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Maximum starting value of X such that repeated replacement of X with X-ceiling(X/7) requires n steps to reach 0.
5

%I #9 Mar 20 2022 18:26:30

%S 0,1,2,3,4,5,6,8,10,12,15,18,22,26,31,37,44,52,61,72,85,100,117,137,

%T 160,187,219,256,299,349,408,477,557,650,759,886,1034,1207,1409,1644,

%U 1919,2239,2613,3049,3558,4152,4845,5653,6596,7696,8979,10476,12223,14261

%N Maximum starting value of X such that repeated replacement of X with X-ceiling(X/7) requires n steps to reach 0.

%C Inspired by A278586.

%C Limit_{n->oo} a(n)/(7/6)^n = 4.03710211215303193642791458111196922950551168987041...

%F a(n) = floor(a(n-1)*7/6) + 1.

%e 10 -> 10-ceiling(10/7) = 8,

%e 8 -> 8-ceiling(8/7) = 6,

%e 6 -> 6-ceiling(6/7) = 5,

%e 5 -> 5-ceiling(5/7) = 4,

%e 4 -> 4-ceiling(4/7) = 3,

%e 3 -> 3-ceiling(3/7) = 2,

%e 2 -> 2-ceiling(2/7) = 1,

%e 1 -> 1-ceiling(1/7) = 0,

%e so reaching 0 from 10 requires 8 steps;

%e 11 -> 11-ceiling(11/7) = 9,

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

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

%e 6 -> 6-ceiling(6/7) = 5,

%e 5 -> 5-ceiling(5/7) = 4,

%e 4 -> 4-ceiling(4/7) = 3,

%e 3 -> 3-ceiling(3/7) = 2,

%e 2 -> 2-ceiling(2/7) = 1,

%e 1 -> 1-ceiling(1/7) = 0,

%e so reaching 0 from 11 (or more) requires 9 (or more) steps;

%e thus, 10 is the largest starting value from which 0 can be reached in 8 steps, so a(8) = 10.

%o (Magma) a:=[0]; aCurr:=0; for n in [1..53] do aCurr:=Floor(aCurr*7/6)+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), (this sequence) (k=7), A279078 (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