A230404 -id:A230404 - cp's OEIS frontend

A055881 a(n) = largest m such that m! divides n.

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

Offset: 1

Leroy Quet and Labos Elemer, Jul 16 2000

Number of factorial divisors of n. - Amarnath Murthy, Oct 19 2002
The sequence may be constructed as follows. Step 1: start with 1, concatenate and add +1 to last term gives: 1,2. Step 2: 2 is the last term so concatenate twice those terms and add +1 to last term gives: 1, 2, 1, 2, 1, 3 we get 6 terms. Step 3: 3 is the last term, concatenate 3 times those 6 terms and add +1 to last term gives: 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, iterates. At k-th step we obtain (k+1)! terms. - Benoit Cloitre, Mar 11 2003
From Benoit Cloitre, Aug 17 2007, edited by M. F. Hasler, Jun 28 2016: (Start)
Another way to construct the sequence: start from an infinite series of 1's:
   1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... Replace every second 1 by a 2 giving:
   1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ... Replace every third 2 by a 3 giving:
   1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, ... Replace every fourth 3 by a 4 etc. (End)
This sequence is the fixed point, starting with 1, of the morphism m, where m(1) = 1, 2, and for k > 1, m(k) is the concatenation of m(k - 1), the sequence up to the first k, and k + 1. Thus m(2) = 1, 2, 1, 3; m(3) = 1, 2, 1, 3, 1, 2, 1, 2, 1, 4; m(4) = 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 5, etc. - Franklin T. Adams-Watters, Jun 10 2009
All permutations of n elements can be listed as follows: Start with the (arbitrary) permutation P(0), and to obtain P(n + 1), reverse the first a(n) + 1 elements in P(n). The last permutation is the reversal of the first, so the path is a cycle in the underlying graph. See example and fxtbook link. - Joerg Arndt, Jul 16 2011
Positions of rightmost change with incrementing rising factorial numbers, see example. - Joerg Arndt, Dec 15 2012
Records appear at factorials. - Robert G. Wilson v, Dec 21 2012
One more than the number of trailing zeros (A230403(n)) in the factorial base representation of n (A007623(n)). - Antti Karttunen, Nov 18 2013
A062356(n) and a(n) coincide quite often. - R. J. Cano, Aug 04 2014
For n>0 and 1<=j<=(n+1)!-1, (n+1)^2-1=A005563(n) is the number of times that a(j)=n-1. - R. J. Cano, Dec 23 2016

			a(12) = 3 because 3! is highest factorial to divide 12.
From _Joerg Arndt_, Jul 16 2011: (Start)
All permutations of 4 elements via prefix reversals:
   n:   permutation  a(n)+1
   0:   [ 0 1 2 3 ]  -
   1:   [ 1 0 2 3 ]  2
   2:   [ 2 0 1 3 ]  3
   3:   [ 0 2 1 3 ]  2
   4:   [ 1 2 0 3 ]  3
   5:   [ 2 1 0 3 ]  2
   6:   [ 3 0 1 2 ]  4
   7:   [ 0 3 1 2 ]  2
   8:   [ 1 3 0 2 ]  3
   9:   [ 3 1 0 2 ]  2
  10:   [ 0 1 3 2 ]  3
  11:   [ 1 0 3 2 ]  2
  12:   [ 2 3 0 1 ]  4
  13:   [ 3 2 0 1 ]  2
  14:   [ 0 2 3 1 ]  3
  15:   [ 2 0 3 1 ]  2
  16:   [ 3 0 2 1 ]  3
  17:   [ 0 3 2 1 ]  2
  18:   [ 1 2 3 0 ]  4
  19:   [ 2 1 3 0 ]  2
  20:   [ 3 1 2 0 ]  3
  21:   [ 1 3 2 0 ]  2
  22:   [ 2 3 1 0 ]  3
  23:   [ 3 2 1 0 ]  2
(End)
From _Joerg Arndt_, Dec 15 2012: (Start)
The first few rising factorial numbers (dots for zeros) with 4 digits and the positions of the rightmost change with incrementing are:
  [ 0]    [ . . . . ]   -
  [ 1]    [ 1 . . . ]   1
  [ 2]    [ . 1 . . ]   2
  [ 3]    [ 1 1 . . ]   1
  [ 4]    [ . 2 . . ]   2
  [ 5]    [ 1 2 . . ]   1
  [ 6]    [ . . 1 . ]   3
  [ 7]    [ 1 . 1 . ]   1
  [ 8]    [ . 1 1 . ]   2
  [ 9]    [ 1 1 1 . ]   1
  [10]    [ . 2 1 . ]   2
  [11]    [ 1 2 1 . ]   1
  [12]    [ . . 2 . ]   3
  [13]    [ 1 . 2 . ]   1
  [14]    [ . 1 2 . ]   2
  [15]    [ 1 1 2 . ]   1
  [16]    [ . 2 2 . ]   2
  [17]    [ 1 2 2 . ]   1
  [18]    [ . . 3 . ]   3
  [19]    [ 1 . 3 . ]   1
  [20]    [ . 1 3 . ]   2
  [21]    [ 1 1 3 . ]   1
  [22]    [ . 2 3 . ]   2
  [23]    [ 1 2 3 . ]   1
  [24]    [ . . . 1 ]   4
  [25]    [ 1 . . 1 ]   1
  [26]    [ . 1 . 1 ]   2
(End)

Antti Karttunen, Table of n, a(n) for n = 1..10080
Joerg Arndt, Matters Computational (The Fxtbook), section 10.4, pp.245-248 (prefix reversals); section 10.5, pp. 248-250 (Heap's method).
R. J. Cano, Alternative sequencer (PARI/GP).
Claude Lenormand, Comments on this sequence.
József Sándor, On Additive Analogues of Certain Arithmetic Smarandache Functions.
Index entries for sequences related to factorial base representation.

Cf. A055874, A055926, A055770, A062356, A073575, A091131, A230403, A230404, A230405, A076733, A232096, A232098, A233285, A233267, A233269, A231719, A232741, A232742, A232743, A232744, A232745, A060832 (partial sums).
This sequence occurs also in the next to middle diagonals of A230415 and as the second rightmost column of triangle A230417.
Other sequences related to factorial base representation (A007623): A034968, A084558, A099563, A060130, A227130, A227132, A227148, A227149, A153880.
Analogous sequence for binary (base-2) representation: A001511.

Table[Length[Intersection[Divisors[n], Range[5]!]], {n, 125}] (* Alonso del Arte, Dec 10 2012 *)
f[n_] := Block[{m = 1}, While[Mod[n, m!] == 0, m++]; m - 1]; Array[f, 105] (* Robert G. Wilson v, Dec 21 2012 *)

PARI
```
See Cano link.
```

n=5; f=n!; x='x+O('x^f); Vec(sum(k=1,n,x^(k!)/(1-x^(k!)))) \\ Joerg Arndt, Jan 28 2014

a(n)=for(k=2,n+1,if(n%k, return(k-1),n/=k)) \\ Charles R Greathouse IV, May 28 2015

(define (A055881 n) (let loop ((n n) (i 2)) (cond ((not (zero? (modulo n i))) (- i 1)) (else (loop (/ n i) (+ 1 i))))))

G.f.: Sum_{k > 0} x^(k!)/(1 - x^(k!)). - Vladeta Jovovic, Dec 13 2002
a(n) = A230403(n)+1. - Antti Karttunen, Nov 18 2013
a(n) = A230415(n-1,n) = A230415(n,n-1) = A230417(n,n-1). - Antti Karttunen, Nov 19 2013
a(m!+n) = a(n) if 1 <= n <= m*m! - 1 = A001563(m) - 1. - R. J. Cano, Jun 27 2016
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = e - 1 (A091131). - Amiram Eldar, Jul 23 2022

A230403 a(n) = the largest k such that (k+1)! divides n; the number of trailing zeros in the factorial base representation of n (A007623(n)).

Original entry on oeis.org

0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1

Offset: 1

Antti Karttunen, Oct 31 2013

Many of the comments given in A055881 apply also here.
From Amiram Eldar, Mar 10 2021: (Start)
The asymptotic density of the occurrences of k is (k+1)/(k+2)!.
The asymptotic mean of this sequence is e - 2 = 0.718281... (A001113 - 2). (End)

			In factorial number base representation (A007623), the numbers from 1 to 9 are represented as:
n  A007623(n)  a(n) (gives the number of trailing zeros)
1        1       0
2       10       1
3       11       0
4       20       1
5       21       0
6      100       2
7      101       0
8      110       1
9      111       0

Antti Karttunen, Table of n, a(n) for n = 1..10080
Tyler Ball, Joanne Beckford, Paul Dalenberg, Tom Edgar, and Tina Rajabi, Some Combinatorics of Factorial Base Representations, J. Int. Seq., Vol. 23 (2020), Article 20.3.3.
Index entries for sequences related to factorial base representation

Cf. A001113, A055881. Bisection: A230404.
A few sequences related to factorial base representation (A007623): A034968, A084558, A099563, A060130, A227130, A227132, A227148, A227149, A153880.
Analogous sequence for binary system: A007814.

With[{b = MixedRadix[Range[12, 2, -1]]}, Array[LengthWhile[Reverse@ IntegerDigits[#, b], # == 0 &] &, 105]] (* Michael De Vlieger, Jun 03 2020 *)

(define (A230403 n) (if (zero? n) 0 (let loop ((n n) (i 2)) (cond ((not (zero? (modulo n i))) (- i 2)) (else (loop (/ n i) (1+ i)))))))

a(n) = A055881(n)-1.

cp's OEIS Frontend

A230405 a(n) = A000217(A230404(n+1)); the first differences of A219650.

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A055881 a(n) = largest m such that m! divides n.

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A230403 a(n) = the largest k such that (k+1)! divides n; the number of trailing zeros in the factorial base representation of n (A007623(n)).

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Mathematica

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Formula