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

20, 28, 34, 35, 40, 45, 46, 47, 50, 51, 56, 60, 63, 65, 69, 75, 77, 80, 82, 84, 90, 91, 102, 110, 112, 116, 117, 120, 123, 124, 133, 135, 144, 147, 148, 150, 152, 156, 159, 160, 165, 167, 171, 172, 194, 206, 208, 209, 216, 217, 222, 223, 234, 236, 239, 240

Offset

1,1

Comments

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

Links

Robert Price, Table of n, a(n) for n = 1..116

Example

NI(1/20) = (3, 3, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, ...)
NI(5/20) = (2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...)
NI(6/20) = (2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...)
NI(10/20) = (2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...).
These 4 equivalence classes represent all the classes for n = 20, so the factorial fractility of 20 is 4.

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[#] == 4 &] (* Robert Price, Sep 19 2019 *)

Prog

(PARI) select( is_A269986(n)=A269982(n)==4, [1..200]) \\ M. F. Hasler, Nov 05 2018

Crossrefs

Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269984, A269985, A269987, A269988 (numbers with factorial fractility 1, 2, ..., 6, respectively).

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

Sequence in context: A122146 A032482 A024411 * A139805 A141823 A367632

Adjacent sequences: A269983 A269984 A269985 * A269987 A269988 A269989

Keyword

nonn

Author

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

Extensions

Edited by M. F. Hasler, Nov 05 2018

Status

approved