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A269984 Numbers k having factorial fractility A269982(k) = 2.

%I A269984 #24 Oct 06 2023 03:28:27
%S A269984 4,5,8,9,12,14,16,18,22,23,24,26,27,32,33,37,38,39,48,49,53,54,57,58,
%T A269984 61,64,66,78,81,83,86,87,96,97,101,107,113,114,121,129,131,139,163,
%U A269984 169,174,178,181,193,218,227,241,257,263,267,277,302,317,327,331
%N A269984 Numbers k having factorial fractility A269982(k) = 2.
%C A269984 See A269982 for a definition of factorial fractility and a guide to related sequences.
%H A269984 Robert Price, <a href="/A269984/b269984.txt">Table of n, a(n) for n = 1..67</a>
%e A269984 NI(1/5) = (2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, ...)
%e A269984 NI(2/5) = (2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...)
%e A269984 NI(3/5) = (1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, ...)
%e A269984 NI(4/5) = (1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, ...)
%e A269984 so there are 2 equivalences classes for n = 5, and the fractility of 5 is 2.
%t A269984 A269982[n_] := CountDistinct[With[{l = NestWhileList[
%t A269984          Rescale[#, {1/(Floor[x] + 1)!, 1/Floor[x]!} /.
%t A269984             FindRoot[1/x! == #, {x, 1}]] &, #, UnsameQ, All]},
%t A269984       Min@l[[First@First@Position[l, Last@l] ;;]]] & /@
%t A269984     Range[1/n, 1 - 1/n, 1/n]]; (* _Davin Park_, Nov 19 2016 *)
%t A269984 Select[Range[2, 500], A269982[#] == 2 &] (* _Robert Price_, Sep 19 2019 *)
%o A269984 (PARI) select( is_A269984(n)=A269982(n)==2, [1..300]) \\ _M. F. Hasler_, Nov 05 2018
%Y A269984 Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269985, A269986, A269987, A269988 (numbers with factorial fractility 1, 3, ..., 6, respectively).
%Y A269984 Cf. A269570 (binary fractility), A270000 (harmonic fractility).
%K A269984 nonn
%O A269984 1,1
%A A269984 _Clark Kimberling_ and _Peter J. C. Moses_, Mar 11 2016
%E A269984 Edited by _M. F. Hasler_, Nov 05 2018