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A270803
Formal inverse of Thue-Morse sequence A010060.
2
0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
OFFSET
0
COMMENTS
Conjecture: a(n) = A151666(floor((n+1)/2)) for n > 0. - Georg Fischer, Nov 29 2022
Conjecture confirmed. See the attached file. The idea is to guess the automaton for this sequence (in base 2) and then verify that it satisfies the identities in the Gawron-Ulas paper. Next we make a (base 2) automaton for A151666. This is easy because it is just the numbers consisting only of 0 and 1 in base 4. Then finally we assert Fischer's identity and Walnut returns TRUE. - Jeffrey Shallit, Nov 30 2022
LINKS
Maciej Gawron and Maciej Ulas, On formal inverse of the Prouhet-Thue-Morse sequence, Discrete Mathematics 339.5 (2016): 1459-1470. Also arXiv:1601.04840 [math.CO], 2016.
FORMULA
a(0)=0, a(1)=a(2)=1, a(3)=0; thereafter
if n mod 4 = 0 then a(n) = a(n-1),
if n mod 4 = 1 then a(n) = a(n-2),
if n mod 4 = 2 then a(n) = a(n-3),
otherwise a(n) = (a(n-4) + a((n-3)/4)) mod 2.
a(n) = A001002(n) mod 2 for n > 0. - John M. Campbell, Jul 17 2016
MAPLE
A270803 := proc(n)
option remember;
if n <=3 then
op(n+1, [0, 1, 1, 0]) ;
else
if n mod 4 = 0 then
procname(n-1)
elif n mod 4 = 1 then
procname(n-2)
elif n mod 4 = 2 then
procname(n-3)
else
(procname(n-4)+procname((n-3)/4)) mod 2;
end if;
end if;
end proc:
seq(A270803(n), n=0..120) ;
MATHEMATICA
a[n_] := a[n] = Which[n <= 3, {0, 1, 1, 0}[[n + 1]], Mod[n, 4] == 0, a[n - 1], Mod[n, 4] == 1, a[n - 2], Mod[n, 4] == 2, a[n - 3], True, Mod[a[n - 4] + a[(n - 3)/4], 2]];
Table[a[n], {n, 0, 120}] (* Jean-François Alcover, Nov 27 2017, from Maple *)
CROSSREFS
See A270804 for positions of 1's
Sequence in context: A095130 A284789 A288736 * A030301 A316341 A284901
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Apr 02 2016
STATUS
approved