A269983 -id:A269983 - cp's OEIS frontend

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

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

nonn

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

			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.

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

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).

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 *)

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)

Edited by M. F. Hasler, Nov 05 2018

A269984 Numbers k having factorial fractility A269982(k) = 2.

Original entry on oeis.org

4, 5, 8, 9, 12, 14, 16, 18, 22, 23, 24, 26, 27, 32, 33, 37, 38, 39, 48, 49, 53, 54, 57, 58, 61, 64, 66, 78, 81, 83, 86, 87, 96, 97, 101, 107, 113, 114, 121, 129, 131, 139, 163, 169, 174, 178, 181, 193, 218, 227, 241, 257, 263, 267, 277, 302, 317, 327, 331

Offset: 1

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

nonn

See A269982 for a definition of factorial fractility and a guide to related sequences.

			NI(1/5) = (2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, ...)
NI(2/5) = (2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...)
NI(3/5) = (1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 3, ...)
NI(4/5) = (1, 1, 2, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, ...)
so there are 2 equivalences classes for n = 5, and the fractility of 5 is 2.

Robert Price, Table of n, a(n) for n = 1..67

Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269985, A269986, A269987, A269988 (numbers with factorial fractility 1, 3, ..., 6, respectively).

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

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[#] == 2 &] (* Robert Price, Sep 19 2019 *)

select( is_A269984(n)=A269982(n)==2, [1..300]) \\ M. F. Hasler, Nov 05 2018

Edited by M. F. Hasler, Nov 05 2018

A269985 Numbers k having factorial fractility A269982(k) = 3.

Original entry on oeis.org

10, 15, 17, 21, 25, 30, 36, 41, 42, 44, 52, 55, 62, 72, 74, 76, 88, 93, 98, 99, 103, 104, 106, 108, 111, 118, 122, 125, 128, 132, 134, 137, 146, 149, 155, 158, 162, 166, 173, 176, 177, 179, 183, 186, 192, 198, 201, 202, 203, 214, 219, 226, 228, 237, 242, 249

Offset: 1

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

nonn

See A269982 for a definition of factorial fractility and a guide to related sequences.

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

Robert Price, Table of n, a(n) for n = 1..91

Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269984, A269986, A269987, A269988 (numbers with factorial fractility 1, 2, ..., 6, respectively).

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

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[#] == 3 &] (* Robert Price, Sep 19 2019 *)

select( is_A269985(n)=A269982(n)==2, [1..200]) \\ M. F. Hasler, Nov 05 2018

Edited by M. F. Hasler, Nov 05 2018

A269986 Numbers k having factorial fractility A269982(k) = 4.

Original entry on oeis.org

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

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

nonn

See A269982 for a definition of factorial fractility and a guide to related sequences.

			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.

Robert Price, Table of n, a(n) for n = 1..116

Cf. A000142 (factorial numbers), A269982 (factorial fractility of n); A269983, A269984, A269985, A269987, A269988 (numbers with factorial fractility 1, 2, ..., 6, respectively).

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

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 *)

select( is_A269986(n)=A269982(n)==4, [1..200]) \\ M. F. Hasler, Nov 05 2018

Edited by M. F. Hasler, Nov 05 2018

A269987 Numbers k having factorial fractility A269982(k) = 5.

Original entry on oeis.org

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

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

nonn

See A269982 for a definition of factorial fractility and a guide to related sequences.

			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.

Robert Price, Table of n, a(n) for n = 1..70

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).

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 *)

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

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

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

Original entry on oeis.org

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

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

nonn

See A269982 for a definition of factorial fractility and a guide to related sequences.

			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.

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

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).

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 *)

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

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

cp's OEIS Frontend

A269982 Factorial fractility of n.

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A269984 Numbers k having factorial fractility A269982(k) = 2.

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A269985 Numbers k having factorial fractility A269982(k) = 3.

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A269986 Numbers k having factorial fractility A269982(k) = 4.

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A269987 Numbers k having factorial fractility A269982(k) = 5.

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

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