A022087 -id:A022087 - cp's OEIS frontend

A156279 4 times the Lucas number A000032(n).

8, 4, 12, 16, 28, 44, 72, 116, 188, 304, 492, 796, 1288, 2084, 3372, 5456, 8828, 14284, 23112, 37396, 60508, 97904, 158412, 256316, 414728, 671044, 1085772, 1756816, 2842588, 4599404, 7441992

Offset: 0

Paul Curtz, Feb 07 2009

This is a second kind "autosequence" whose first kind companion is A022087. - Jean-François Alcover, Aug 20 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
OEIS Wiki, Autosequence
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045, A014217.

[4*Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[4*LucasL[n], {n,0,30}] (* G. C. Greubel, Dec 21 2017 *)

a(n)=4*(fibonacci(n-1)+fibonacci(n+1)) \\ Charles R Greathouse IV, Oct 16 2015

a(n) = 4*A000032(n).
a(n) = a(n-1) + a(n-2).
a(n) = A014217(n+3) - A014217(n-3), with A014217(-5) = -11, A014217(-4) = 6, A014217(-3) = -4, A014217(-2) = 2, A014217(-1) = -1 extended as proposed in A153263.
G.f. 4*(-2 + x) / (-1 + x + x^2). - R. J. Mathar, Mar 11 2011
a(n) = Lucas(n+3) - Lucas(n-3), where Lucas(i) for i = 0..2 gives -4, 3, -1. - Bruno Berselli, Jul 27 2017

A230216 Number of binary strings of length n avoiding "squares" (that is, repeated blocks of the form xx) with |x| = 3.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 56, 104, 192, 352, 648, 1192, 2192, 4032, 7416, 13640, 25088, 46144, 84872, 156104, 287120, 528096, 971320, 1786536, 3285952, 6043808, 11116296, 20446056, 37606160, 69168512, 127220728, 233995400, 430384640, 791600768, 1455980808

Offset: 0

Nathaniel Johnston, Oct 11 2013

			For n = 6 there are 8 strings omitted, namely 000000, 001001, ..., 111111, so a(6) = 64-8 = 56.

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

Cf. A000073, A022087, A229614, A230127, A230177.

Vec((1 + x + x^2 + x^3 + 2*x^4 + 4*x^5) / (1 - x - x^2 - x^3) + O(x^40)) \\ Colin Barker, Aug 09 2019

a(n) = 8*A000073(n) for n >= 3.
From Colin Barker, Aug 09 2019: (Start)
G.f.: (1 + x + x^2 + x^3 + 2*x^4 + 4*x^5) / (1 - x - x^2 - x^3).
a(n) = a(n-1) + a(n-2) + a(n-3) for n>5.
(End)

A022403 a(0)=a(1)=3; thereafter a(n) = a(n-1) + a(n-2) + 1.

Original entry on oeis.org

3, 3, 7, 11, 19, 31, 51, 83, 135, 219, 355, 575, 931, 1507, 2439, 3947, 6387, 10335, 16723, 27059, 43783, 70843, 114627, 185471, 300099, 485571, 785671, 1271243, 2056915, 3328159, 5385075, 8713235, 14098311, 22811547, 36909859, 59721407, 96631267, 156352675, 252983943, 409336619, 662320563

Offset: 0

N. J. A. Sloane

nonn

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

See A022406 for a similar sequence.

[4*Fibonacci(n+1) - 1: n in [0..40]]; // G. C. Greubel, Mar 01 2018

Table[4*Fibonacci[n+1] -1,{n, 0, 31}] (* Vladimir Joseph Stephan Orlovsky, Apr 03 2011 *)
RecurrenceTable[{a[0]==a[1]==3,a[n]==a[n-1]+a[n-2]+1},a,{n,40}] (* or *) LinearRecurrence[{2,0,-1},{3,3,7},50] (* Harvey P. Dale, Jan 10 2021 *)

for(n=0, 40, print1(4*fibonacci(n+1) -1, ", ")) \\ G. C. Greubel, Mar 01 2018

From R. J. Mathar, Mar 11 2011:  (Start)
a(n+1) - a(n) = A022087(n).
G.f.: ( 3-3*x+x^2 ) / ( (x-1)*(x^2+x-1) ). (End)
a(n) = 4*Fibonacci(n+1) - 1. - G. C. Greubel, Mar 01 2018
a(n) = (-1 + (2^(1-n)*(-(1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n))) / sqrt(5)). - Colin Barker, Mar 02 2018

Terms a(32) onward added by G. C. Greubel, Mar 01 2018

A022406 a(0)=3, a(1)=7; thereafter a(n) = a(n-1) + a(n-2) + 1.

Original entry on oeis.org

3, 7, 11, 19, 31, 51, 83, 135, 219, 355, 575, 931, 1507, 2439, 3947, 6387, 10335, 16723, 27059, 43783, 70843, 114627, 185471, 300099, 485571, 785671, 1271243, 2056915, 3328159, 5385075, 8713235, 14098311, 22811547, 36909859, 59721407, 96631267, 156352675

Offset: 0

N. J. A. Sloane

a(n) is the minimum number of nodes required for a full binary AVL tree of height n+1 whose root node has a balance factor of 0. - Sumukh Patel, Jun 24 2022

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

Cf. A000045, A022087, A122195. See A022403 for a very similar sequence.

[4*Fibonacci(n+2) - 1: n in [0..40]]; // G. C. Greubel, Mar 01 2018

Table[4*Fibonacci[n + 2] - 1, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
CoefficientList[Series[(3+x-3*x^2)/((1-x)*(1-x-x^2)), {x, 0, 50}], x] (* G. C. Greubel, Mar 01 2018 *)

a(n) = 4*fibonacci(n+2)-1; \\ G. C. Greubel, Mar 01 2018

a(n) = 4*A000045(n+2) - 1. - Ron Knott, Aug 25 2006
From R. J. Mathar, May 28 2008: (Start)
a(n) = A022403(n+1).
O.g.f.: (3+x-3*x^2)/((1-x)*(1-x-x^2)).
a(n+1) - a(n) = A022087(n+1). (End)
a(n) = (2^(-n)*(-5*2^n + (10-6*sqrt(5))*(1-sqrt(5))^n + 2*(1+sqrt(5))^n*(5+3*sqrt(5)))) / 5. - Colin Barker, Mar 02 2018
E.g.f.: 4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - exp(x). - Stefano Spezia, Feb 01 2025

A210209 GCD of all sums of n consecutive Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 4, 1, 3, 2, 11, 1, 8, 1, 29, 2, 21, 1, 76, 1, 55, 2, 199, 1, 144, 1, 521, 2, 377, 1, 1364, 1, 987, 2, 3571, 1, 2584, 1, 9349, 2, 6765, 1, 24476, 1, 17711, 2, 64079, 1, 46368, 1, 167761, 2, 121393, 1, 439204, 1, 317811, 2, 1149851, 1, 832040

Offset: 0

Alonso del Arte, Mar 18 2012

Early on in the Posamentier & Lehmann (2007) book, the fact that the sum of any ten consecutive Fibonacci numbers is a multiple of 11 is presented as an interesting property of the Fibonacci numbers. Much later in the book a proof of this fact is given, using arithmetic modulo 11. An alternative proof could demonstrate that 11*F(n + 6) = Sum_{i=n..n+9} F(i).

			a(3) = 2 because all sums of three consecutive Fibonacci numbers are divisible by 2 (F(n) + F(n-1) + F(n-2) = 2F(n)), but since the GCD of 3 + 5 + 8 = 16 and 5 + 8 + 13 = 26 is 2, no number larger than 2 divides all sums of three consecutive Fibonacci numbers.
a(4) = 1 because the GCD of 1 + 1 + 2 + 3 = 7 and 1 + 2 + 3 + 5 = 11 is 1, so the sums of four consecutive Fibonacci numbers have no factors in common.

Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, New York (2007) p. 33.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Dan Guyer and aBa Mbirika, GCD of sums of k consecutive Fibonacci, Lucas, and generalized Fibonacci numbers, Journal of Integer Sequences, 24 No.9, Article 21.9.8 (2021), 25pp; arXiv preprint, arXiv:2104.12262 [math.NT], 2021.
I. D. Ruggles, Elementary problem B-1, Fibonacci Quarterly, Vol. 1, No. 1, February 1963, p. 73; Solution by Marjorie R. Bicknell, published in Vol. 1, No. 3, October 1963, pp. 76-77.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Cf. A000045, A000071, sum of the first n Fibonacci numbers, A001175 (Pisano periods). Cf. also A229339.
Bisections give: A005013 (even part), A131534 (odd part).
Sums of m consecutive Fibonacci numbers: A055389 (m = 3, ignoring the initial 1); A000032 (m = 4, these are the Lucas numbers); A013655 (m = 5); A022087 (m = 6); A022096 (m = 7); A022379 (m = 8).

a:= n-> (Matrix(7, (i, j)-> `if`(i=j-1, 1, `if`(i=7, [1, 0, -3, -1, 1, 3, 0][j], 0)))^iquo(n, 2, 'r'). `if`(r=0, <<0, 1, 1, 4, 3, 11, 8>>, <<1, 2, 1, 1, 2, 1, 1>>))[1, 1]: seq(a(n), n=0..80);  # Alois P. Heinz, Mar 18 2012

Table[GCD[Fibonacci[n + 1] - 1, Fibonacci[n]], {n, 1, 50}] (* Horst H. Manninger, Dec 19 2021 *)

a(n)=([0,1,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,1,0,0,0,0,0,0,0,0,0,0,0; 0,0,0,1,0,0,0,0,0,0,0,0,0,0; 0,0,0,0,1,0,0,0,0,0,0,0,0,0; 0,0,0,0,0,1,0,0,0,0,0,0,0,0; 0,0,0,0,0,0,1,0,0,0,0,0,0,0; 0,0,0,0,0,0,0,1,0,0,0,0,0,0; 0,0,0,0,0,0,0,0,1,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,1,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,1,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,1,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,1; 1,0,0,0,-3,0,-1,0,1,0,3,0,0,0]^n*[0;1;1;2;1;1;4;1;3;2;11;1;8;1])[1,1] \\ Charles R Greathouse IV, Jun 20 2017

G.f.: -x*(x^12-x^11+2*x^10-x^9-2*x^8-x^7-6*x^6+x^5-2*x^4+x^3+2*x^2+x+1) / (x^14-3*x^10-x^8+x^6+3*x^4-1) = -1/(x^4+x^2-1) + (x^2+1)/(x^4-x^2-1) + (x+2)/(6*(x^2+x+1)) + (x-2)/(6*(x^2-x+1)) - 2/(3*(x+1)) - 2/(3*(x-1)). - Alois P. Heinz, Mar 18 2012
a(n) = gcd(Fibonacci(n+1)-1, Fibonacci(n)). - Horst H. Manninger, Dec 19 2021
From Aba Mbirika, Jan 21 2022: (Start)
a(n) = gcd(F(n+1)-1, F(n+2)-1).
a(n) = Lcm_{A001175(m) divides n} m.
Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)

More terms from Alois P. Heinz, Mar 18 2012

A356993 a(n) = b(n - b(n - b(n - b(n)))) for n >= 2, where b(n) = A356988(n).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 10, 11, 11, 11, 11, 11, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 18, 18, 18, 18, 18, 18, 18, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 29, 29, 29

Offset: 2

Peter Bala, Sep 09 2022

The sequence is slow, that is, for n >= 2, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.
The line graph of the sequence {a(n)} thus consists of a series of plateaus (where the value of the ordinate a(n) remains constant as n increases) joined by lines of slope 1.
The sequence of plateau heights beginning 3, 4, 5, 7, 8, 11, 13, 18, 21, 29, 34, 47, 55, ..., consists of alternating Fibonacci numbers A000045 and Lucas numbers A000032.

Peter Bala, Notes on A356993

Cf. A000032, A000045, A000285, A022087, A356988, A356991 - A356999.

# b(n) = A356988
b := proc(n) option remember; if n = 1 then 1 else n - b(b(n - b(b(b(n-1))))) end if; end proc:
seq( b(n - b(n - b(n - b(n)))), n = 2..100 );

a(2) = a(3) = a(4) = a(5) = 1 and then for k >= 3 there holds
a(3*F(k) + j) = F(k) for 0 <= j <= F(k-1) (local plateau)
a(L(k+1) + j) = F(k) + j for 0 <= j <= F(k-2) (ascent to plateau of height L(k-1))
a(4*F(k) + j) = L(k-1) for 0 <= j <= F(k-1) (local plateau)
a(4*F(k) + F(k-1) + j) = L(k-1) + j for 0 <= j <= F(k-3) (ascent to next plateau of height F(k+1)).

A168674 a(n) = 2*A001610(n).

Original entry on oeis.org

0, 4, 6, 12, 20, 34, 56, 92, 150, 244, 396, 642, 1040, 1684, 2726, 4412, 7140, 11554, 18696, 30252, 48950, 79204, 128156, 207362, 335520, 542884, 878406, 1421292, 2299700, 3720994, 6020696, 9741692, 15762390, 25504084, 41266476, 66770562, 108037040

Offset: 0

Roger L. Bagula and Gary W. Adamson, Jun 01 2010

This sequence has a golden mean ratio limit.

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

Cf. A022087, A019274, A006355, A001610.

I:=[0,4,6]; [n le 3 select I[n] else 2*Self(n-1)-Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jul 30 2016

M = {{0, 1}, {1, 1}} v[0] = {0, 1}; v[n_] := v[n] = M.v[n - 1] + {3, 2} a = Table[v[n][[1]], {n, 0, 30}]
LinearRecurrence[{2, 0, -1}, {0, 4, 6}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2012 *)
RecurrenceTable[{a[0] == 0, a[1] == 4, a[2] == 6, a[n] == 2 a[n-1] - a[n-3]}, a, {n, 50}] (* Vincenzo Librandi, Jul 30 2016 *)

a(n)=([0,1,0; 0,0,1; -1,0,2]^n*[0;4;6])[1,1] \\ Charles R Greathouse IV, Jul 30 2016

a(n) = 2*a(n-1) - a(n-3). [Dec 03 2009]

G.f.: 2*x*(2 - x)/((1-x)*(1 -x -x^2)). [Dec 03 2009]

Definition simplified and notation in formulas set to OEIS standards by the Assoc. Editors of the OEIS, Dec 03 2009

cp's OEIS Frontend

A156279 4 times the Lucas number A000032(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A230216 Number of binary strings of length n avoiding "squares" (that is, repeated blocks of the form xx) with |x| = 3.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

Formula

A022403 a(0)=a(1)=3; thereafter a(n) = a(n-1) + a(n-2) + 1.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A022406 a(0)=3, a(1)=7; thereafter a(n) = a(n-1) + a(n-2) + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A210209 GCD of all sums of n consecutive Fibonacci numbers.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A356993 a(n) = b(n - b(n - b(n - b(n)))) for n >= 2, where b(n) = A356988(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

A168674 a(n) = 2*A001610(n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs