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

4, 5, 8, 9, 12, 14, 16, 18, 22, 23, 24, 26, 27, 32, 33, 37, 38, 39, 48, 49, 53, 54, 57, 58, 61, 64, 66, 78, 81, 83, 86, 87, 96, 97, 101, 107, 113, 114, 121, 129, 131, 139, 163, 169, 174, 178, 181, 193, 218, 227, 241, 257, 263, 267, 277, 302, 317, 327, 331

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..67

Example

NI(1/5) = (2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, ...)
NI(2/5) = (2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...)
NI(3/5) = (1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, ...)
NI(4/5) = (1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, ...)
so there are 2 equivalences classes for n = 5, and the fractility of 5 is 2.

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[#] == 2 &] (* Robert Price, Sep 19 2019 *)

Prog

(PARI) select( is_A269984(n)=A269982(n)==2, [1..300]) \\ M. F. Hasler, Nov 05 2018

Crossrefs

Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269985, A269986, A269987, A269988 (numbers with factorial fractility 1, 3, ..., 6, respectively).

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

Sequence in context: A321333 A333384 A334992 * A188085 A316097 A206554

Adjacent sequences: A269981 A269982 A269983 * A269985 A269986 A269987

Keyword

nonn

Author

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

Extensions

Edited by M. F. Hasler, Nov 05 2018

Status

approved