%I #8 Sep 08 2022 08:46:18

%S 0,1,2,3,4,5,6,7,8,10,12,14,16,19,22,25,29,33,38,43,49,56,64,73,83,94,

%T 106,120,136,154,174,196,221,249,281,317,357,402,453,510,574,646,727,

%U 818,921,1037,1167,1313,1478,1663,1871,2105,2369,2666,3000,3376,3799

%N Maximum starting value of X such that repeated replacement of X with X-ceiling(X/9) requires n steps to reach 0.

%C Inspired by A278586.

%C Lim_{n->inf} a(n)/(9/8)^n = 5.19544896392362185906460915572195169945039729234281...

%F a(n) = floor(a(n-1)*9/8) + 1.

%e 12 -> 12-ceiling(12/9) = 10,

%e 10 -> 10-ceiling(10/9) = 8,

%e 8 -> 8-ceiling(8/9) = 7,

%e 7 -> 7-ceiling(7/9) = 6,

%e ...

%e 1 -> 1-ceiling(1/9) = 0,

%e so reaching 0 from 12 requires 10 steps;

%e 13 -> 13-ceiling(13/9) = 11,

%e 11 -> 11-ceiling(11/9) = 9,

%e 9 -> 9-ceiling(9/9) = 8,

%e 8 -> 8-ceiling(8/9) = 7,

%e 7 -> 7-ceiling(7/9) = 6,

%e ...

%e 1 -> 1-ceiling(1/9) = 0,

%e so reaching 0 from 13 (or more) requires 11 (or more) steps;

%e thus, 12 is the largest starting value from which 0 can be reached in 10 steps, so a(10) = 12.

%o (Magma) a:=[0]; aCurr:=0; for n in [1..56] do aCurr:=Floor(aCurr*9/8)+1; a[#a+1]:=aCurr; end for; a;

%Y Cf. A278586.

%Y 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), (this sequence) (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, ...}?

%K nonn

%O 0,3

%A _Jon E. Schoenfield_, Dec 06 2016