Hello all,
What do you think of the following sequence?
a(n) is the smallest positive integer not present as a substring of any term of the first n terms of the Fibonacci sequence:
[2, 2, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 47, 47, 47, 47, 47, 47, 66, 66, 66, 66, 66, 66, 66, 66, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 104, 104, 104, 104, 104, 104, 104]
Example: a(9) = 6 because the first 9 terms of the Fibonacci sequence are [1,1,2,3,5,8,13,21,34] and 1, 2, 3, 4, and 5 all appear as substrings of at least one term in this subset of Fibonacci sequence.
This sequence grows very slowly (a(10000) = 10010). Is it infinite?
If we concatenate the terms of the Fibonacci sequence, we instead get:
[2, 2, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 15, 15, 16, 16, 16, 17, 17, 19, 19, 19, 19, 19, 20, 20, 20, 24, 24, 24, 27, 27, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 38, 47, 47, 47, 47, 47, 47, 47, 66, 66, 66, 66, 66, 66, 66, 66, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 82, 100, 100, 100, 103, 103, 103, 104, 104, 104, 104, 104, 104, 104, 104, 104, 104]
Note that 11 is no longer in the sequence because the concatenated Fibonacci sequence is
112358132134... and '11' is present in the first two digits.
This could also be extended to other sequences. For example, the digits of Pi:
[2, 2, 2, 2, 2, 2, 6, 7, 7, 7, 7, 7, 7, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12]
Do you think sequences of this type merit inclusion in the OEIS? I've searched for several and found nothing like this there.
Thanks,
Dave Consiglio