cp's OEIS Frontend

This constant can generate for all exponents k > 0 a prime number if the following conjecture is true: Let p be a prime > 2 and q = nexprime(p+1) then if there is always at least one prime inside the interval nextprime(p*q) to nextprime((p+1)*q)). But if this constant can generate prime numbers for all k, it is not directly a proof of this conjecture. If we would strengthen this further by omitting "nextprime" and allowing natural numbers for p and q, we will obtain essentially Oppermann's conjecture.

Conjecture: All terms are prime numbers. For details see A382260.

Inverse binomial transform gives the factorial numbers up to sign.

The main entry for this sequence of fractions is in A381931.

This is the main entry for this sequence of fractions.

Convergence and analytic continuation of this series representation are interesting research topics with many unsolved problems and open questions.

Evaluating the polynomial of row n P(x) = Sum_{k=1..n} x^(n+1-k)*A381932(n, k)/T(n, k) gives A144150(n+1, x-1)/(n+1)!.

Conjecture: All terms are powers of two.

Shifts left under COMPOSE transform with itself.

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.

A trivial but not lexicographically earliest solution for this problem is given by a(m) = ceiling(m/2) (A110654).

We assume a(<0) = 0 in this definition but conjecture that a(n) < n for n > 1.

If we iterate recursively k_{n+1} = a(k_{n})+k_{n} starting with k{0}=2, then we will observe a monotone increasing trajectory. An interesting consequence of the existence of this sequence is that if we start a trajectory at some k{m} > 2, such a trajectory would in the majority of cases ( or in all such cases? ) merge into the trajectory which is starting at 2.