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

1,1

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..70

Example

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.

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 *)

Prog

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

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).

Sequence in context: A043854 A043862 A043871 * A058906 A130694 A269748

Adjacent sequences: A269984 A269985 A269986 * A269988 A269989 A269990

Keyword

nonn

Author

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

Extensions

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

Status

approved