OFFSET
0,3
COMMENTS
Inspired by A278586.
Limit_{n->oo} a(n)/(6/5)^n = 3.24387249751177521384734853905517802618171089570674...
LINKS
FORMULA
a(n) = floor(a(n-1)*6/5) + 1.
EXAMPLE
7 -> 7-ceiling(7/6) = 5,
5 -> 5-ceiling(5/6) = 4,
4 -> 4-ceiling(4/6) = 3,
3 -> 3-ceiling(3/6) = 2,
2 -> 2-ceiling(2/6) = 1,
1 -> 1-ceiling(1/6) = 0,
so reaching 0 from 7 requires 6 steps;
8 -> 8-ceiling(8/6) = 6,
6 -> 6-ceiling(6/6) = 5,
5 -> 5-ceiling(5/6) = 4,
4 -> 4-ceiling(4/6) = 3,
3 -> 3-ceiling(3/6) = 2,
2 -> 2-ceiling(2/6) = 1,
1 -> 1-ceiling(1/6) = 0,
so reaching 0 from 8 (or more) requires 7 (or more) steps;
thus, 7 is the largest starting value from which 0 can be reached in 6 steps, so a(6) = 7.
PROG
(Magma) a:=[0]; aCurr:=0; for n in [1..51] do aCurr:=Floor(aCurr*6/5)+1; a[#a+1]:=aCurr; end for; a;
CROSSREFS
Cf. A278586.
See the following sequences for maximum starting value of X such that repeated replacement of X with X-ceiling(X/k) requires n steps to reach 0: A000225 (k=2), A006999 (k=3), A155167 (k=4, apparently; see Formula entry there), A279075 (k=5), (this sequence) (k=6), A279077 (k=7), A279078 (k=8), A279079 (k=9), A279080 (k=10). For each of these values of k, is the sequence the L-sieve transform of {k-1, 2k-1, 3k-1, ...}?
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Dec 06 2016
STATUS
approved