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A279077
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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.
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5
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0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 26, 31, 37, 44, 52, 61, 72, 85, 100, 117, 137, 160, 187, 219, 256, 299, 349, 408, 477, 557, 650, 759, 886, 1034, 1207, 1409, 1644, 1919, 2239, 2613, 3049, 3558, 4152, 4845, 5653, 6596, 7696, 8979, 10476, 12223, 14261
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OFFSET
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0,3
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COMMENTS
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Limit_{n->oo} a(n)/(7/6)^n = 4.03710211215303193642791458111196922950551168987041...
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LINKS
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FORMULA
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a(n) = floor(a(n-1)*7/6) + 1.
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EXAMPLE
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10 -> 10-ceiling(10/7) = 8,
8 -> 8-ceiling(8/7) = 6,
6 -> 6-ceiling(6/7) = 5,
5 -> 5-ceiling(5/7) = 4,
4 -> 4-ceiling(4/7) = 3,
3 -> 3-ceiling(3/7) = 2,
2 -> 2-ceiling(2/7) = 1,
1 -> 1-ceiling(1/7) = 0,
so reaching 0 from 10 requires 8 steps;
11 -> 11-ceiling(11/7) = 9,
9 -> 9-ceiling(9/7) = 7,
7 -> 7-ceiling(7/7) = 6,
6 -> 6-ceiling(6/7) = 5,
5 -> 5-ceiling(5/7) = 4,
4 -> 4-ceiling(4/7) = 3,
3 -> 3-ceiling(3/7) = 2,
2 -> 2-ceiling(2/7) = 1,
1 -> 1-ceiling(1/7) = 0,
so reaching 0 from 11 (or more) requires 9 (or more) steps;
thus, 10 is the largest starting value from which 0 can be reached in 8 steps, so a(8) = 10.
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PROG
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(Magma) a:=[0]; aCurr:=0; for n in [1..53] do aCurr:=Floor(aCurr*7/6)+1; a[#a+1]:=aCurr; end for; a;
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CROSSREFS
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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, ...}?
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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