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
Keywords
Examples
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.
Links
- Robert Price, Table of n, a(n) for n = 1..116
Crossrefs
Programs
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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 *)
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PARI
select( is_A269986(n)=A269982(n)==4, [1..200]) \\ M. F. Hasler, Nov 05 2018
Extensions
Edited by M. F. Hasler, Nov 05 2018
Comments