A269988 Numbers k having factorial fractility A269982(k) = 6.

94, 105, 115, 141, 142, 153, 170, 175, 182, 184, 187, 189, 196, 205, 207, 210, 212, 213, 215, 221, 225, 235, 245, 252, 254, 255, 260, 265, 275, 276, 282, 290, 299, 306, 314, 325, 367, 368, 370, 378, 381, 388, 392, 399, 414, 424, 425, 426, 434, 435, 446, 450

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(1/94) = (4, 3, 2, 3, 1, 1, 1, 1, 3, 2, 1, 2, 1, 1, 3, 1, 4, 1, 1, 2, ...),
NI(2/94) = (4, 2, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, ...),
NI(4/94) = (3, 5, 1, 1, 2, 2, 1, 3, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 3, ...),
NI(7/94) = (3, 2, 2, 2, 1, 2, 4, 3, 2, 3, 1, 1, 1, 1, 3, 2, 1, 2, 1, 1, ...),
NI(11/94) = (3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, ...),
NI(47/94) = (2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...):
These 6 equivalence classes represent all the classes for k = 94, so the factorial fractility of 94 is 6.
		

Links

• Robert Price, Table of n, a(n) for n = 1..60

Crossrefs

Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269984, A269985, A269986, A269987 (numbers with factorial fractility 1, 2, ..., 5, 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[#] == 6 &] (* Robert Price, Sep 19 2019 *)

PARI

select( is_A269988(n)=A269982(n)==6, [1..400]) \\ M. F. Hasler, Nov 05 2018

Extensions

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