OFFSET
1,1
COMMENTS
For n > 2, a(n) is not the smallest k such that prime(n) divides A090252(k), but it is the smallest k such that prime(n) divides both A090252(k) and A090252(2*k+1). If k_(0) = a(n) we may find either an infinite or finite range of indices where prime(n) divides A090252 using the recurrence k_(n) = 2*k_(n-1)+1, but there is a caveat: in very rare cases, some k values of this recurrence may be wrong by +-1, and the next iteration will then fit again. This uncertainty is caused by the fact that two terms of A090252 will be governed by the same floor(n/2) history. For yet unknown reasons, there may be an upper limit where such a recurrence may break.
This works because in A090252 the number of primes which do not divide the last floor(n/2) terms is growing faster than they are used up by this sequence. For each prime p then there exists an index k into A090252 where the supply of unused factors is so large that, when p becomes coprime to the last floor(n/2) terms, we can always immediately find a matching second prime to build a yet-unused semiprime or use p as a yet-unused power of itself.
LINKS
Michael S. Branicky, Table of n, a(n) for n = 1..476
FORMULA
EXAMPLE
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Scheuerle, Jun 22 2022
EXTENSIONS
STATUS
approved