0,5

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 (kociemba(AT)t-online.de), Jun 16 2004

a(n)=4a(n-1)-3a(n-2)-2a(n-3)+a(n-4), a(0)=a(1)=a(2)=0, a(3)=1. - Asher Auel (asher.auel(AT)reed.edu), Jun 06 2000

Convolution of Fibonacci numbers (F(n)) with (F(2n)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 07 2004

G.f.: x^3/(1-4x+3x^2+2x^3-x^4) - Herbert Kociemba (kociemba(AT)t-online.de), 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 (pbarry(AT)wit.ie), 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 (g_m(AT)mcraefamily.com), Jun 28 2006

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

Cf. A000045, A056015.

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

Sequence in context: A049611 A084851 A094706 this_sequence A058693 A027076 A105693

Adjacent sequences: A056011 A056012 A056013 this_sequence A056015 A056016 A056017

nonn

Asher Auel (asher.auel(AT)reed.edu), Jun 06 2000