OFFSET
0,3
COMMENTS
Inspired by A278586.
Limit_{n->oo} a(n)/(5/4)^n = 2.68723058270145442816383476567331957329199286146873...
LINKS
FORMULA
a(n) = floor(a(n-1)*5/4) + 1.
EXAMPLE
8 -> 8-ceiling(8/5) = 6,
6 -> 6-ceiling(6/5) = 4,
4 -> 4-ceiling(4/5) = 3,
3 -> 3-ceiling(3/5) = 2,
2 -> 2-ceiling(2/5) = 1,
1 -> 1-ceiling(1/5) = 0,
so reaching 0 from 8 requires 6 steps;
9 -> 9-ceiling(9/5) = 7,
7 -> 7-ceiling(7/5) = 5,
5 -> 5-ceiling(5/5) = 4,
4 -> 4-ceiling(4/5) = 3,
3 -> 3-ceiling(3/5) = 2,
2 -> 2-ceiling(2/5) = 1,
1 -> 1-ceiling(1/5) = 0,
so reaching 0 from 9 (or more) requires 7 (or more) steps;
thus, 8 is the largest starting value from which 0 can be reached in 6 steps, so a(6) = 8.
MATHEMATICA
RecurrenceTable[{a[1] == 0, a[n] == Floor[a[n-1] 5/4] + 1}, a, {n, 50}] (* Vincenzo Librandi, Dec 06 2016 *)
PROG
(Magma) a:=[0]; aCurr:=0; for n in [1..48] do aCurr:=Floor(aCurr*5/4)+1; a[#a+1]:=aCurr; end for; a;
(Magma) [n eq 1 select n-1 else Floor(Self(n-1)*5/4)+1: n in [1..70]]; // Vincenzo Librandi, Dec 06 2016
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), (this sequence) (k=5), A279076 (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