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A279079 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/9) requires n steps to reach 0.

%I A279079 #8 Sep 08 2022 08:46:18
%S A279079 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 A279079 106,120,136,154,174,196,221,249,281,317,357,402,453,510,574,646,727,
%U A279079 818,921,1037,1167,1313,1478,1663,1871,2105,2369,2666,3000,3376,3799
%N A279079 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/9) requires n steps to reach 0.
%C A279079 Inspired by A278586.
%C A279079 Lim_{n->inf} a(n)/(9/8)^n = 5.19544896392362185906460915572195169945039729234281...
%F A279079 a(n) = floor(a(n-1)*9/8) + 1.
%e A279079   12 -> 12-ceiling(12/9) = 10,
%e A279079   10 -> 10-ceiling(10/9) = 8,
%e A279079    8 ->  8-ceiling(8/9)  = 7,
%e A279079    7 ->  7-ceiling(7/9)  = 6,
%e A279079 ...
%e A279079    1 ->  1-ceiling(1/9)  = 0,
%e A279079 so reaching 0 from 12 requires 10 steps;
%e A279079   13 -> 13-ceiling(13/9) = 11,
%e A279079   11 -> 11-ceiling(11/9) = 9,
%e A279079    9 ->  9-ceiling(9/9)  = 8,
%e A279079    8 ->  8-ceiling(8/9)  = 7,
%e A279079    7 ->  7-ceiling(7/9)  = 6,
%e A279079 ...
%e A279079    1 ->  1-ceiling(1/9)  = 0,
%e A279079 so reaching 0 from 13 (or more) requires 11 (or more) steps;
%e A279079 thus, 12 is the largest starting value from which 0 can be reached in 10 steps, so a(10) = 12.
%o A279079 (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 A279079 Cf. A278586.
%Y A279079 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 A279079 nonn
%O A279079 0,3
%A A279079 _Jon E. Schoenfield_, Dec 06 2016