OFFSET
0,3
COMMENTS
The sequence b(n) = floor((4/3)*(n+2)) appears as an upper bound in Fijavz and Wood.
Binary expansion does not end in 11.
From Guenther Schrack, May 04 2023: (Start)
LINKS
Daniel Starodubtsev, Table of n, a(n) for n = 0..10000
Gasper Fijavz and David R. Wood, Graph Minors and Minimum Degree, arXiv:0812.1064 [math.CO], 2008.
Niall Graham and Frank Harary, Edge Sums of Hypercubes, Bull. Irish Math. Soc., Vol. 21 (1988), pp. 8-12.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
FORMULA
G.f.: x*(1+x+2*x^2)/((1-x)*(1-x^3)).
a(0) = 0, a(n+1) = a(n) + a(n) mod 4 + 0^(a(n) mod 4). - Reinhard Zumkeller, Mar 23 2003
a(n) = floor(n/3) + n. - Gary Detlefs, Mar 20 2010
a(n) = (12*n-3+3*cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/9. - Wesley Ivan Hurt, Sep 30 2017
E.g.f.: (3*exp(x)*(4*x - 1) + exp(-x/2)*(3*cos((sqrt(3)*x)/2) + sqrt(3)*sin((sqrt(3)*x)/2)))/9. - Stefano Spezia, Jun 09 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(2)-1)*Pi/8 + sqrt(2)*log(sqrt(2)+2)/4 + (2-sqrt(2))*log(2)/8. - Amiram Eldar, Dec 05 2021
From Guenther Schrack, May 04 2023: (Start)
a(n) = (12*n - 3 + w^(2*n)*(w + 2) - w^n*(w - 1))/9 where w = (-1 + sqrt(-3))/2.
a(n) = 2*floor(n/3) + floor((n+1)/3) + floor((n+2)/3).
a(n) = (4*n - n mod 3)/3.
a(n) = a(n-3) + 4.
a(n) = a(n-1) + a(n-3) - a(n-4).
a(n) = A042968(n+1) - 1.
(End)
MAPLE
seq(floor(n/3)+n, n=0..68); # Gary Detlefs, Mar 20 2010
MATHEMATICA
f[n_] := Floor[4 n/3]; Array[f, 69, 0] (* Robert G. Wilson v, Dec 24 2010 *)
fQ[n_] := Mod[n, 4] != 3; Select[ Range[0, 90], fQ] (* Robert G. Wilson v, Dec 24 2010 *)
a[0] = 0; a[n_] := a[n] = a[n - 1] + 2 - If[ Mod[a[n - 1], 4] < 2, 1, 0]; Array[a, 69, 0] (* Robert G. Wilson v, Dec 24 2010 *)
CoefficientList[ Series[x (1 + x + 2 x^2)/((1 - x) (1 - x^3)), {x, 0, 68}], x] (* Robert G. Wilson v, Dec 24 2010 *)
PROG
(Magma) [n: n in [0..100] | n mod 4 in [0..2]]; // Vincenzo Librandi, Dec 23 2010
(PARI) a(n)=4*n\3 \\ Charles R Greathouse IV, Sep 27 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved