A269570 -id:A269570 - cp's OEIS frontend

A070940 Number of digits that must be counted from left to right to reach the last 1 in the binary representation of n.

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 4, 2, 4, 3, 4, 1, 5, 4, 5, 3, 5, 4, 5, 2, 5, 4, 5, 3, 5, 4, 5, 1, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 2, 6, 5, 6, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 1, 7, 6, 7, 5, 7, 6, 7, 4, 7, 6, 7, 5, 7, 6, 7, 3, 7, 6, 7, 5, 7, 6, 7, 4, 7, 6, 7, 5, 7

Offset: 1

N. J. A. Sloane, May 18 2002

Length of longest carry sequence when adding numbers <= n to n in binary representation: a(n) = T(n, A080079(n)) and T(n,k) <= a(n) for 1 <= k <= n, with T defined as in A080080. - Reinhard Zumkeller, Jan 26 2003
a(n+1) is the number of distinct values of gcd(2^n, binomial(n,j)) (or, equivalently, A007814(binomial(n,j))) arising for j=0..n-1. Proof using Kummer's Theorem given by Marc Schwartz. - Labos Elemer, Apr 23 2003
E.g., n=10: 10th row of Pascal's triangle = {1,10,45,120,210,252,210,120,45,10,1}, largest powers of 2 dividing binomial coefficients is: {1,2,1,8,2,4,2,8,1,2,1}; including distinct powers of 2, thus a(10)=4. If m=-1+2^k, i.e., m=0,1,3,7,15,31,... then a(m)=1. This corresponds to "odd rows" of Pascal's triangle. - Labos Elemer
Smallest x > 0 for which a(x)=n equals 2^n. - Labos Elemer
a(n) <= A070939(n), a(n) = A070939(n) iff n is odd, where A070939(n) = floor(log_2(n)) + 1. - Reinhard Zumkeller, Jan 26 2003
Can be regarded as a table with row n having 2^(n-1) columns, with odd columns repeating the previous row, and even columns containing the row number. - Franklin T. Adams-Watters, Nov 08 2011
It appears that a(n) is the greatest number in a periodicity equivalence class defined at A269570; e.g., the 5 classes for n = 35 are (1, 1, 2, 2, 6), (1, 1, 1, 1, 4, 2, 2), (3), (1, 3), (1, 2); in these the greatest number is 6, so that a(35) = 6. - Clark Kimberling, Mar 01 2016
Number of binary digits of the largest odd factor of n. - Andres Cicuttin, May 18 2017

			a(10)=3 is the number of digits that must be counted from left to right to reach the last 1 in 1010, the binary representation of 10.
The table starts:
  1
  1 2
  1 3 2 3
  1 4 3 4 2 4 3 4

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A070939, A001511. Differs from A002487 around 11th term.
Cf. A000005, A007318, A000079, A082907, A082908.
Bisections give A070941 and this sequence (again).
Cf. A002064 (row sums), A199570.

a070940 = maximum . a080080_row  -- Reinhard Zumkeller, Apr 22 2013

A070940 := n -> if n mod 2 = 0 then A070939(n)-A001511(n/2) else A070939(n); fi;

Table[Length[Union[Table[GCD[2^n, Binomial[n, j]], {j, 0, n}]]], {n, 0, 256}]
f[n_] := Position[ IntegerDigits[n, 2], 1][[ -1, 1]]; Table[ f[n], {n, 105}] (* Robert G. Wilson v, Dec 01 2004 *)
(* By exploiting the "positional" regularity of the sequence *)
b = {}; a = {1, 1};
Do[a = Riffle[a, j];
  b = AppendTo[b, a[[1 ;; Floor[Length[a]/2]]]] // Flatten, {j, 1, 10}];
Print[b[[1 ;; 100]]] (* Andres Cicuttin, May 18 2017 *)
(* By following the alternative definition "Number of binary digits of the largest integer odd factor of n" *)
c = Table[IntegerDigits[n/(2^IntegerExponent[n, 2]), 2] // Length , {n,
    2^10 - 1}];
Print[c[[1 ;; 100]]] (* Andres Cicuttin, May 18 2017 *)
lidn[n_]:=Module[{idn=IntegerDigits[n,2]},idn=If[Last[idn]==0,Flatten[ Most[ Split[ idn]]],idn];Length[idn]]; Array[lidn,100] (* Harvey P. Dale, Oct 18 2020 *)
Table[IntegerLength[FromDigits[Reverse[IntegerDigits[n,2]]]],{n,100}] (* Harvey P. Dale, Jan 26 2025 *)

def A070940(n):
    while n%2 == 0:
        n = n//2
    a = 0
    while n != 0:
        n, a = n//2, a+1
    return a
n = 0
while n < 100:
    n = n+1
    print(n,A070940(n)) # A.H.M. Smeets, Aug 19 2019

def A070940(n): return n.bit_length()-(~n&n-1).bit_length() # Chai Wah Wu, Jul 13 2022

blocklevel <- 7  # by choice
a <- 1
for(m in 0:blocklevel)
  for(k in 0:(2^m-1)){
    a[2^(m+1)+2*k  ] <-  a[2^m+k]
    a[2^(m+1)+2*k+1] <-  m + 2
}
a
# Yosu Yurramendi, Aug 08 2019

a(n) = floor(log_2(n)) - A007814(n) = A070939(n) - A007814(n).
a(n) = f(n, 1), f(n, k) = if n=1 then k else f(floor(n/2), k+(if k>1 then 1 else n mod 2)). - Reinhard Zumkeller, Feb 01 2003
G.f.: Sum_{k>=0} (t/(1-t^2)) * (1 + Sum_{L>=1} t^2^L), where t=x^2^k. - Ralf Stephan, Mar 15 2004
a(n) = A070939(A000265(n)). - Andres Cicuttin, May 19 2017
a(1) = 1 and for m >= 0, 0 <= k < 2^m, a(2^(m+1)+2*k) = a(2^m+k), a(2^(m+1)+2*k+1) = m+2. - Yosu Yurramendi, Aug 08 2019

Entry revised by Ralf Stephan, Nov 29 2004

A270000 Harmonic fractility of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 3, 1, 2, 1, 1, 1, 3, 2, 1, 3, 3, 1, 3, 3, 1, 1, 3, 2, 1, 1, 3, 1, 3, 1, 2, 3, 3, 2, 4, 1, 2, 3, 2, 3, 3, 1, 1, 3, 1, 3, 3, 1, 5, 1, 3, 3, 2, 2, 2, 1, 1, 1, 5, 3, 3, 1, 3, 3, 4, 1, 2, 2, 3, 3, 6, 3, 3, 2, 1, 4, 3, 1, 2, 2, 3, 3

Offset: 2

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

In order to define (harmonic) fractility of an integer m > 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 index n such that r(n(1)+1) + L(1)*r(n+1) < 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 identical. For m > 1, the r-fractility of m is the number of equivalence classes of sequences NI(k/m) for 0 < k < m. Taking r = (1/1, 1/2, 1/3, 1/4, ... ) gives harmonic fractility.
For harmonic fractility, r(n) = 1/n, n(j+1) = floor(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(1/11) = (11, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...),
NI(2/11) = (5, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...),
NI(3/11) = (3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, ...),
NI(4/11) = (2, 5, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...),
NI(5/11) = (2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...),
NI(6/11) = (1, 11, 1, 1, 1, 1, 1, 1, 1, 1, ...),
NI(7/11) = (1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, ...),
NI(8/11) = (1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...),
NI(9/11) = (1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, ...),
NI(10/11) = (1, 1, 1, 3, 3, 3, 3, 3, 3, 3, ...),
so that there are 3 equivalence classes for n = 11, and that the harmonic fractility of 11 is 3.

Jack W Grahl, Table of n, a(n) for n = 2..999
Jack W Grahl, Python code to generate this sequence

Guide to related sequences:
  k - numbers with harmonic fractility k:
  1 - A269804
  2 - A269805
  3 - A269806
  4 - A269807
  5 - A269808
  6 - A269809
Cf. A269570 (binary fractility), A269982 (factorial fractility).

A270000[n_] := CountDistinct[With[{l = NestWhileList[Rescale[#, {1/(Floor[1/#] + 1), 1/Floor[1/#]}] &, #, UnsameQ, All]}, Min@l[[First@FirstPosition[l, Last@l] ;;]]] & /@ Range[1/n, 1 - 1/n, 1/n]] (* Davin Park, Nov 09 2016 *)

A270000(n)=#Set(vector(n-1,k,NIR(k/n))) \\ where:
NIR(x, n, L=1, S=[], c=0)={for(i=2, oo, n=L\x; S=setunion(S, [x/L]); x-=L/(n+1); L/=n*(n+1); setsearch(S, x/L)&& if(c, break, c=!S=[])); S[1]} \\ variant of the function NI() below; returns just a unique representative (smallest x/L occurring within the period) of the equivalence class.
NI(x, n=[], L=1, S=[], c=0)={for(i=2, oo, n=concat(n, L\x); c|| S=setunion(S, [x/L]); x-=L/(n[#n]+1); L/=n[#n]*(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 the harmonic nested interval sequence for x in the form [transition, period]. (End)

Definition corrected by Jack W Grahl, Jun 27 2018

Edited by M. F. Hasler, Nov 05 2018

A269982 Factorial fractility of n.

Original entry on oeis.org

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

A269983 Numbers k having factorial fractility A269982(k) = 1.

Original entry on oeis.org

2, 3, 6, 7, 11, 13, 19, 29, 31, 43, 59, 67, 73, 79, 89, 109, 151, 197, 199, 211, 229, 233, 269, 281, 283, 293, 337, 373, 379, 389, 397, 419, 421, 439, 449, 463, 487, 503, 509, 547, 557, 619, 673, 701, 727, 733, 797, 809, 811, 827, 877, 883, 887, 937, 941, 947, 953, 983

Offset: 1

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

nonn

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

Is 6 the largest even term of this sequence? - M. F. Hasler, Nov 05 2018

			NI(1/7) = (3, 1, 1, 2, 2, 3, 1, 1, 2, 2, 3, 1, 1, 2, 2, ...),
NI(2/7) = (2, 2, 1, 3, 1, 1, 2, 2, 3, 1, 1, 2, 2, 3, 1, ...),
NI(3/7) = (2, 1, 1, 3, 1, 1, 2, 2, 3, 1, 1, 2, 2, 3, 1, ...),
NI(4/7) = (1, 3, 1, 1, 2, 2, 3, 1, 1, 2, 2, 3, 1, 1, 2, ...),
NI(5/7) = (1, 2, 1, 1, 3, 1, 1, 2, 2, 3, 1, 1, 2, 2, 3, ...),
NI(6/7) = (1, 1, 2, 1, 1, 3, 1, 1, 2, 2, 3, 1, 1, 2, 2, ...):
all are eventually periodic with period (1, 1, 2, 2, 3), so there is only one equivalence class for n = 7, and the fractility of 7 is 1.

Cf. A269982 (factorial fractility of n); A269984, A269985, A269986, A269987, A269988 (numbers with factorial fractility 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, 1000], A269982[#] == 1 &] (* Robert Price, Sep 19 2019 *)

select( is_A269983(n)=A269982(n)==1, [1..300]) \\ M. F. Hasler, Nov 05 2018

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

a(54)-a(58) from Robert Price, Sep 19 2019

A269571 Numbers having binary fractility 1.

Original entry on oeis.org

2, 3, 4, 5, 8, 11, 13, 16, 19, 29, 32, 37, 53, 59, 61, 64, 67, 83, 101, 107, 128, 131, 139, 149, 163, 173, 179, 181, 197, 211, 227, 256, 269, 293, 317, 347, 349, 373, 379, 389, 419, 421, 443, 461, 467, 491, 509, 512, 523, 541, 547, 557, 563, 587, 613, 619

Offset: 1

Clark Kimberling, Mar 01 2016

nonn

For each x in (0,1], let 1/2^p(1) + 1/2^p(2) + ... be the infinite binary representation of x. Let d(1) = p(1) and d(i) = p(i) - p(i-1) for i >=2. Call (d(i)) the powerdifference sequence of x, and denote it by D(x). Call m/n and u/v equivalent if every period of D(m/n) is a period of D(u/v). Define the binary fractility of n to be the number of distinct equivalence classes of {m/n: 0 < m < n}.

			D(1/5) = (3,1,3,1,3,1,3,1,...)
D(2/5) = (2,1,3,1,3,1,3,1,...)
D(3/5) = (1,3,1,3,1,3,1,3,...)
D(4/5) = (1,1,3,1,3,1,3,1,...).
This shows that all m/5, for 0<m<5 are equivalent to 1/5, so that there is only 1 equivalence class.

Robert Price, Table of n, a(n) for n = 1..76

Cf. A269570, A269572.

A269570[n_] := CountDistinct[With[{l = NestWhileList[
        Rescale[#, {1/2^(Floor[-Log[2, #]] + 1),
           1/2^(Floor[-Log[2, #]])}] &, #, UnsameQ, All]},
     Min@l[[First@First@Position[l, Last@l] ;;]]] & /@
   Range[1/n, 1 - 1/n, 1/n]] (* from Davin Park, Nov 19 2016 *)
Select[Range[1000], A269570[#] == 1 &] (* Robert Price, Sep 20 2019 *)

Corrected Offset by Robert Price, Sep 20 2019

a(37)-a(56) from Robert Price, Sep 20 2019

cp's OEIS Frontend

A070940 Number of digits that must be counted from left to right to reach the last 1 in the binary representation of n.

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A270000 Harmonic fractility of n.

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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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