OFFSET
1,1
COMMENTS
Let b(1)= 2, b(n+1)= b(n)^(1 + 1/b(n)) for n > 0 and c(n) = round(b(n)). Let d(n) = a(n) - c(n). The d(n) sequence has a very interesting course: at the beginning of zero and the long cycles of positive and negative terms.
Firoozbakht conjecture: prime(n+1) < prime(n)^(1+1/n) for all n.
Let p(n+1) = p(n)^(1 + y(n)/p(n)). Conjecture: y(n) = O(log p(n)). The average value of y(n) is 1.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
FORMULA
a(n+1) = a(n) + log a(n) + O(1). a(n) ~ n log n. - Charles R Greathouse IV, Jul 30 2012
a(n) = n log n + n log log n + O(n) so a(n) = prime(n) + O(n). - Thomas Ordowski, May 07 2013
PROG
(PARI) a=[2]; for(n=2, 100, a=concat(a, round(a[n-1]^(1+1/a[n-1])))); a \\ Colin Barker, Jul 22 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Ordowski, Jul 28 2012
STATUS
approved