A289037 Positions of 1 in A289035; complement of A289036.

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

Offset: 1

Author

Clark Kimberling, Jun 27 2017

Keywords

• nonn
• easy

Comments

Conjecture: a(n)/n -> 2 + sqrt(2), and 0 < 2 + sqrt(2) - a(n)/n < -1 + sqrt(2) for n >= 1.

From Michel Dekking, Mar 20 2022: (Start)

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 final-letter-removed version of the String-Replace map has the same fixed point A289035.

Let N1(n) be the number of 1's in the n-th iterate theta(n) of the final-letter-removed mapping defined in A289035. It was observed in A293077 that N1(n) = L(n-1)/2, where L(n) = A293077(n) is the length of the n-th iterate theta(n) of the final-letter-removed 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(n-1)/2L(n) -> 1 - sqrt(2)/2 as n->oo.

We know from A293077 that

L(n) = 2 L(n-1) - L(n-2) + 2 floor(L(n-2)/4).

For the determination of the limit we replace the term 2 floor(L(n-2)/4) by L(n-2)/2, as we may.

So we have L(n) ~ 2 L(n-1) - L(n-2)/2, which leads to

L(n)/2L(n+1) ~ 2 L(n-1)/2L(n+1) - L(n-2)/4L(n+1).

Replacing L(n-1)/L(n+1) by L(n-1)/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 1-sqrt(2)/2. Since 0

Note: The full conjecture of Kimberling would follow from the conjecture A289035(n) = A171588(n-3) for n>3. (End)

Conjecture: a(n+1) - a(n) = A276864(n) for all n. - Michel Dekking, Mar 20 2022

From Michel Dekking, Mar 22 2022: (Start)

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 n-th 1 in these sequences.

Note that this does not apply to the sequence A289165 where the String-Replace map is not equal to a two-block map because the 2-block 11 occurs at position 85 in A289165. (End)

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A289035, A289036.

Programs

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 *)

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