OFFSET
1,1
COMMENTS
Conjecture: let r(z)= (1/2) *(z+sqrt(z^2+4*z)), for any integral z>=1. Then the sequence a(n)-4n (where a(n) is the sequence of odd numbers m such that the sequence defined by b(1) = m; for k>1, b(k) = floor(r(z)*b(k-1)) contains only odd numbers) becomes ultimately periodic. Benoit Cloitre, Aug 10 2003
FORMULA
Observation: a(n+1)-a(n) = 2, 4 or 6 for every n, a(n)=4n+O(1) and more precisely it seems that abs(a(n)-4n)<=9. Is the sequence a(n)-4n ultimately periodic ? Benoit Cloitre, Aug 10 2003
EXAMPLE
For m = 5 we get 5, 13, 35, 95, 259, 707, 1931, 5275, 14411, 39371, ... (cf. A057960).
CROSSREFS
KEYWORD
nonn
AUTHOR
Philippe Deléham, Aug 09 2003
STATUS
approved