A286940 Length of the word appended at the n-th step when trying to construct the earliest infinite cubefree word A282317 by appending the "incremented" longest possible subsequence A282317(0..a(n)-1) with the last term set to 1, for all n > 1.

1, 1, 1, 3, 2, 7, 15, 13, 1, 42, 39, 42, 39, 42, 28, 1, 42, 39, 42, 39, 42, 28, 1, 42, 39, 42, 39, 35, 7, 32, 7, 35, 7, 32, 7, 35, 7, 32, 4, 39, 42, 39, 42, 39, 35, 7, 32, 7, 35, 7, 32, 7, 35, 7, 32, 4, 39, 42, 39, 42, 39, 35, 7, 32, 7, 35, 7, 32, 7, 35, 7, 32, 4, 28, 1, 42, 39, 42, 39

Offset: 0

Author

M. F. Hasler, May 20 2017

Keywords

• nonn

Comments

The initial a(0) = 1 corresponds to the initial term S(0) = 0, and a(1) = 1 refers to the 1-element sequence S(0) = 0 which is appended at the first step (S = A282317). This is the only instance where the appended chunk does not end in 1. All other chunks end in 1, i.e., A282317(Sum_{k=1..n} a(k)) = 1 for all n > 1.

We can see a pseudo-periodic pattern arise in this sequence (cf. Example), but it cannot be periodic, and must even be cubefree*, else S would not be cubefree. (*Here and below we ignore the initial terms.)

It also seems that a(n) is in { 1, 4, 7, 28, 32, 35, 39, 42 } for all n > 7, and furthermore, that 1 is always preceded by 28 and followed by (42, 39, 42, 39), and similarly 4 is always preceded by the same 15 elements (42, ..., 32). (Observation based on 10^4 terms of S, under further investigation.)

So far it is not proved that this algorithm never produces a maximal cubefree word (cf. A282133), then necessarily of the form S.S'.1 where S' is an initial segment of S), and then is unable to continue. At least this does not happen before reaching 2*10^4 terms.

Examples

			Start with S = (0); we set a(0) = 1.
Append W = S, of length a(1) = 1, to yields S = (0, 0).
Now we can't append W (or even S) again, "increment" it to get W = (1) of length a(2) = 1, and S = (0, 0, 1).
Append W = S, of length a(3) = 3, to yields S = (0, 0, 1, 0, 0, 1).
We can't append W (or even S) again, a cube would appear at the 3rd appended digit which is already 1. So we must "increment" this (discarding all from the 3rd digit on) to get W = (0, 1) of length a(4) = 2, to yields S = (0, 0, 1, 0, 0, 1, 0, 1).
If we try to append W = S we see that there is a cube ending at the penultimate digit, namely three times (0, 1, 0). So we must increase this 7th digit of S to get W = (0, 0, 1, 0, 0, 1, 1) of length a(5) = 7, and so on.
The sequence starts
1, 1, 1, 3, 2, 7, 15, 13,
1, 42, 39, 42, 39, 42, 28, 1, 42, 39, 42, 39, 42, 28,            /* =: A,A */
1, 42, 39, 42, 39, 35, 7, 32, 7, 35, 7, 32, 7, 35, 7, 32, 4,     /* =: 1,B */
39, 42, 39, 42, 39, 35, 7, 32, 7, 35, 7, 32, 7, 35, 7, 32, 4,    /* = 39,B */
39, 42, 39, 42, 39, 35, 7, 32, 7, 35, 7, 32, 7, 35, 7, 32, 4, 28 /* = 39,B,28 */.
Then this repeats: A,A, 1,B, 39,B, 39,B,28,   /* =: C */
        and again: A,A, 1,B, 39,B, 39,B',28,  /* =: C' */
where B' is B with its last two elements removed.
Then it continues again C, C, C', C, C, C", C, C, C', etc.
I conjecture that this goes on with C, C', C"... appearing in a "ternary ruler sequence" pattern (A007949) until the length of the successively truncated B-part reaches length 0. This would be after approximately 12*10^6 terms of the sequence S.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 0..742 (Lengths of words until the total length exceeds 2*10^4.)

Programs

PARI

#A282317_vec(2e4,2^4) \\ verbosity flag 2^4 does print1(#w",") near a=concat(a,w) in the code for A282317.

PARI

A286940_vec(Nmin)={my(a=[1,1,1,3,2,7,15,13], A=[1,42,39,42,39,42,28], B=concat(A[2..5],[35,7,32,7,35,7,32,7,35,7,32,4]), C=concat([A,A,1,B,39,B,39,B]),n=0);while(#a