login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A055881 a(n) = largest m such that m! divides n. 31
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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:
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
LINKS
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.
FORMULA
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
EXAMPLE
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)
MATHEMATICA
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 *)
PROG
(Scheme)
(define (A055881 n) (let loop ((n n) (i 2)) (cond ((not (zero? (modulo n i))) (- i 1)) (else (loop (/ n i) (+ 1 i))))))
(PARI) See Cano link.
(PARI) n=5; f=n!; x='x+O('x^f); Vec(sum(k=1, n, x^(k!)/(1-x^(k!)))) \\ Joerg Arndt, Jan 28 2014
(PARI) a(n)=for(k=2, n+1, if(n%k, return(k-1), n/=k)) \\ Charles R Greathouse IV, May 28 2015
CROSSREFS
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.
Sequence in context: A078380 A062356 A257993 * A332202 A204917 A232098
KEYWORD
easy,nonn
AUTHOR
Leroy Quet and Labos Elemer, Jul 16 2000
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified April 30 03:24 EDT 2024. Contains 372118 sequences. (Running on oeis4.)