A279080 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/10) requires n steps to reach 0.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 19, 22, 25, 28, 32, 36, 41, 46, 52, 58, 65, 73, 82, 92, 103, 115, 128, 143, 159, 177, 197, 219, 244, 272, 303, 337, 375, 417, 464, 516, 574, 638, 709, 788, 876, 974, 1083, 1204, 1338, 1487, 1653, 1837, 2042, 2269

Offset: 0

Author

Jon E. Schoenfield, Dec 06 2016

Keywords

• nonn

Comments

Inspired by A278586.

Limit_{n->oo} a(n)/(10/9)^n = 5.60655601136196116133057876294687807265035051745268...

Examples

			  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.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

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, ...}?

Programs

Magma

a:=[0]; aCurr:=0; for n in [1..57] do aCurr:=Floor(aCurr*10/9)+1; a[#a+1]:=aCurr; end for; a;

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 *)

Formula

a(n) = floor(a(n-1)*10/9) + 1.