OFFSET
0,2
COMMENTS
In binary there is a pattern in how the zeros and ones appear:
a(0) = 01
a(1) = 101
a(2) = 1001
a(3) = 11001
a(4) = 110001
a(5) = 1110001
a(6) = 11100001
a(7) = 111100001
a(8) = 1111000001
a(9) = 11111000001
a(10) = 111110000001
a(11) = 1111110000001
a(12) = 11111100000001
a(13) = 111111100000001
a(14) = 1111111000000001
a(15) = 11111111000000001
Graphically, each term can be obtained by successively and alternately forming squares and centered squares as shown in the illustration.
LINKS
Daniel Poveda Parrilla, Table of n, a(n) for n = 0..1000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (3,0,-6,4).
FORMULA
a(n) = 1 + 2^(n+2) - 2^(1 + n/2) + (-1)^(n+1)*2^(1 + n/2) - 2^((n+1)/2) + (-1)^(n+2)*2^((n+1)/2).
a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4) for n>3.
G.f.: (-1-2*x+6*x^2-4*x^3)/(-1+3*x-6*x^3+4*x^4).
MATHEMATICA
Table[1+2^(n+2)-2^(1+n/2)+(-1)^(n+1) 2^(1+n/2)-2^((n+1)/2)+(-1)^(n+2) 2^((n+1)/2), {n, 0, 28}] (*or*)
CoefficientList[Series[(-1 - 2 x + 6 x^2 - 4 x^3)/(-1 + 3 x - 6 x^3 + 4 x^4), {x, 0, 28}], x] (*or*)
LinearRecurrence[{3, 0, -6, 4}, {1, 5, 9, 25}, 29]
PROG
(PARI) Vec((-1-2*x+6*x^2-4*x^3) / (-1+3*x-6*x^3+4*x^4) + O(x^29))
CROSSREFS
KEYWORD
nonn
AUTHOR
Daniel Poveda Parrilla, Jan 26 2017
STATUS
approved