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A269986 Numbers k having factorial fractility A269982(k) = 4.

%I A269986 #18 Oct 06 2023 03:27:33
%S A269986 20,28,34,35,40,45,46,47,50,51,56,60,63,65,69,75,77,80,82,84,90,91,
%T A269986 102,110,112,116,117,120,123,124,133,135,144,147,148,150,152,156,159,
%U A269986 160,165,167,171,172,194,206,208,209,216,217,222,223,234,236,239,240
%N A269986 Numbers k having factorial fractility A269982(k) = 4.
%C A269986 See A269982 for a definition of factorial fractility and a guide to related sequences.
%H A269986 Robert Price, <a href="/A269986/b269986.txt">Table of n, a(n) for n = 1..116</a>
%e A269986 NI(1/20) = (3, 3, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, ...)
%e A269986 NI(5/20) = (2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...)
%e A269986 NI(6/20) = (2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...)
%e A269986 NI(10/20) = (2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...).
%e A269986 These 4 equivalence classes represent all the classes for n = 20, so the factorial fractility of 20 is 4.
%t A269986 A269982[n_] := CountDistinct[With[{l = NestWhileList[
%t A269986         Rescale[#, {1/(Floor[x] + 1)!, 1/Floor[x]!} /.
%t A269986            FindRoot[1/x! == #, {x, 1}]] &, #, UnsameQ, All]},
%t A269986      Min@l[[First@First@Position[l, Last@l] ;;]]] & /@
%t A269986    Range[1/n, 1 - 1/n, 1/n]]; (* _Davin Park_, Nov 19 2016 *)
%t A269986 Select[Range[2, 500], A269982[#] == 4 &] (* _Robert Price_, Sep 19 2019 *)
%o A269986 (PARI) select( is_A269986(n)=A269982(n)==4, [1..200]) \\ _M. F. Hasler_, Nov 05 2018
%Y A269986 Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269984, A269985, A269987, A269988 (numbers with factorial fractility 1, 2, ..., 6, respectively).
%Y A269986 Cf. A269570 (binary fractility), A270000 (harmonic fractility).
%K A269986 nonn
%O A269986 1,1
%A A269986 _Clark Kimberling_ and _Peter J. C. Moses_, Mar 11 2016
%E A269986 Edited by _M. F. Hasler_, Nov 05 2018