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

Author

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

Keywords

• nonn

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

Examples

			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.
		

Links

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

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).

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 *)

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). (End)

Extensions

Edited by M. F. Hasler, Nov 05 2018