OFFSET
0,1
COMMENTS
Conjecture 1: Let m(n) be the position in A268865 corresponding to a(n). Then m(n) = (2/3)*(4^n-1), if n is even, and m(n) = (2/3)*(4^(n-1)-1) + 3*4^(n-1), if n is odd.
Conjecture 2: a(n) = 2*m(n) + 2, if n is even, and a(n) = (7*m+12)/11, if n is odd.
If Conjectures 1 and 2 are true, then we easily have for even n >= 0, m(n) == 0 (mod 10), a(n) == 2 (mod 10), while a(1) = m(1) = 3 and for odd n >= 3, m(n), a(n) == 8 (mod 10). The sequence {m(n)} begins: 0, 3, 10, 58, 170, 938, 2730, 15018, 43690, ...
LINKS
Jeffrey Shallit, Sonja Linghui Shan, and Kai Hsiang Yang, Automatic Sequences in Negative Bases and Proofs of Some Conjectures of Shevelev, arXiv:2208.06025 [cs.FL], 2022.
Vladimir Shevelev, Two analogs of Thue-Morse sequence, arXiv:1603.04434 [math.NT], 2016.
MATHEMATICA
nn = 10^5; f[n_, b_] := Most@ Mod[NestWhileList[-(#1 - Mod[#1, b])/b &, n, #1 != 0 &], b]; Union@ FoldList[Max, Array[(k = 0; While[Mod[Total@ f[k, 2], 2] == ThueMorse[# + k], k++]; k) &, nn - 1, 0]] (* Michael De Vlieger, Aug 25 2022 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Vladimir Shevelev and Peter J. C. Moses, Feb 15 2016
STATUS
approved