A056015 -id:A056015 - cp's OEIS frontend

A056014 a(n) = (Fibonacci(2n-1) - Fibonacci(n+1))/2.

0, 0, 0, 1, 4, 13, 38, 106, 288, 771, 2046, 5401, 14212, 37324, 97904, 256621, 672336, 1760997, 4611642, 12075526, 31617520, 82781215, 216732890, 567428401, 1485570024, 3889310328, 10182407328, 26657986681, 69791674108

Offset: 0

Asher Auel, Jun 06 2000

With a(0)=0, a(1)=1, a(2)=1, a(3)=2, this recurrence produces a(n)=A000045(n) (Fibonacci numbers).

Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 4. - Herbert Kociemba, Jun 16 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Éva Czabarka, Rigoberto Flórez, and Leandro Junes, A Discrete Convolution on the Generalized Hosoya Triangle, Journal of Integer Sequences, 18 (2015), Article 15.1.6.
Henrik Eriksson and Markus Jonsson, Level sizes of the Bulgarian solitaire game tree, Fib. Q., 35:3 (2017), 243-251.
Hideyuki Othsuka, Problem B-1345, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 1 (2024), p. 85.
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

Cf. A000045, A000079, A051450, A056015, A094966.

a(1-2n)=A059512(2n), a(-2n)=A027994(2n-1).

I:=[0, 0, 0, 1]; [n le 4 select I[n] else 4*Self(n-1)-3*Self(n-2)-2*Self(n-3)+Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 23 2012

Table[(Fibonacci[2n-1]-Fibonacci[n+1])/2,{n,0,40}]  (* Harvey P. Dale, Mar 24 2011 *)
LinearRecurrence[{4,-3,-2,1},{0,0, 0,1},40] (* Vincenzo Librandi, Jun 23 2012 *)

a(n)=(fibonacci(2*n-1)-fibonacci(n+1))/2

a(n) = 4*a(n-1) - 3*a(n-2) - 2*a(n-3) + a(n-4), a(0)=a(1)=a(2)=0, a(3)=1.
Convolution of Fibonacci numbers F(n) with F(2n). - Benoit Cloitre, Jun 07 2004
G.f.: x^3/((1 - x - x^2)*(1 - 3*x + x^2)). - Herbert Kociemba, Jun 16 2004
Binomial transform of x^3/(1-3x^2+x^4), or (essentially) F(2n) with interpolated zeros. a(n)=sum{k=0..n, binomial(n, k)((3/2-sqrt(5)/2)^(k/2)((sqrt(5)/20+1/4)(-1)^k-sqrt(5)/20-1/4)+ (sqrt(5)/2+3/2)^(k/2)((sqrt(5)/20-1/4)(-1)^k-sqrt(5)/20+1/4))}. - Paul Barry, Jul 26 2004
Convolution of the powers of 2 (A000079) with the number of positive rational knots with 2n+1 crossings (A051450), with three leading zeros. - Graeme McRae, Jun 28 2006
a(n) = (A001519(n) - A000045(n+1))/2. - R. J. Mathar, Jun 24 2011
a(n) = Sum_{k=1..n-1} binomial(n-1, k) * A094966(k-1) (Othsuka, 2024). - Amiram Eldar, Feb 29 2024

A108140 a(n) = 4a(n-1) -3a(n-2) -2*a(n-3) +a(n-4), n>8.

Original entry on oeis.org

1, 1, 1, 1, 0, 4, 17, 55, 161, 449, 1220, 3266, 8667, 22879, 60203, 158107, 414728, 1087064, 2848061, 7459703, 19535229, 51152749, 133933964, 350666854, 918095255, 2403665279, 6292975607, 16475382935, 43133369616, 112925043724

Offset: 0

Roger L. Bagula, Jun 05 2005

Roger Bagula, Factoring Double Fibonacci Sequences, 2000.
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

Cf. A000045, A056015, A056016.

F[1] = 1; F[2] = 1; F[3] = 1; F[4] = 1; F[n__] := F[n] = 4*F[n - 1] - 3*F[n - 2] - 2*F[n - 3] + F[n - 4] a = Table[Abs[F[n]], {n, 1, 50}]

a(n) = -A000045(n+2)/2 + A001906(n-1)/2, n>3. [Sep 28 2009]

G.f.: (-1+2*x^7-2*x^6-8*x^5-2*x^3+3*x)/((x^2+x-1)*(x^2-3*x+1)). [Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009]

Definition replaced by recurrence by the Associate Editors of the OEIS, Sep 28 2009

A108142 a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 1; a[5] = 1; a[6] = 1; for n >= 7, a[n] = 6a[n - 1] - 5a[n - 2] - 4a[n - 3] - 3a[ n - 4] + 2*a[n - 5] + a[n - 6]; then take absolute values.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 27, 151, 759, 3679, 17599, 83767, 397943, 1889059, 8964891, 42539855, 201849743, 957752095, 4544385823, 21562354767, 102309686479, 485441784803, 2303337053819, 10928934112423, 51855892302151

Offset: 1

Roger L. Bagula, Jun 05 2005

nonn

The 2nd countdown sequence.

Roger Bagula, Factoring Double Fibonacci Sequences, 2000

Index entries for linear recurrences with constant coefficients, signature (6, -5, -4, -3, 2, 1).

Cf. A056015, A056016.

F[1] = 1; F[2] = 1; F[3] = 1; F[4] = 1; F[5] = 1; F[6] = 1; F[n__] := F[n] = 6*F[n - 1] - 5*F[n - 2] - 4*F[n - 3] - 3*F[ n - 4] + 2*F[n - 5] + F[n - 6] a = Table[Abs[F[n]], {n, 1, 50}]
LinearRecurrence[{6,-5,-4,-3,2,1},{1,1,1,1,1,1,3,27,151,759,3679,17599},30] (* Harvey P. Dale, Apr 25 2018 *)

Edited by N. J. A. Sloane, Jun 08 2007

A108143 a(n)= 5*a(n-1) -a(n-2) -a(n-3).

Original entry on oeis.org

1, 1, 1, 3, 13, 61, 289, 1371, 6505, 30865, 146449, 694875, 3297061, 15643981, 74227969, 352198803, 1671122065, 7929183553, 37622596897, 178512678867, 847011613885, 4018922793661, 19069089675553, 90479513970219

Offset: 0

Roger L. Bagula, Jun 05 2005

Roger Bagula, Factoring Double Fibonacci Sequences, 2000

Index entries for linear recurrences with constant coefficients, signature (5,-1,-1)

Cf. A056015, A056016.

F[1] = 1; F[2] = 1; F[3] = 1; F[n__] := F[n] = 5*F[n - 1] - F[n - 2] - F[n - 3] a = Table[Abs[F[n]], {n, 1, 50}]
LinearRecurrence[{5,-1,-1},{1,1,1},30] (* Harvey P. Dale, Jan 21 2023 *)

G.f.: (1-4*x-3*x^2)/(1-5*x+x^2+x^3) [Sep 28 2009]

Definition replaced by recurrence by the Associate Editors of the OEIS, Sep 28 2009

cp's OEIS Frontend

A056014 a(n) = (Fibonacci(2n-1) - Fibonacci(n+1))/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A108140 a(n) = 4*a(n-1) -3*a(n-2) -2*a(n-3) +a(n-4), n>8.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A108142 a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 1; a[5] = 1; a[6] = 1; for n >= 7, a[n] = 6*a[n - 1] - 5*a[n - 2] - 4*a[n - 3] - 3*a[ n - 4] + 2*a[n - 5] + a[n - 6]; then take absolute values.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Extensions

A108143 a(n)= 5*a(n-1) -a(n-2) -a(n-3).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A108140 a(n) = 4a(n-1) -3a(n-2) -2*a(n-3) +a(n-4), n>8.

A108142 a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 1; a[5] = 1; a[6] = 1; for n >= 7, a[n] = 6a[n - 1] - 5a[n - 2] - 4a[n - 3] - 3a[ n - 4] + 2*a[n - 5] + a[n - 6]; then take absolute values.