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URL: https://oeis.org/A282087

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A282087
Number of length-n ternary strings that do not contain both "00" and "11".
1
1, 3, 9, 27, 79, 229, 657, 1871, 5295, 14909, 41801, 116783, 325287, 903741, 2505377, 6932479, 19151519, 52833853, 145578265, 400705135, 1101936119, 3027902045, 8314284721, 22816209855, 62579270191, 171559358493, 470132335209, 1287861941487, 3526800739399
OFFSET
0,2
FORMULA
a(n) = 4*a(n-1) - a(n-2) - 6*a(n-3) - 2*a(n-4) for n >= 4 (derived in the math.stackexchange.com link).
From Colin Barker, Feb 07 2017: (Start)
a(n) = -(1-u)^(1+n)/2 - (1+u)^(1+n)/2 + (1-v)^n - (2*(1-v)^n)/v + (1+v)^n + (2*(1+v)^n) / v where u=sqrt(2) and v=sqrt(3).
G.f.: (1 + x)*(1 - 2*x) / ((1 - 2*x - x^2)*(1 - 2*x - 2*x^2)).
(End)
EXAMPLE
for n=5 the 229 acceptable ternary strings are all length 5 strings of '0', '1', and '2' _except_ '00011', '00110', '00111', '00112', '00211', '01100', '10011', '11000', '11001', '11002', '11100', '11200', '20011', '21100'.
MATHEMATICA
Table[3^n, {n, 0, 3}]~Join~LinearRecurrence[{4, -1, -6, -2}, {79, 229, 657, 1871}, 24] (* or *)
Table[Count[Tuples[Range[0, 2], n], w_ /; Boole[SequenceCount[w, {0, 0}] > 0] Boole[SequenceCount[w, {1, 1}] > 0] == 0], {n, 0, 12}] (* Michael De Vlieger, Feb 05 2017, latter program version 10.1 *)
PROG
(Python)
import itertools
# Not feasible on most machines for large numbers
def find_a_sub_n(n):
c = 0
for q in itertools.product(*([['0', '1', '2']]*n)):
h = ''.join(q)
if not (('11' in h) and ('00' in h)):
c = c+1
return c
(PARI) a(n)=([0, 1, 0, 0; 0, 0, 1, 0; 0, 0, 0, 1; -2, -6, -1, 4]^n*[1; 3; 9; 27])[1, 1] \\ Charles R Greathouse IV, Feb 05 2017
(PARI) Vec((1 + x)*(1 - 2*x) / ((1 - 2*x - x^2)*(1 - 2*x - 2*x^2)) + O(x^30)) \\ Colin Barker, Feb 07 2017
CROSSREFS
Cf. A078057 (number of length-n ternary strings that contain neither "00" nor "11").
Sequence in context: A269650 A266497 A291020 * A238440 A269578 A026289
KEYWORD
nonn,easy
AUTHOR
Daniel T. Martin, Feb 05 2017
STATUS
approved