%I #23 Mar 20 2022 17:24:43
%S 0,1,2,3,4,6,8,11,14,18,23,29,37,47,59,74,93,117,147,184,231,289,362,
%T 453,567,709,887,1109,1387,1734,2168,2711,3389,4237,5297,6622,8278,
%U 10348,12936,16171,20214,25268,31586,39483,49354,61693,77117,96397,120497
%N Maximum starting value of X such that repeated replacement of X with X-ceiling(X/5) requires n steps to reach 0.
%C Inspired by A278586.
%C Limit_{n->oo} a(n)/(5/4)^n = 2.68723058270145442816383476567331957329199286146873...
%H A.H.M. Smeets, <a href="/A279075/a279075.txt">Constants related to repeated replacement of X with X-ceiling(X/d)</a>
%F a(n) = floor(a(n-1)*5/4) + 1.
%e 8 -> 8-ceiling(8/5) = 6,
%e 6 -> 6-ceiling(6/5) = 4,
%e 4 -> 4-ceiling(4/5) = 3,
%e 3 -> 3-ceiling(3/5) = 2,
%e 2 -> 2-ceiling(2/5) = 1,
%e 1 -> 1-ceiling(1/5) = 0,
%e so reaching 0 from 8 requires 6 steps;
%e 9 -> 9-ceiling(9/5) = 7,
%e 7 -> 7-ceiling(7/5) = 5,
%e 5 -> 5-ceiling(5/5) = 4,
%e 4 -> 4-ceiling(4/5) = 3,
%e 3 -> 3-ceiling(3/5) = 2,
%e 2 -> 2-ceiling(2/5) = 1,
%e 1 -> 1-ceiling(1/5) = 0,
%e so reaching 0 from 9 (or more) requires 7 (or more) steps;
%e thus, 8 is the largest starting value from which 0 can be reached in 6 steps, so a(6) = 8.
%t RecurrenceTable[{a[1] == 0, a[n] == Floor[a[n-1] 5/4] + 1}, a, {n, 50}] (* _Vincenzo Librandi_, Dec 06 2016 *)
%o (Magma) a:=[0]; aCurr:=0; for n in [1..48] do aCurr:=Floor(aCurr*5/4)+1; a[#a+1]:=aCurr; end for; a;
%o (Magma) [n eq 1 select n-1 else Floor(Self(n-1)*5/4)+1: n in [1..70]]; // _Vincenzo Librandi_, Dec 06 2016
%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), (this sequence) (k=5), A279076 (k=6), A279077 (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