

A279078


Maximum starting value of X such that repeated replacement of X with Xceiling(X/8) requires n steps to reach 0.


5



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, 121, 139, 159, 182, 209, 239, 274, 314, 359, 411, 470, 538, 615, 703, 804, 919, 1051, 1202, 1374, 1571, 1796, 2053, 2347, 2683, 3067, 3506, 4007, 4580, 5235, 5983, 6838
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OFFSET

0,3


COMMENTS

Inspired by A278586.
Limit_{n>oo} a(n)/(8/7)^n = 4.42210347959393228709604412445802201220907917744900...


LINKS

Table of n, a(n) for n=0..55.


FORMULA

a(n) = floor(a(n1)*8/7) + 1.


EXAMPLE

11 > 11ceiling(11/8) = 9,
9 > 9ceiling(9/8) = 7,
7 > 7ceiling(7/8) = 6,
6 > 6ceiling(6/8) = 5,
...
1 > 1ceiling(1/8) = 0,
so reaching 0 from 11 requires 9 steps;
12 > 12ceiling(12/8) = 10,
10 > 10ceiling(10/8) = 8,
8 > 8ceiling(8/8) = 7,
7 > 7ceiling(7/8) = 6,
...
1 > 1ceiling(1/8) = 0,
so reaching 0 from 12 (or more) requires 10 (or more) steps;
thus, 11 is the largest starting value from which 0 can be reached in 9 steps, so a(9) = 11.


PROG

(Magma) a:=[0]; aCurr:=0; for n in [1..55] do aCurr:=Floor(aCurr*8/7)+1; a[#a+1]:=aCurr; end for; a;


CROSSREFS

Cf. A278586.
See the following sequences for maximum starting value of X such that repeated replacement of X with Xceiling(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 Lsieve transform of {k1, 2k1, 3k1, ...}?
Sequence in context: A275673 A347327 A026445 * A308627 A330500 A030151
Adjacent sequences: A279075 A279076 A279077 * A279079 A279080 A279081


KEYWORD

nonn


AUTHOR

Jon E. Schoenfield, Dec 06 2016


STATUS

approved



