OFFSET
0,1
COMMENTS
Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iterate of the mapping 00->1000, 10->01, starting with 00; see A288729.
From Michel Dekking, Mar 25 2018: (Start)
Note that a(n) - a(n-1) = 2*[a(n-4) - a(n-5)] for n>4.
It follows that this sequence is a union of four simple sequences:
a(4k-4) = 4*2^k - 6 = A131130(k) for k = 1,2,3,...
a(4k-3) = 5*2^k - 6 = A020714(k) - 6 for k = 1,2,3...
a(4k-2) = 6*2^k - 6 = A007283(k+1) - 6 for k = 1,2,3, ...
a(4k-1) = 7*2^k - 6 = A048489(k) for k = 1,2,3...
(End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 2, -2).
FORMULA
a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5) for n >= 5, where a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8, a(4) = 10.
G.f.: -((2*(-1 - x - x^2 - x^3 + x^4))/(1 - x - 2*x^4 + 2*x^5)).
MAPLE
f:= gfun:-rectoproc({a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5),
a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8, a(4) = 10}, a(n), remember):
map(f, [$0..50]); # Robert Israel, Mar 25 2018
MATHEMATICA
LinearRecurrence[{1, 0, 0, 2, -2}, {2, 4, 8, 8, 10}, 40]
PROG
(GAP) a:=[2, 4, 6, 8, 10];; for n in [6..45] do a[n]:=a[n-1]+2*a[n-4]-2*a[n-5]; od; a; # Muniru A Asiru, Mar 22 2018
(PARI) x='x+O('x^99); Vec(2*(1+x+x^2+x^3-x^4)/(1-x-2*x^4+2*x^5)) \\ Altug Alkan, Mar 22 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 16 2017
EXTENSIONS
a(41)-a(49) from Muniru A Asiru, Mar 22 2018
STATUS
approved