cp's OEIS Frontend

Each term of the Q-sequence is the sum of two previous terms. This sequence gives the Q-sequence terms which appear never to be used as one of those two. In other words, a(n) != x - A005185(x-1), and a(n) != x - A005185(x-2).

The first 65 generations of A306211 were computed explicitly. Since the tail of each generation is the RUNS of the tail of the previous generation, if we start with a complete generation then we can get just the tail of some further generations by iterating RUNS on it. This allowed the computation of the remaining 55 terms.

Generations 11, 16, 25, 42, and 75 end in a 4. The number of generations in between those which end in a 4 is thus 4, 8, 16, 32.

If A277558 is a permutation, this is the full inverse of it.

After 10^11 terms of A277558, the smallest number which has not appeared is 609790506. The largest number in the first 600 million terms of this sequence is a(597249348) = 97840303230.

Is it ever impossible to extend the sequence -- meaning there is no number less than a(n-1)+n which has not appeared?

After 10^11 terms, the smallest number which has not appeared is 609790506.

If A274648 is proved to be a permutation, then this is the full inverse of it.

The "curling number" k = k(S) of a string of numbers S = s(1), ..., s(m) is defined as follows. Write S as XY^k for strings X and Y (where Y has positive length) and k is maximized, i.e. k = the maximal number of repeating blocks at the end of S.

The "tail length" tail(S) of S is defined as follows: start with S and repeatedly append the curling number (recomputing it at each step) until a 1 is reached; tail(S) is the number of terms that are appended to S before a 1 is reached.

No example is known where both tail(2 S) > tail(S) and tail(3 S) > tail(S) hold.

See A216730 for definition.

The "curling number" k = k(S) of a string of numbers S = s(1), ..., s(m) is defined as follows. Write S as XY^k for strings X and Y (where Y has positive length) and k is maximized, i.e., k = the maximal number of repeating blocks at the end of S.

The "tail length" t(S) of S is defined as follows: start with S and repeatedly append the curling number (recomputing it at each step) until a 1 is reached; t(S) is the number of terms that are appended to S before a 1 is reached.

If a 1 is never reached, set t(S)=oo (the Curling Number Conjecture says this will never happen).

A sequence S in {2,3}* is called "rotten" if either of t(2S) or t(3S) (or both) is strictly less than t(S).

Example: S = 32323 has curling number k=2, so we get 323232; now k=3, so we get 3232323; now k=3, so we get 32323233; now k=2, so we get 323232332; now k=1 so we stop. We added 4 terms before reaching 1, so t(S)=4.

On the other hand, 2S = 232323 only extends to 232323321..., so t(2S)=2 which means S is rotten.

The subword 01010101 (corresponding to X = 01) for example cannot occur.

The next three terms are conjectured to be 68, 76 and 77.