A269982 Factorial fractility of n.

1, 1, 2, 2, 1, 1, 2, 2, 3, 1, 2, 1, 2, 3, 2, 3, 2, 1, 4, 3, 2, 2, 2, 3, 2, 2, 4, 1, 3, 1, 2, 2, 4, 4, 3, 2, 2, 2, 4, 3, 3, 1, 3, 4, 4, 4, 2, 2, 4, 4, 3, 2, 2, 3, 4, 2, 2, 1, 4, 2, 3, 4, 2, 4, 2, 1, 5, 4, 5, 5, 3, 1, 3, 4, 3, 4, 2, 1, 4, 2, 4, 2, 4, 5, 2, 2

Offset

2,3

Comments

In order to define (factorial) fractility of an integer n > 1, we first define nested interval sequences. Suppose that r = (r(n)) is a sequence satisfying (i) 1 = r(1) > r(2) > r(3) > ... and (ii) r(n) -> 0. For x in (0,1], let n(1) be the index n such that r(n+1) < x <= r(n), and let L(1) = r(n(1))-r(n(1)+1). Let n(2) be the largest index n such that x <= r(n(1)+1) + L(1)*r(n), and let L(2) = (r(n(2)) - r(n(2)+1))*L(1). Continue inductively to obtain the sequence (n(1), n(2), n(3), ...) =: NI(x), the r-nested interval sequence of x.

For fixed r, call x and y equivalent if NI(x) and NI(y) are eventually equal (up to an offset). For n > 1, the r-fractility of n is the number of equivalence classes of sequences NI(m/n) for 0 < m < n. Taking r = (1/1, 1/2!, 1/3!, 1/4!, ... ) gives factorial fractility.

For factorial fractility, r(n) = 1/n!, n(j+1) = A084558(L(j)/(x - Sum_{i=1..j} L(i-1)/(n(i)+1)!)) for all j >= 0, L(0) = 1. - M. F. Hasler, Nov 05 2018

Links

Robert Price, Table of n, a(n) for n = 2..500

Example

NI(1/10) = (3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, ...)
NI(2/10) = (2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, ...)
NI(3/10) = (2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...)
NI(4/10) = (2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...)
NI(5/10) = (2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...)
NI(6/10) = (1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, ...)
NI(7/10) = (1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...)
NI(8/10) = (1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, ...)
NI(9/10) = (1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, ...),
so that there are 3 equivalence classes for n = 10, and the factorial fractility of 10 is 3.

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 *)

Prog

(PARI) A269982(n)=#Set(vector(n-1, k, NIFR(k/n))) \\ where:
NIFR(x, n, L=1, S=[], c=0)={for(i=2, oo, n=A084558(L\x); S=setunion(S, [x/L]); x-=L/(n+1)!; L/=(n+1)!\n; setsearch(S, x/L)&& if(c, break, c=!S=[])); S[1]} \\ variant of the function NIF() below; returns just a unique representative (smallest x/L occurring within the period) of the equivalence class.
NIF(x, n=[], L=1, S=[], c=0)={for(i=2, oo, n=concat(n, A084558(L\x)); c|| S=setunion(S, [x/L]); x-=L/(n[#n]+1)!; L/=(n[#n]-1)!*(n[#n]+1); if(!c, setsearch(S, x/L)&& [c, S]=[i, x/L], x/L==S, c-=i; break)); [n[1..2*c-1], n[c..-1]]} \\ Returns [transition, period] of "factorial" NI(x). \\ M. F. Hasler, Nov 05 2018

Crossrefs

Cf. A000142 (factorial numbers), A084558 (largest m: m! < n).

Cf. A269983, A269984, A269985, A269986, A269987, A269988: numbers with factorial fractility k = 1, 2, ..., 6, respectively.

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

Sequence in context: A160418 A168115 A337530 * A329462 A388737 A153864

Adjacent sequences: A269979 A269980 A269981 * A269983 A269984 A269985

Keyword

nonn

Author

Clark Kimberling and Peter J. C. Moses, Mar 10 2016

Extensions

Edited by M. F. Hasler, Nov 05 2018

Status

approved