A269987 Numbers k having factorial fractility A269982(k) = 5.

68, 70, 71, 85, 92, 100, 126, 127, 130, 136, 138, 145, 154, 157, 161, 164, 168, 180, 185, 195, 200, 204, 220, 224, 232, 247, 253, 266, 272, 288, 291, 300, 304, 310, 318, 324, 328, 333, 334, 336, 341, 342, 348, 360, 365, 369, 371, 390, 395, 400, 404, 407, 408, 412, 418, 433, 440, 441, 443, 444, 447

Offset: 1

Author

Clark Kimberling and Peter J. C. Moses, Mar 11 2016

Keywords

• nonn

Comments

See A269982 for a definition of factorial fractility and a guide to related sequences.

Examples

			NI(1/68) = (4, 2, 3, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, ...)
NI(4/68) = (3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, ...)
NI(6/68) = (3, 2, 1, 2, 2, 3, 1, 2, 3, 1, 1, 2, 1, 2, 2, 3, 1, 2, 3, ...)
NI(17/68) = (2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...)
NI(34/68) = (2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...).
These 5 equivalence classes represent all the classes for n = 68, so the factorial fractility of 68 is 5.
		

Links

• Robert Price, Table of n, a(n) for n = 1..70

Crossrefs

Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269984, A269985, A269986, A269988 (numbers with factorial fractility 1, 2, ..., 6, respectively).

Cf. A269570 (binary fractility), A270000 (harmonic fractility).

Programs

Mathematica

A269982[n_] := CountDistinct[With[{l = NestWhileList[
        Rescale[#, {1/(Floor[x] + 1)!, 1/Floor[x]!} /.
           FindRoot[1/x! == #, {x, 1}]] &, #, UnsameQ, All]},
     Min@l[[First@First@Position[l, Last@l] ;;]]] & /@
   Range[1/n, 1 - 1/n, 1/n]]; (* Davin Park, Nov 19 2016 *)
Select[Range[2, 500], A269982[#] == 5 &] (* Robert Price, Sep 19 2019 *)

PARI

select( is_A269987(n)=A269982(n)==5, [1..400]) \\ M. F. Hasler, Nov 05 2018

Extensions

Edited and more terms added by M. F. Hasler, Nov 05 2018