A279079 Maximum starting value of X such that repeated replacement of X with X-ceiling(X/9) requires n steps to reach 0.
0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 19, 22, 25, 29, 33, 38, 43, 49, 56, 64, 73, 83, 94, 106, 120, 136, 154, 174, 196, 221, 249, 281, 317, 357, 402, 453, 510, 574, 646, 727, 818, 921, 1037, 1167, 1313, 1478, 1663, 1871, 2105, 2369, 2666, 3000, 3376, 3799
Offset: 0
Keywords
Examples
12 -> 12-ceiling(12/9) = 10, 10 -> 10-ceiling(10/9) = 8, 8 -> 8-ceiling(8/9) = 7, 7 -> 7-ceiling(7/9) = 6, ... 1 -> 1-ceiling(1/9) = 0, so reaching 0 from 12 requires 10 steps; 13 -> 13-ceiling(13/9) = 11, 11 -> 11-ceiling(11/9) = 9, 9 -> 9-ceiling(9/9) = 8, 8 -> 8-ceiling(8/9) = 7, 7 -> 7-ceiling(7/9) = 6, ... 1 -> 1-ceiling(1/9) = 0, so reaching 0 from 13 (or more) requires 11 (or more) steps; thus, 12 is the largest starting value from which 0 can be reached in 10 steps, so a(10) = 12.
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), A279078 (k=8), (this sequence) (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..56] do aCurr:=Floor(aCurr*9/8)+1; a[#a+1]:=aCurr; end for; a;
Formula
a(n) = floor(a(n-1)*9/8) + 1.
Comments