OFFSET
0,3
COMMENTS
Inspired by A278586.
Limit_{n->oo} a(n)/(10/9)^n = 5.60655601136196116133057876294687807265035051745268...
LINKS
Robert Israel, Table of n, a(n) for n = 0..10000
FORMULA
a(n) = floor(a(n-1)*10/9) + 1.
EXAMPLE
13 -> 13-ceiling(13/10) = 11,
11 -> 11-ceiling(11/10) = 9,
9 -> 9-ceiling(9/10) = 8,
8 -> 8-ceiling(8/10) = 7,
...
1 -> 1-ceiling(1/10) = 0,
so reaching 0 from 13 requires 11 steps;
14 -> 14-ceiling(14/10) = 12,
12 -> 12-ceiling(12/10) = 10,
10 -> 10-ceiling(10/10) = 9,
9 -> 9-ceiling(9/10) = 8,
8 -> 8-ceiling(8/10) = 7,
...
1 -> 1-ceiling(1/10) = 0,
so reaching 0 from 14 (or more) requires 12 (or more) steps;
thus, 13 is the largest starting value from which 0 can be reached in 11 steps, so a(11) = 13.
MAPLE
H:= proc(y) local u, v;
v:= -y-1 mod 9+1;
(10*y+v)/9
end proc:
A:= Array(0..100):
A[0]:= 0:
for i from 1 to 100 do A[i]:= H(A[i-1]) od:
convert(A, list); # Robert Israel, Jun 23 2020
MATHEMATICA
With[{s = Array[-1 + Length@ NestWhileList[# - Ceiling[#/10] &, #, # > 0 &] &, 2400, 0]}, Array[-1 + Position[s, #][[-1, 1]] &, Max@ s, 0]] (* Michael De Vlieger, Jun 23 2020 *)
PROG
(Magma) a:=[0]; aCurr:=0; for n in [1..57] do aCurr:=Floor(aCurr*10/9)+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), A279076 (k=6), A279077 (k=7), A279078 (k=8), A279079 (k=9), (this sequence) (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