ((x-1)*(1+x)*(x^2+x-1)) + O(x^50)) \\ Colin Barker, Jul 12 2017

a(n) = Sum (L(2i+e), (i=0, 1, .., Floor(n/2))), where L(n) are Lucas numbers and e=2(n/2 - Floor(n/2)).
Convolution of L(n) with the sequence (1, 0, 1, 0, 1, 0, ...)
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>3.
G.f.: ( 2-x ) / ( (x-1)*(1+x)*(x^2+x-1) ).
a(n) = 2*A052952(n)-A052952(n-1). - R. J. Mathar, Oct 04 2013
a(n) = 2^(-1-n) * (3*(-2)^n - 2^n + (1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n)). - Colin Barker, Jul 12 2017

A099573 Reverse of number triangle A054450.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 5, 1, 1, 5, 5, 8, 8, 1, 1, 6, 6, 12, 12, 13, 1, 1, 7, 7, 17, 17, 21, 21, 1, 1, 8, 8, 23, 23, 33, 33, 34, 1, 1, 9, 9, 30, 30, 50, 50, 55, 55, 1, 1, 10, 10, 38, 38, 73, 73, 88, 88, 89, 1, 1, 11, 11, 47, 47, 103, 103, 138, 138, 144, 144, 1, 1, 12, 12, 57, 57, 141, 141, 211, 211, 232, 232, 233

Offset: 0

Paul Barry, Oct 23 2004

			First few rows of the array:
  1, 1, 2, 3,  5,  8, ... (A000045)
  1, 1, 3, 4,  8, 12, ... (A052952)
  1, 1, 4, 5, 12, 17, ... (A054451)
  1, 1, 5, 6, 17, 23, ... (A099571)
  1, 1, 6, 7, 23, 30, ... (A099572)
  ...
Triangle begins as:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 3, 3;
  1, 1, 4, 4,  5;
  1, 1, 5, 5,  8,  8;
  1, 1, 6, 6, 12, 12, 13;
  1, 1, 7, 7, 17, 17, 21, 21;
  1, 1, 8, 8, 23, 23, 33, 33, 34;
  1, 1, 9, 9, 30, 30, 50, 50, 55, 55;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000045, A029907 (row sums), A052951, A052952, A054450, A054451, A052952.

Cf. A099571, A099572, A099574 (diagonal sums), A099575.

[(&+[Binomial(n-j,j): j in [0..Floor(k/2)]]): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

T[n_, k_]:= Sum[Binomial[n-j,j], {j,0,Floor[k/2]}];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

def A099573(n,k): return sum(binomial(n-j, j) for j in (0..(k//2)))
flatten([[A099573(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
T(n, n) = A000045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A099574(n).
Sum_{k=0..n} T(n, k) = A029907(n+1).
Antidiagonals of the following array: the first row equals the Fibonacci numbers, (1, 1, 2, 3, 5, ...), and the (n+1)-st row is obtained by the matrix-vector product A128174 * n-th row. - Gary W. Adamson, Jan 19 2011
From G. C. Greubel, Jul 25 2022: (Start)
T(n, n-1) = A052952(n-1), n >= 1.
T(n, n-2) = A054451(n-2), n >= 2.
T(n, n-3) = A099571(n-3), n >= 3.
T(n, n-4) = A099572(n-4), n >= 4. (End)

A131132 a(n) = a(n-1) + a(n-2) + 1 if n is a multiple of 6, otherwise a(n) = a(n-1) + a(n-2).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 14, 22, 36, 58, 94, 152, 247, 399, 646, 1045, 1691, 2736, 4428, 7164, 11592, 18756, 30348, 49104, 79453, 128557, 208010, 336567, 544577, 881144, 1425722, 2306866, 3732588, 6039454, 9772042, 15811496, 25583539, 41395035, 66978574, 108373609

Offset: 0

N. J. A. Sloane, May 25 2008

nonn

Also: convolution of A000045 with the period-6 sequence (0,0,0,0,0,0, 1,...). - R. J. Mathar, May 30 2008
Sequences of the form s(0)=a, s(1)= b, s(n) = s(n-1) + s(n-2) + k if n mod m = p, else s(n) = s(n-1) + s(n-2) have a form s(n) = fibonacci(n-1)*a + fibonacci(n)*b + P(x)*k. a(n) is the P(x) sequence for m=6. s(n) = fib(n)*a + fib(n-1)*b + a(n-6-p)*k. - Gary Detlefs, Dec 05 2010
a(n) is the number of compositions of n where the order of the 2 and the 3 does not matter. - Gregory L. Simay, May 18 2017

			Since 5 is not a multiple of 6, a(5) = a(4) + a(3) = 5 + 3 = 8. Since 6 is a multiple of 6, a(6) = a(5) + a(4) + 1 = 8 + 5 + 1 = 14. - _Michael B. Porter_, Jul 26 2016

H. Matsui et al., Problem B-1035, Fibonacci Quarterly, Vol. 45, Number 2; 2007; p. 182.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,0,1,-1,-1).

Cf. A052952, A004695, A080239, A124502, A066983.

A131132:=proc(n) option remember; local t1; if n <= 2 then RETURN(1); fi: if n mod 6 = 1 then t1:=1 else t1:=0; fi: procname(n-1)+procname(n-2)+t1; end; [seq(A131132(n), n=1..100)]; # N. J. A. Sloane, May 25 2008; Typo corrected by R. J. Mathar, May 30 2008

Print[Table[Round[(1 + Sqrt[5])/8 Fibonacci[n + 3]], {n, 0, 50}]] ;
Print[RecurrenceTable[{c[n] == c[n - 1] + c[n - 2] + c[n - 6] - c[n - 7] - c[n - 8], c[0] == 1, c[1] == 1, c[2] == 2, c[3] == 3, c[4] == 5, c[5] == 8, c[6] == 14, c[7] == 22}, c, {n, 0, 50}]] ;  (* John M. Campbell, Jul 07 2016 *)
LinearRecurrence[{1, 1, 0, 0, 0, 1, -1, -1}, {1, 1, 2, 3, 5, 8, 14, 22}, 40] (* Vincenzo Librandi, Jul 07 2016 *)

O.g.f.: 1/((1-x^6)(1 - x - x^2)). - R. J. Mathar, May 30 2008
a(n) = ((-1)^n-1)/6 + A099837(n+3)/12 + A000045(n+4)/4 + A057079(n)/12. - R. J. Mathar, Dec 07 2010
a(n) = floor(A066983(n+4)/3). - Gary Detlefs, Dec 19 2010
a(n) = round((1 + sqrt(5))/8 A000045(n+3)). - John M. Campbell, Jul 06 2016
a(n) = (number of compositions of n consisting of only 1 or 2 or 6) - (number of compositions with only 7 or ((1 or 2) and 7)) - (number of compositions with only 8 or ((1 or 2) and 8)). The "or" is inclusive. - Gregory L. Simay, May 25 2017

More specific name from R. J. Mathar, Dec 09 2009

A245738 Number of compositions of n into parts 1 and 2 with both parts present.

Original entry on oeis.org

2, 3, 7, 11, 20, 32, 54, 87, 143, 231, 376, 608, 986, 1595, 2583, 4179, 6764, 10944, 17710, 28655, 46367, 75023, 121392, 196416, 317810, 514227, 832039, 1346267, 2178308, 3524576, 5702886, 9227463, 14930351, 24157815, 39088168, 63245984, 102334154, 165580139, 267914295, 433494435, 701408732, 1134903168, 1836311902

Offset: 3

David Neil McGrath, Jul 31 2014

			a(9) = 54. The tuples are (22221) = 5!/4! = 5, (222111) = 6!/3!/3! = 20, (2211111) = 7!/5!/2! = 21, (21111111) = 8!/7! = 8.

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

Cf. A245332, A245492, A245487, A245527.

Column k=2 of A373118.

LinearRecurrence[{1,2,-1,-1},{2,3,7,11},50] (* Harvey P. Dale, Dec 20 2014 *)

Vec(1+1/(1-x-x^2)-1/(1-x)-1/(1-x^2)+O(x^66)) \\ Joerg Arndt, Aug 04 2014

G.f.: 1+1/(1-x-x^2)-1/(1-x)-1/(1-x^2).
a(n) = A052952(n-4)+2*A052952(n-3). - R. J. Mathar, Aug 05 2014
From Colin Barker, Jul 13 2017: (Start)
a(n) = (-20 + sqrt(5)*(-(1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n))/2^n) / 10 for n even.
a(n) = (-10 + sqrt(5)*(-(1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n))/2^n) / 10 for n odd.
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>6. (End)
a(n) = Sum_{i=1..floor((n-1)/2)} C(n-i,i). - Wesley Ivan Hurt, Sep 19 2017
a(n) = A000045(n+1) - A000034(n+1). - J. M. Bergot and Robert Israel, Oct 11 2021

A128618 Triangle read by rows: A128174 * A127647 as infinite lower triangular matrices.

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 1, 0, 3, 1, 0, 2, 0, 5, 0, 1, 0, 3, 0, 8, 1, 0, 2, 0, 5, 0, 13, 0, 1, 0, 3, 0, 8, 0, 21, 1, 0, 2, 0, 5, 0, 13, 0, 34, 0, 1, 0, 3, 0, 8, 0, 21, 0, 55, 1, 0, 2, 0, 5, 0, 13, 0, 34, 0, 89, 0, 1, 0, 3, 0, 8, 0, 21, 0, 55, 0, 144, 1, 0, 2, 0, 5, 0, 13, 0, 34, 0, 89, 0, 233

Offset: 1

Gary W. Adamson, Mar 14 2007

This triangle is different from A128619, which is A128619 = A127647 * A128174.

			First few rows of the triangle are:
  1;
  0, 1;
  1, 0, 2;
  0, 1, 0, 3;
  1, 0, 2, 0, 5;
  0, 1, 0, 3, 0, 8;
  1, 0, 2, 0, 5, 0, 13;
  0, 1, 0, 3, 0, 8,  0, 21;
  1, 0, 2, 0, 5, 0, 13,  0, 34;
  0, 1, 0, 3, 0, 8,  0, 21,  0, 55;
  1, 0, 2, 0, 5, 0, 13,  0, 34,  0, 89;
  ...

G. C. Greubel, Rows n = 1..100 of the triangle, flattened

Cf. A000045, A052952, A127647, A128174, A128619, A355020.

[((n+k+1) mod 2)*Fibonacci(k): k in [1..n], n in [1..15]]; // G. C. Greubel, Mar 17 2024

Table[Fibonacci[k]*Mod[n-k+1,2], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Mar 17 2024 *)

flatten([[((n-k+1)%2)*fibonacci(k) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Mar 17 2024

By columns, Fibonacci(k) interspersed with alternate zeros in every column, k=1,2,3,...
Sum_{k=1..n} T(n, k) = A052952(n-1) (row sums).
From G. C. Greubel, Mar 17 2024: (Start)
T(n, k) = (1/2)*(1 + (-1)^(n+k))*Fibonacci(k).
T(n, n) = A000045(n).
T(2*n-1, n) = (1/2)*(1-(-1)^n)*A000045(n).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (-1)^(n-1)*A052952(n-1).
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = (1/2)*(1 - (-1)^n)*(Fibonacci((n+ 5)/2) - 1).
Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*T(n-k+1, k) = (1/2)*(1-(-1)^n) *  A355020(floor((n-1)/2)). (End)

a(6) corrected and more terms from Georg Fischer, May 30 2023

A131328 Row sums of triangle A131327.

Original entry on oeis.org

1, 4, 5, 12, 17, 32, 49, 84, 133, 220, 353, 576, 929, 1508, 2437, 3948, 6385, 10336, 16721, 27060, 43781, 70844, 114625, 185472, 300097, 485572, 785669, 1271244, 2056913, 3328160, 5385073, 8713236, 14098309, 22811548, 36909857, 59721408, 96631265, 156352676

Offset: 0

Gary W. Adamson, Jun 28 2007

a(n)/a(n-1) tends to phi. (Cf. A062114).

			a(3) = 12 = sum of row 3 terms of A131327: (3 + 5 + 3 + 1).
a(3) = (9 + 3) since we add terms of A131326: (1, 3, 4, 9, 13,...) to A052952: (0, 1, 1, 3, 4,...), getting (9 + 3 ) = 12.

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

Cf. A062114, A052952, A131324, A131325, A131326, A131327.

Vec((1 + 3*x - x^2) / ((1 - x)*(1 + x)*(1 - x - x^2)) + O(x^50)) \\ Colin Barker, Jul 12 2017

a(n+1) = A131326(n) + A052952(n+1).
a(n) = -3*(1+(-1)^n)/2 +4*A000045(n+1). - R. J. Mathar, Aug 13 2012
G.f.: ( 1+3*x-x^2 ) / ( (x-1)*(1+x)*(x^2+x-1) ). - R. J. Mathar, Aug 13 2012
From Colin Barker, Jul 12 2017: (Start)
a(n) = (2^(1-n)*((1+sqrt(5))^(n+1) - (1-sqrt(5))^(n+1))) / sqrt(5) - 3 for n even.
a(n) = (2^(1-n)*((1+sqrt(5))^(n+1) - (1-sqrt(5))^(n+1))) / sqrt(5) for n odd.
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>3.
(End)

More terms from Colin Barker, Jul 12 2017

cp's OEIS Frontend

A214656 Floor of the imaginary part of the zeros of the complex Fibonacci function on the left half-plane.

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A099571 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+3, k).

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A124502 a(1)=a(2)=1; thereafter, a(n+1) = a(n) + a(n-1) + 1 if n is a multiple of 5, otherwise a(n+1) = a(n) + a(n-1).

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A214315 Floor of the real part of the zeros of the complex Fibonacci function on the right half-plane.

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A074392 a(n) = Lucas(n+1) + (3*(-1)^n - 1)/2.

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A099573 Reverse of number triangle A054450.

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A131132 a(n) = a(n-1) + a(n-2) + 1 if n is a multiple of 6, otherwise a(n) = a(n-1) + a(n-2).

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