A279078 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/8) requires n steps to reach 0.
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
Offset: 0
Keywords
Examples
11 -> 11-ceiling(11/8) = 9, 9 -> 9-ceiling(9/8) = 7, 7 -> 7-ceiling(7/8) = 6, 6 -> 6-ceiling(6/8) = 5, ... 1 -> 1-ceiling(1/8) = 0, so reaching 0 from 11 requires 9 steps; 12 -> 12-ceiling(12/8) = 10, 10 -> 10-ceiling(10/8) = 8, 8 -> 8-ceiling(8/8) = 7, 7 -> 7-ceiling(7/8) = 6, ... 1 -> 1-ceiling(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.
Crossrefs
Cf. A278586.
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, ...}?
Programs
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Magma
a:=[0]; aCurr:=0; for n in [1..55] do aCurr:=Floor(aCurr*8/7)+1; a[#a+1]:=aCurr; end for; a;
Formula
a(n) = floor(a(n-1)*8/7) + 1.
Comments