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%I #11 Mar 20 2022 18:26:41
%S 0,1,2,3,4,5,6,7,9,11,13,15,18,21,25,29,34,39,45,52,60,69,79,91,105,
%T 121,139,159,182,209,239,274,314,359,411,470,538,615,703,804,919,1051,
%U 1202,1374,1571,1796,2053,2347,2683,3067,3506,4007,4580,5235,5983,6838
%N Maximum starting value of X such that repeated replacement of X with X-ceiling(X/8) requires n steps to reach 0.
%C Inspired by A278586.
%C Limit_{n->oo} a(n)/(8/7)^n = 4.42210347959393228709604412445802201220907917744900...
%F a(n) = floor(a(n-1)*8/7) + 1.
%e 11 -> 11-ceiling(11/8) = 9,
%e 9 -> 9-ceiling(9/8) = 7,
%e 7 -> 7-ceiling(7/8) = 6,
%e 6 -> 6-ceiling(6/8) = 5,
%e ...
%e 1 -> 1-ceiling(1/8) = 0,
%e so reaching 0 from 11 requires 9 steps;
%e 12 -> 12-ceiling(12/8) = 10,
%e 10 -> 10-ceiling(10/8) = 8,
%e 8 -> 8-ceiling(8/8) = 7,
%e 7 -> 7-ceiling(7/8) = 6,
%e ...
%e 1 -> 1-ceiling(1/8) = 0,
%e so reaching 0 from 12 (or more) requires 10 (or more) steps;
%e thus, 11 is the largest starting value from which 0 can be reached in 9 steps, so a(9) = 11.
%o (Magma) a:=[0]; aCurr:=0; for n in [1..55] do aCurr:=Floor(aCurr*8/7)+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), (this sequence) (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, ...}?
%K nonn
%O 0,3
%A _Jon E. Schoenfield_, Dec 06 2016
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