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%I #16 Jun 23 2017 00:59:20
%S 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,
%T 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,
%U 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
%N a(1) = 2; a(n) = a(floor(n/a(n-1))) + 1 for n > 1.
%C Least values of k such that a(k) = n are 1, 2, 6, 24, 120, 720, 5040, ... (n > 1).
%C These appear to be (n-1)!. Verified for 2 <= n <= 11. - _Robert Israel_, Jun 22 2017
%H Robert Israel, <a href="/A288914/b288914.txt">Table of n, a(n) for n = 1..10000</a>
%p f:= proc(n) option remember;
%p procname(floor(n/procname(n-1)))+1
%p end proc:
%p f(1):= 2:
%p map(f, [$1..200]);# _Robert Israel_, Jun 22 2017
%t a = {2}; Do[AppendTo[a, a[[Floor[n/a[[n - 1]] ] ]] + 1], {n, 2, 105}]; a (* _Michael De Vlieger_, Jun 21 2017 *)
%o (PARI) q=vector(10000); q[1]=2; for(n=2, #q, q[n] = q[n\q[n-1]]+1); q
%Y Cf. A000142, A002024, A130147.
%K nonn,easy
%O 1,1
%A _Altug Alkan_, Jun 19 2017
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