cp's OEIS Frontend

There is a second version of this sequence possible if we change the definition to a(1) = 2 and a(n) > 1, then the sequence will start 2, 4, 5, 8, 10, 7, 14, ... . It will after this continue in the same way as our actual sequence does (and would also extend the valid range of the recurrence formulas).

Start a(1) = 2 and value 1 allowed is A257794.

While extending the sequence at a(k) we will check if k equals a previous term in the sequence. If such a term a(m) = k is found a(k) is determined as a(k) = a(m)*m. If no previous term matches k we may choose a(k) = k+c with the least c such that c > 0 and k+c does not equal any previous term in the sequence. It is conjectured that this check is sufficient. Reasoning behind this conjecture:

The greatest common divisor of two consecutive Fibonacci numbers is 1, thus we know that (k-1)^F(m)*k^F(m+1) and (t-1)^F(n)*t^F(n+1) are all different for some m,n > 1 if k and t are chosen such that for m or n < 2 no solution for (k-1)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) exist, because this cannot be equal if (k-1)*k and (t-1)*t have different prime numbers as divisors and if the only difference is the exponent of the prime factors, then the distribution of these between (t-1) and t and thus their progression F(n) or F(n+1) is individually distinct. In this sequence we need also to consider the more general case (k-c)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) because sometimes we need to set a(k) = k+c. It is conjectured that in this case c is bounded to be < 3.