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A288914
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a(1) = 2; a(n) = a(floor(n/a(n-1))) + 1 for n > 1.
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3
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2, 3, 3, 3, 3, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 4, 4, 4, 4, 5, 5, 5, 4, 5, 4, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
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OFFSET
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1,1
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COMMENTS
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Least values of k such that a(k) = n are 1, 2, 6, 24, 120, 720, 5040, ... (n > 1).
These appear to be (n-1)!. Verified for 2 <= n <= 11. - Robert Israel, Jun 22 2017
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LINKS
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MAPLE
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f:= proc(n) option remember;
procname(floor(n/procname(n-1)))+1
end proc:
f(1):= 2:
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MATHEMATICA
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a = {2}; Do[AppendTo[a, a[[Floor[n/a[[n - 1]] ] ]] + 1], {n, 2, 105}]; a (* Michael De Vlieger, Jun 21 2017 *)
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PROG
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(PARI) q=vector(10000); q[1]=2; for(n=2, #q, q[n] = q[n\q[n-1]]+1); q
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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