A060832 -id:A060832 - 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

A062071 a(n) = [n/1] + [n/(2^2)] + [n/(3^3)] + [n/(4^4)] + ... + [n/(k^k)] + ..., up to infinity, where [ ] is the floor function.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 34, 36, 37, 38, 39, 41, 42, 43, 44, 46, 47, 48, 49, 51, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 64, 66, 67, 69, 70, 72, 73, 74, 75, 77, 78, 79, 80, 82, 83, 84, 85, 87, 88, 89, 90

Offset: 1

Amarnath Murthy, Jun 13 2001

			a(7) = [7/1] + [7/4] + [7/27] + ... = 7 + 1 + 0 + 0 + ... = 8.
a(8) = [8/1] + [8/4] + [8/27] + [8/256] + ... = 8 + 2 + 0 + 0 + ... = 10.

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)
Vaclav Kotesovec, Plot of a(n)/n for n = 1..100000

Cf. A006218, A011371, A060832, A347397.

Flatten[{1, Table[Sum[Floor[n/k^k], {k, 1, Floor[N[Log[n]/LambertW[Log[n]]]] + 1}], {n, 2, 100}]}] (* Vaclav Kotesovec, Aug 30 2021 *)

\p 10 v=[]; for(n=1,120,v=concat(v,suminf(k=1,floor(n/k^k)))); v

for (n=1, 1000, write("b062071.txt", n, " ", suminf(k=1, n\k^k)\1) ) \\ Harry J. Smith, Jul 31 2009

a(n)=sum(k=1,exp(lambertw(log(n)))+1,n\k^k) \\ Charles R Greathouse IV, May 28 2015

[sum( floor(n/j^j) for j in (1..1+log(n)) ) for n in (1..100)] # G. C. Greubel, May 06 2022

a(n) = Sum_{i=1..n} floor(n/i^i). - Wesley Ivan Hurt, Sep 15 2017
G.f.: (1/(1 - x)) * Sum_{k>=1} x^(k^k)/(1 - x^(k^k)). - Seiichi Manyama, Aug 30 2021
Conjecture: a(n) ~ c * n, where c = A073009. - Vaclav Kotesovec, Aug 30 2021

More terms from Jason Earls, Jun 21 2001

A355986 a(n) = Sum_{k=1..n} floor(n/k)!.

Original entry on oeis.org

1, 3, 8, 28, 125, 731, 5052, 40352, 362917, 3628935, 39916936, 479002360, 6227021561, 87178296283, 1307674373184, 20922789928484, 355687428136485, 6402373706091651, 121645100409195652, 2432902008180269688, 51090942171713074013, 1124000727777647602015

Offset: 1

Seiichi Manyama, Jul 22 2022

			a(6) = 6! + 3! + 2! + 1! + 1! + 1! = 731.

Cf. A007489, A060832, A081123.

a[n_] := Sum[Floor[n/k]!, {k, 1, n}]; Array[a, 22] (* Amiram Eldar, Jul 22 2022 *)

PARI
```
a(n) = sum(k=1, n, (n\k)!);
```

A380408 a(n) = Sum_{k>=0} floor(n/(2k)!).

Original entry on oeis.org

0, 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 21, 22, 24, 25, 27, 28, 30, 31, 33, 34, 37, 38, 40, 41, 43, 44, 46, 47, 49, 50, 52, 53, 55, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 74, 75, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104, 105, 107

Offset: 0

Akiva Weinberger, Jan 23 2025

nonn

Partial sum of A060832 except for the first term in the sum.
Congruent to A034968(n) mod 2. Therefore, the parity of a(n) is the parity of the n-th permutation of k elements (k>=n) in lexicographic order.
For even n, a(n) equals A059563(n/2) whenever cosh(1)*n - a(n) < 1. The first time this fails is n=70, as a(70)=107 but A059563(35)=108. For small n, such failures appear to be very rare; however, the asymptotic density of these failures approaches 1.

Cf. A060832, A034968, A013936, A013939, A038663.

a(n) = round(sumpos(k=0, n\(2*k)!)); \\ Michel Marcus, Jan 24 2025

a(n) = cosh(1)*n - f(n) where f(n) = Sum_{k>=0} fract(n/(2k)!). Here, fract() is the fractional part. The error term f(n) is unbounded above, and the greatest lower bound is 0 (even excluding n=0). The first values for which f(n) > s for s=1,2,3 are f(13)=1.06005, f(407) = 2.03382, and f(22319) = 3.01669. The error is almost periodic: for large m, f(n) is approximately f(n+(2m)!). If n is odd, f(n) > 1/2. f(n) alternately rises and descends, that is, f(2*n)f(2*n+2) for all n.

cp's OEIS Frontend

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

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A062071 a(n) = [n/1] + [n/(2^2)] + [n/(3^3)] + [n/(4^4)] + ... + [n/(k^k)] + ..., up to infinity, where [ ] is the floor function.

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A355986 a(n) = Sum_{k=1..n} floor(n/k)!.

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A380408 a(n) = Sum_{k>=0} floor(n/(2k)!).

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