OFFSET
0,4
COMMENTS
Note that a(2,3,4,5) = {1,4,13,121} coincide with A003462(1,2,3,5). - Zak Seidov, Dec 19 2014
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..90
FORMULA
a(n) = a(n-1) + 3^(n-2) * a(n-2).
Associated constant: C_3 = lim_{n->infinity} a(n)*a(n-2)/a(n-1)^2 = 1.147262608254535257774121586... . - Benoit Cloitre, Aug 30 2003
a(n)*a(n+3) - a(n)*a(n+2) - 3*a(n+1)*a(n+2) + 3*a(n+1)^2 = 0. - Emanuele Munarini, Dec 05 2017
MAPLE
q:=3; seq(add((product((1-q^(n-j-1-k))/(1-q^(k+1)), k=0..j-1))*q^(j^2), j = 0..floor((n-1)/2)), n = 0..20); # G. C. Greubel, Dec 16 2019
MATHEMATICA
RecurrenceTable[{a[0]==0, a[1]==1, a[n]==a[n-1]+a[n-2]*3^(n-2)}, a, {n, 30}] (* Vincenzo Librandi, Nov 08 2012 *)
F[n_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^(j^2), {j, 0, Floor[(n-1)/2]}];
Table[F[n, 3], {n, 0, 20}] (* G. C. Greubel, Dec 16 2019 *)
PROG
(Magma) [0] cat[n le 2 select 1 else Self(n-1) + Self(n-2)*(3^(n-2)): n in [1..20]]; // Vincenzo Librandi, Nov 08 2012
(PARI) q=3; m=20; v=concat([0, 1], vector(m-2)); for(n=3, m, v[n]=v[n-1]+q^(n-3)*v[n-2]); v \\ G. C. Greubel, Dec 16 2019
(Sage)
def F(n, q): return sum( q_binomial(n-j-1, j, q)*q^(j^2) for j in (0..floor((n-1)/2)))
[F(n, 3) for n in (0..20)] # G. C. Greubel, Dec 16 2019
(GAP) q:=3;; a:=[0, 1];; for n in [3..30] do a[n]:=a[n-1]+q^(n-3)*a[n-2]; od; a; # G. C. Greubel, Dec 16 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved