



3, 6, 9, 13, 16, 20, 23, 26, 30, 33, 37, 40, 43, 47, 50, 54, 57, 61, 64, 67, 71, 74, 78, 81, 84, 88, 91, 95, 98, 102, 105, 108, 112, 115, 119, 122, 125, 129, 132, 136, 139, 142, 146, 149, 153, 156, 160, 163, 166, 170, 173, 177, 180, 183, 187, 190, 194, 197
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OFFSET

1,1


COMMENTS

Conjecture: a(n)/n > 2 + sqrt(2), and 0 < 2 + sqrt(2)  a(n)/n < 1 + sqrt(2) for n >= 1.
Proof of (a part of) Kimberling's conjecture. This is based on the formula for the recursion in the FORMULA section of A293077, where we use that the finalletterremoved version of the StringReplace map has the same fixed point A289035.
Let N1(n) be the number of 1's in the nth iterate theta(n) of the finalletterremoved mapping defined in A289035. It was observed in A293077 that N1(n) = L(n1)/2, where L(n) = A293077(n) is the length of the nth iterate theta(n) of the finalletterremoved mapping. As usual, proving a(n)/n > 2 + sqrt(2) is the same as proving that the frequency of 1's in A293077 is equal to 1/(2 + sqrt(2)) = 1  sqrt(2)/2 =: mu. Ignoring a proof of the existence of the limit, this means that we have to show that
N1(n) /L(n) > 1  sqrt(2)/2 as n>oo.
By the observation above this is the same as proving that
L(n1)/2L(n) > 1  sqrt(2)/2 as n>oo.
L(n) = 2 L(n1)  L(n2) + 2 floor(L(n2)/4).
For the determination of the limit we replace the term 2 floor(L(n2)/4) by L(n2)/2, as we may.
So we have L(n) ~ 2 L(n1)  L(n2)/2, which leads to
L(n)/2L(n+1) ~ 2 L(n1)/2L(n+1)  L(n2)/4L(n+1).
Replacing L(n1)/L(n+1) by L(n1)/2L(n)*2L(n)/L(n+1), and similarly for the last term, we obtain an equation for mu as n>oo:
mu = 4mu^2  2mu^3.
The solutions to this equation are 0, 1+sqrt(2)/2, and 1sqrt(2)/2. Since 0<mu<1, it follows that mu = 1  sqrt(2)/2.
Note: The full conjecture of Kimberling would follow from the conjecture A289035(n) = A171588(n3) for n>3. (End)
The arguments given in the proof above carry over to the positions of 1 in the sequences A288997, A289001, A289011, A289025, A289035, A289071, A289239, and A289242. It follows that for all these sequences x(n)/n > 2 + sqrt(2), where x(n) is the position of the nth 1 in these sequences.
Note that this does not apply to the sequence A289165 where the StringReplace map is not equal to a twoblock map because the 2block 11 occurs at position 85 in A289165. (End)


LINKS



MATHEMATICA

z = 10; (* number of iterates *)
s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n  1], {"00" > "0010", "01" > "010", "10" > "010"}]
TableForm[Table[w[n], {n, 0, 10}]]
st = ToCharacterCode[w[z]]  48 (* A289035 *)
Flatten[Position[st, 0]] (* A289036 *)
Flatten[Position[st, 1]] (* A289037 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



