1,3

Convolution of Fibonacci sequence with sequence (1,0,0,0,1,0,0,0,1, ...)

There is an interesting relation between a(n) and the Fibonacci sequence f(n). Sqrt(a(4n-2)) = f(2n), where we denote by Sqrt() the square root. By using this fact we can calculate the value of a(n) by the following (1),(2),(3),(4) and (5). (1) a(1) = 1. (2) If n = 2 (mod 4), then a(n) = f((n+2)/2)^2. (3) If n = 3 (mod 4), then a(n) = (f((n+5)/2)^2-2f((n+1)/2)^2-1)/3. (4) If n = 0 (mod 4), then a(n) = (f((n+4)/2)^2+f(n/2)^2-1)/3. (5) If n = 1 (mod 4), then a(n) = (2f((n+3)/2)^2-f((n-1)/2)^2+1)/3. - Hiroshi Matsui and Ryohei Miyadera (miyadera1272000(AT)yahoo.co.jp), Aug 08 2006

H. Matsui et al., Problem B-1019, Fibonacci Quarterly, Vol. 45, Number 2; 2007; p. 182.

G.f.: 1/((1-x^4)(1-x-x^2)) = 1/(1-x-x^2-x^4+x^5+x^6); a(n)=a(n-1)+a(n-2)+a(n-4)-a(n-5)-a(n-6) a(n)=Sum{j=0..floor(n/2) Sum{k=0..floor((n-j)/2) binomial(n-j-2k, j-2k}}

Another recurrence is given in the Maple code.

f:=proc(n) option remember; local t1; if n <= 2 then RETURN(1); fi: if n mod 4 = 1 then t1:=1 else t1:=0; fi: f(n-1)+f(n-2)+t1; end; [seq(f(n), n=1..100)]; - N. J. A. Sloane (njas(AT)research.att.com), May 25 2008

(*f[n] is the Fibonacci sequence and a[n] is the sequence of A080239*) f[n_] := f[n] = f[n - 1] + f[n - 2]; f[1] = 1; f[2] = 1; a[n_] := Which[n == 1, 1, Mod[n, 4] == 2, f[(n + 2)/2]^2, Mod[n, 4] == 3, (f[(n + 5)/2]^2 - 2f[(n + 1)/2]^2 - 1)/3, Mod[n, 4] == 0, (f[(n + 4)/2]^2 + f[n/2]^2 - 1)/3, Mod[n, 4] == 1, (2f[(n + 3)/2]^2 - f[(n - 1)/2]^2 + 1)/3] - Hiroshi Matsui and Ryohei Miyadera (miyadera1272000(AT)yahoo.co.jp), Aug 08 2006

Cf. A035317, A000045, A026636, A026647, A004695.

Sequence in context: A086642 A014214 A094993 this_sequence A018158 A057928 A026735

Adjacent sequences: A080236 A080237 A080238 this_sequence A080240 A080241 A080242

easy,nonn

Paul Barry (pbarry(AT)wit.ie), Feb 11 2003