|
| |
|
|
A279075
|
|
Maximum starting value of X such that repeated replacement of X with X-ceiling(X/5) requires n steps to reach 0.
|
|
6
|
|
|
|
0, 1, 2, 3, 4, 6, 8, 11, 14, 18, 23, 29, 37, 47, 59, 74, 93, 117, 147, 184, 231, 289, 362, 453, 567, 709, 887, 1109, 1387, 1734, 2168, 2711, 3389, 4237, 5297, 6622, 8278, 10348, 12936, 16171, 20214, 25268, 31586, 39483, 49354, 61693, 77117, 96397, 120497
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
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
|
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
|
|
|
|
STATUS
|
approved
|
| |
|
|
