A060130 -id:A060130 - 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

A060502 a(n) = number of occupied digit slopes in the factorial base representation of n (see comments for the definition); number of drops in the n-th permutation of list A060117.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 2, 3, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 2, 2, 2, 3, 3, 4, 3, 3, 2, 3, 2, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1

Offset: 0

Antti Karttunen, Mar 22 2001

From Antti Karttunen, Aug 11-24 2016: (Start)
a(n) gives the number of occupied "digit slopes" in the factorial base representation of n, or more formally, the number of distinct elements in a multiset [(i_x - d_x) | where d_x ranges over each nonzero digit present in factorial base representation of n and i_x is that digit's position from the right]. Here one-based indexing is used, thus the least significant digit is in position 1. Each value {digit's position} - {digit's value} determines on which slope that particular nonzero digit is. The nonzero digits for which (position - digit) = 0, are said to be on the "maximal slope" (see A260736), those with value 1 on "sub-maximal", etc.
The number of occupied digit slopes translates directly to the number of drops in the n-th permutation as given in the list A060117 because only the largest (and thus leftmost) of all nonzero digits on any particular slope adds a (single) drop to the permutation, when constructed by the unranking algorithm employed in A060117.
The original definition of this sequence is (essentially):
  a(n) = the average of digits (where "digits" may eventually obtain also any values > 9) in each siteswap pattern A060498(n) constructed from each permutation in list A060117, which is equal to number of balls used in that pattern.
The equivalence of the old and the new definitions is seen from the following (as kindly pointed by Olivier Gérard in personal mail): For any permutation p of [1..n], Sum(i=1..n) p(i)-i = 0 (whether taken modulo n or not), thus Sum(i=1..n) (p(i)-i modulo n) = Sum(i={set of nondrops}) (p(i)-i) + Sum(i={set of drops}) (n + (p(i)-i)) = 0 + n * #{set of drops}, where drops is the set of those i where p[i] < i and nondrops are those i for which p[i] >= 1.
Involution A225901 maps this metric to another metric A275806 which gives the number of distinct nonzero digits in factorial base representation of n. See also A275811.
A007489 (repunits in this context) gives the positions where a(n) = A084558(n) (the length of factorial base representation of n). These are also the positions of records.
(End)

			For n=23 ("321" in factorial base representation, A007623), all the digits are maximal for their positions (they occur on the "maximal slope"), thus there is only one distinct digit slope present and a(23)=1. Also, for the 23rd permutation in the ordering A060117, [2341], there is just one drop, as p[4] = 1 < 4.
For n=29 ("1021"), there are three nonzero digits, where both 2 and the rightmost 1 are on the maximal slope, while the most significant 1 is on the "sub-sub-sub-maximal", thus there are two occupied slopes in total, and a(29) = 2. In the 29th permutation of A060117, [23154], there are two drops as p[3] = 1 < 3 and p[5] = 4 < 5.
For n=37 ("1201"), there are three nonzero digits, where the rightmost 1 is on the maximal slope, 2 is on the submaximal, and the most significant 1 is on the "sub-sub-sub-maximal", thus there are three occupied slopes in total, and a(37) = 3. In the 37th permutation of A060117, [51324], there are three drops at indices 2, 4 and 5.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A000120, A001221, A007489, A007623, A033312, A060117, A060118, A060130, A060498, A225901, A275734, A275806.
Cf. A007489 (positions of records, the first occurrence of each n).
Cf. A276001, A276002, A276003 (positions where a(n) obtains values 1, 2, 3).
Cf. also A060500, A060501, A060125, A084558, A153880, A255411, A260736, A273670, A275804, A275811, A275946, A275947, A275849, A275956, A276004,
A276005, A276010.

# The following program follows the original 2001 interpretation of this sequence:
A060502 := n -> avg(Perm2SiteSwap3(PermUnrank3R(n)));
with(group);
permul := (a, b) -> mulperms(b, a);
# factorial_base(n) gives the digits of A007623(n) as a list, uncorrupted even when there are digits > 9:
factorial_base := proc(nn) local n, a, d, j, f; n := nn; if(0 = n) then RETURN([0]); fi; a := []; f := 1; j := 2; while(n > 0) do d := floor(`mod`(n, (j*f))/f); a := [d, op(a)]; n := n - (d*f); f := j*f; j := j+1; od; RETURN(a); end;
# PermUnrank3R(r) gives the permutation with rank r in list A060117:
PermUnrank3R := proc(r) local n; n := nops(factorial_base(r)); convert(PermUnrank3Raux(n+1, r, []), 'permlist', 1+(((r+2) mod (r+1))*n)); end;
PermUnrank3Raux := proc(n, r, p) local s; if(0 = r) then RETURN(p); else s := floor(r/((n-1)!)); RETURN(PermUnrank3Raux(n-1, r-(s*((n-1)!)), permul(p, [[n, n-s]]))); fi; end;
Perm2SiteSwap3 := proc(p) local ip,n,i,a; n := nops(p); ip := convert(invperm(convert(p,'disjcyc')),'permlist',n); a := []; for i from 1 to n do if(0 = ((ip[i]-i) mod n)) then a := [op(a),0]; else a := [op(a), n-((ip[i]-i) mod n)]; fi; od; RETURN(a); end;
avg := a -> (convert(a, `+`)/nops(a));

From Antti Karttunen, Aug 11-21 2016: (Start)
The following formula reflects the original definition of computing the average, with a few unnecessary steps eliminated:
a(n) = 1/s * Sum_{i=1..s} ((p[i]-i) modulo s), where p is the permutation of rank n as ordered in the list A060117, and s is its size (the number of its elements) computed as s = 1+A084558(n).
a(n) = Sum_{i=1..s} [p[i]a(n) = 1/s * Sum_{i=1..s} ((i-p[i]) modulo s). [If inverse permutations from list A060118 are used, then we just flip the order of difference that is used in the first formula].
Following formulas do not need intermediate construction of permutation lists:
a(n) = A001221(A275734(n)).
a(n) = A275806(A225901(n)).
a(n) = A000120(A276010(n)).
Other identities and observations. For all n >= 0:
a(n) = A275946(n) + A275947(n).
a(n) = A060500(A060125(n)).
a(n) = A060128(n) + A276004(n).
a(n) = A060129(n) - A060500(n).
a(n) = A084558(n) - A275849(n) = 1 + A084558(n) - A060501(n).
a(A007489(n)) = n. [Particularly, A007489(n) gives the position of the first occurrence of each n.]
A060128(n) <= a(n) <= A060129(n).
a(n!) = 1.
a(A033312(n)) = 1 for all n > 1.
a(A059590(n)) = A000120(n).
a(A060112(n)) = A007895(n).
a(n) = a(A153880(n)) = a(A255411(n)). [The shift-operations do not change the number of distinct slopes.]
a(A275804(n)) = A060130(A275804(n)). [A275804 gives all the positions where this coincides with A060130.]
(End)

Entry revised, with a new interpretation and formulas. Maple-code cleaned up. - Antti Karttunen, Aug 11 2016

Another new interpretation added and the original definition moved to the comments - Antti Karttunen, Aug 24 2016

A275734 Prime-factorization representations of "factorial base slope polynomials": a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).

Original entry on oeis.org

1, 2, 3, 6, 2, 4, 5, 10, 15, 30, 10, 20, 3, 6, 9, 18, 6, 12, 2, 4, 6, 12, 4, 8, 7, 14, 21, 42, 14, 28, 35, 70, 105, 210, 70, 140, 21, 42, 63, 126, 42, 84, 14, 28, 42, 84, 28, 56, 5, 10, 15, 30, 10, 20, 25, 50, 75, 150, 50, 100, 15, 30, 45, 90, 30, 60, 10, 20, 30, 60, 20, 40, 3, 6, 9, 18, 6, 12, 15, 30, 45, 90, 30, 60, 9, 18, 27

Offset: 0

Antti Karttunen, Aug 08 2016

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of nonzero digits that occur on the slope (k-1) levels below the "maximal slope" in the factorial base representation of n. See A275811 for the definition of the "digit slopes" in this context.

			For n=23 ("321" in factorial base representation, A007623), all three nonzero digits are maximal for their positions (they all occur on "maximal slope"), thus a(23) = prime(1)^3 = 2^3 = 8.
For n=29 ("1021"), there are three nonzero digits, where both 2 and the rightmost 1 are on the "maximal slope", while the most significant 1 is on the "sub-sub-sub-maximal", thus a(29) = prime(1)^2 * prime(4)^1 = 2*7 = 28.
For n=37 ("1201"), there are three nonzero digits, where the rightmost 1 is on the maximal slope, 2 is on the sub-maximal, and the most significant 1 is on the "sub-sub-sub-maximal", thus a(37) = prime(1) * prime(2) * prime(4) = 2*3*7 = 42.
For n=55 ("2101"), the least significant 1 is on the maximal slope, and the digits "21" at the beginning are together on the sub-sub-maximal slope (as they are both two less than the maximal digit values 4 and 3 allowed in those positions), thus a(55) = prime(1)^1 * prime(3)^2 = 2*25 = 50.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A001221, A001222, A002110, A007489, A007814, A048675, A051903, A056169, A056170, A060130, A060502, A225901.
Cf. A257684, A275732.
Cf. A275811.
Cf. A260736, A275728, A275808, A275812, A275946, A275947, A275962.
Cf. A275804 (indices of squarefree terms), A275805 (of terms not squarefree).
Cf. also A275725, A275733, A275735, A276076 for other such prime factorization encodings of A060117/A060118-related polynomials.

from operator import mul
from sympy import prime, factorial as f
def a007623(n, p=2): return n if n0 else '0' for i in x)[::-1]
    return 0 if n==1 else sum(int(y[i])*f(i + 1) for i in range(len(y)))
def a(n): return 1 if n==0 else a275732(n)*a(a257684(n))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 19 2017

a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).
Other identities and observations. For all n >= 0:
a(n) = A275735(A225901(n)).
a(A007489(n)) = A002110(n).
A001221(a(n)) = A060502(n).
A001222(a(n)) = A060130(n).
A007814(a(n)) = A260736(n).
A051903(a(n)) = A275811(n).
A048675(a(n)) = A275728(n).
A248663(a(n)) = A275808(n).
A056169(a(n)) = A275946(n).
A056170(a(n)) = A275947(n).
A275812(a(n)) = A275962(n).

A275735 Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

Original entry on oeis.org

1, 2, 2, 4, 3, 6, 2, 4, 4, 8, 6, 12, 3, 6, 6, 12, 9, 18, 5, 10, 10, 20, 15, 30, 2, 4, 4, 8, 6, 12, 4, 8, 8, 16, 12, 24, 6, 12, 12, 24, 18, 36, 10, 20, 20, 40, 30, 60, 3, 6, 6, 12, 9, 18, 6, 12, 12, 24, 18, 36, 9, 18, 18, 36, 27, 54, 15, 30, 30, 60, 45, 90, 5, 10, 10, 20, 15, 30, 10, 20, 20, 40, 30, 60, 15, 30, 30, 60, 45, 90, 25, 50, 50, 100, 75

Offset: 0

Antti Karttunen, Aug 09 2016

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of times a nonzero digit k occurs in the factorial base representation of n. See the examples.

			For n = 0 whose factorial base representation (A007623) is also 0, there are no nonzero digits at all, thus there cannot be any prime present in the encoding, and a(0) = 1.
For n = 1 there is just one 1, thus a(1) = prime(1) = 2.
For n = 2 ("10"), there is just one 1-digit, thus a(2) = prime(1) = 2.
For n = 3 ("11") there are two 1-digits, thus a(3) = prime(1)^2 = 4.
For n = 18 ("300") there is just one 3, thus a(18) = prime(3) = 5.
For n = 19 ("301") there is one 1 and one 3, thus a(19) = prime(1)*prime(3) = 2*5 = 10.
For n = 141 ("10311") there are three 1's and one 3, thus a(141) = prime(1)^3 * prime(3) = 2^3 * 5^1 = 40.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A000079, A001221, A001222, A003961, A007623, A008683, A181819, A225901, A257511, A257684, A265349, A265350, A264990, A275729, A275806, A351954.
Cf. also A275725, A275733, A275734 for other such prime factorization encodings of A060117/A060118-related polynomials, and also A276076.
Differs from A227154 for the first time at n=18, where a(18) = 5, while A227154(18) = 4.

A276076(n) = { my(i=0,m=1,f=1,nextf); while((n>0),i=i+1; nextf = (i+1)*f; if((n%nextf),m*=(prime(i)^((n%nextf)/f));n-=(n%nextf));f=nextf); m; };
A181819(n) = factorback(apply(e->prime(e),(factor(n)[,2])));
A275735(n) = A181819(A276076(n)); \\ Antti Karttunen, Apr 03 2022

from sympy import prime
from operator import mul
import collections
def a007623(n, p=2): return n if nIndranil Ghosh, Jun 19 2017

a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).
Other identities and observations. For all n >= 0:
a(n) = A275734(A225901(n)).
A001221(a(n)) = A275806(n).
A001222(a(n)) = A060130(n).
A048675(a(n)) = A275729(n).
A051903(a(n)) = A264990(n).
A008683(a(A265349(n))) = -1 or +1 for all n >= 0.
A008683(a(A265350(n))) = 0 for all n >= 1.
From Antti Karttunen, Apr 03 2022: (Start)
A342001(a(n)) = A351954(n).
a(n) = A181819(A276076(n)). (End)

A275725 a(n) = A275723(A002110(1+A084558(n)), n); prime factorization encodings of cycle-polynomials computed for finite permutations listed in the order that is used in tables A060117 / A060118.

Original entry on oeis.org

2, 4, 18, 8, 12, 8, 150, 100, 54, 16, 24, 16, 90, 40, 54, 16, 36, 16, 60, 40, 36, 16, 24, 16, 1470, 980, 882, 392, 588, 392, 750, 500, 162, 32, 48, 32, 270, 80, 162, 32, 108, 32, 120, 80, 72, 32, 48, 32, 1050, 700, 378, 112, 168, 112, 750, 500, 162, 32, 48, 32, 450, 200, 162, 32, 72, 32, 300, 200, 108, 32, 48, 32, 630, 280, 378, 112, 252, 112, 450, 200

Offset: 0

Antti Karttunen, Aug 09 2016

nonn

In this context "cycle-polynomials" are single-variable polynomials where the coefficients (encoded with the exponents of prime factorization of n) are equal to the lengths of cycles in the permutation listed with index n in tables A060117 or A060118. See the examples.

			Consider the first eight permutations (indices 0-7) listed in A060117:
  1 [Only the first 1-cycle explicitly listed thus a(0) = 2^1 = 2]
  2,1 [One transposition (2-cycle) in beginning, thus a(1) = 2^2 = 4]
  1,3,2 [One fixed element in beginning, then transposition, thus a(2) = 2^1 * 3^2 = 18]
  3,1,2 [One 3-cycle, thus a(3) = 2^3 = 8]
  3,2,1 [One transposition jumping over a fixed element, a(4) = 2^2 * 3^1 = 12]
  2,3,1 [One 3-cycle, thus a(5) = 2^3 = 8]
  1,2,4,3 [Two 1-cycles, then a 2-cycle, thus a(6) = 2^1 * 3^1 * 5^2 = 150].
  2,1,4,3 [Two 2-cycles, not crossed, thus a(7) = 2^2 * 5^2 = 100]
and also the seventeenth one at n=16 [A007623(16)=220] where we have:
  3,4,1,2 [Two 2-cycles crossed, thus a(16) = 2^2 * 3^2 = 36].

Antti Karttunen, Table of n, a(n) for n = 0..40319
Indranil Ghosh, Python program for computing this sequence
Index entries for sequences related to factorial base representation

Cf. A000040, A001222, A001221, A002110, A007814, A046660, A048675, A051903, A056169, A056170, A084558, A243054, A257510, A275723, A275803, A275832.
Cf. A275807 (terms divided by 2).
Cf. also A060129, A060128, A060130, A060131, A072411, A275726, A275812, A275851.
Cf. also A275733, A275734, A275735 for other such prime factorization encodings of A060117/A060118-related polynomials.

(define (A275725 n) (A275723bi (A002110 (+ 1 (A084558 n))) n)) ;; Code for A275723bi given in A275723.

a(n) = A275723(A002110(1+A084558(n)), n).
Other identities:
A001221(a(n)) = 1+A257510(n) (for all n >= 1).
A001222(a(n)) = 1+A084558(n).
A007814(a(n)) = A275832(n).
A048675(a(n)) = A275726(n).
A051903(a(n)) = A275803(n).
A056169(a(n)) = A275851(n).
A046660(a(n)) = A060130(n).
A072411(a(n)) = A060131(n).
A056170(a(n)) = A060128(n).
A275812(a(n)) = A060129(n).
a(n!) = 2 * A243054(n) = A000040(n)*A002110(n) for all n >= 1.

A257511 Number of 1's in factorial base representation of n (A007623).

Original entry on oeis.org

0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, 4, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1

Offset: 0

Antti Karttunen, Apr 27 2015

Antti Karttunen, Table of n, a(n) for n = 0..10080
Index entries for sequences related to factorial base representation

Cf. A001221, A007623, A007814, A034968, A056169, A060130, A225901, A257687, A265333, A275732, A275735, A260736, A276076.
Cf. A255411 (numbers n such that a(n) = 0), A255341 (such that a(n) = 1), A255342 (such that a(n) = 2), A255343 (such that a(n) = 3).
Positions of records: A007489.
Cf. also A257510.

factBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > i!, i++]; m = n; len = i; dList = Table[0, {len}]; Do[currDigit = 0; While[m >= j!, m = m - j!; currDigit++]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; dList]; s = Table[FromDigits[factBaseIntDs@ n], {n, 0, 120}];
First@ DigitCount[#] & /@ s (* Michael De Vlieger, Apr 27 2015, after Alonso del Arte at A007623 *)
nn = 120; b = Module[{m = 1}, While[Factorial@ m < nn, m++]; MixedRadix[Reverse@ Range[2, m]]]; Table[Count[IntegerDigits[n, b], 1], {n, 0, nn}] (* Michael De Vlieger, Aug 29 2016, Version 10.2 *)

(define (A257511 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (floor->exact (/ n i)) (+ 1 i) (+ s (if (= 1 (modulo n i)) 1 0)))))))

a(0) = 0; for n >= 1, a(n) = A265333(n) + a(A257687(n)). - Antti Karttunen, Aug 29 2016
Other identities and observations. For all n >= 0:
a(n) = A260736(A225901(n)).
a(n) = A001221(A275732(n)) = A001222(A275732(n)).
a(n) = A007814(A275735(n)).
a(n) = A056169(A276076(n)).
a(A007489(n)) = n. [Particularly, A007489(n) gives the position where n first appears.]
a(n) <= A060130(n) <= A034968(n).

A257687 Discard the most significant digit from factorial base representation of n, then convert back to decimal: a(n) = n - A257686(n).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0

Offset: 0

Antti Karttunen, May 04 2015

A060130(n) gives the number of steps needed to reach zero, when starting iterating as a(k), a(a(k)), etc., from the starting value k = n.

			Factorial base representation (A007623) of 1 is "1", discarding the most significant digit leaves nothing, taken to be zero, thus a(1) = 0.
Factorial base representation of 2 is "10", discarding the most significant digit leaves "0", thus a(2) = 0.
Factorial base representation of 3 is "11", discarding the most significant digit leaves "1", thus a(3) = 1.
Factorial base representation of 4 is "20", discarding the most significant digit leaves "0", thus a(4) = 0.

Antti Karttunen, Table of n, a(n) for n = 0..10080

Cf. A007623, A257686.
Can be used (together with A099563) to define simple recurrences for sequences like A034968, A060130, A227153, A246359, A257511, A257679, A257680.
Cf. also A257684.

f[n_] := Block[{m = p = 1}, While[p*(m + 1) <= n, p = p*m; m++]; Mod[n, p]]; Array[f, 101, 0] (* Robert G. Wilson v, Jul 21 2015 *)

from sympy import factorial as f
def a007623(n, p=2): return n if nIndranil Ghosh, Jun 21 2017

Scheme
```
(define (A257687 n) (- n (A257686 n)))
```

a(n) = n - A257686(n).

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.

A276091 Numbers obtained by reinterpreting base-2 representation of n in A001563-base (A276326): a(n) = Sum_{k>=0} A030308(n,k)*A001563(k+1).

Original entry on oeis.org

0, 1, 4, 5, 18, 19, 22, 23, 96, 97, 100, 101, 114, 115, 118, 119, 600, 601, 604, 605, 618, 619, 622, 623, 696, 697, 700, 701, 714, 715, 718, 719, 4320, 4321, 4324, 4325, 4338, 4339, 4342, 4343, 4416, 4417, 4420, 4421, 4434, 4435, 4438, 4439, 4920, 4921, 4924, 4925, 4938, 4939, 4942, 4943, 5016, 5017, 5020, 5021, 5034, 5035, 5038, 5039, 35280, 35281

Offset: 0

Antti Karttunen, Aug 19 2016

Numbers that are sums of distinct terms of A001563.
A number is included if and only if all the nonzero digits in its factorial base representation (A007623) are maximal allowed in those digit positions, thus this sequence gives all numbers n for which A060130(n) = A260736(n).
Numbers n for which A276328(n) = A276337(n), thus from 1 onward the positions of ones in A276336.
Conjectured also to give all numbers n for which A255411(n) = A276340(n) (thus zeros of A276339).

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences related to factorial base representation

Cf. A000120, A000142, A001563, A030308, A059590, A060130, A260736, A225901, A255411, A275959, A276082, A276083, A276090, A276326, A276328, A276336, A276337, A276339, A276340.

Table[Total[Times @@@ Transpose@ {Map[# #! &, Range@ Length@ #], Reverse@ #}] &@ IntegerDigits[n, 2], {n, 64}] (* Michael De Vlieger, Aug 31 2016 *)

from sympy import factorial as f
def a007623(n, p=2): return n if n0 else '0' for i in x)[::-1]
    return 0 if n==0 else sum(int(y[i])*f(i + 1) for i in range(len(y)))
def a(n): return 0 if n==0 else a255411(a(n//2)) if n%2==0 else 1 + a255411(a((n - 1)//2))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 20 2017

;; This is a standalone program:
(define (A276091 n) (let loop ((n n) (s 0) (f 1) (i 2)) (cond ((zero? n) s) ((even? n) (loop (/ n 2) s (* i f) (+ 1 i))) (else (loop (/ (- n 1) 2) (+ s (* (- i 1) f)) (* i f) (+ 1 i))))))
;; This implements one of the given recurrences:
(definec (A276091 n) (cond ((zero? n) n) ((even? n) (A255411 (A276091 (/ n 2)))) (else (+ 1 (A255411 (A276091 (/ (- n 1) 2)))))))
;; Alternatively, we can use A276340 in place of A255411:
(definec (A276091 n) (cond ((zero? n) n) ((even? n) (A276340 (A276091 (/ n 2)))) (else (+ 1 (A276340 (A276091 (/ (- n 1) 2)))))))

a(0) = 0, a(2n) = A255411(a(n)), a(2n+1) = 1+A255411(a(n)).
a(0) = 0, a(2n) = A276340(a(n)), a(2n+1) = 1+A276340(a(n)).
Other identities. For all n >= 0:
a(n) = A225901(A059590(n)).
a(n) = A276090(A275959(n)).
A276328(a(n)) = A276337(a(n)) = A000120(n).

Name changed (to emphasize the functional nature of the sequence) with the original definition moved to the comments by Antti Karttunen, Sep 01 2016

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

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A267263 Number of nonzero digits in representation of n in primorial base.

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A060502 a(n) = number of occupied digit slopes in the factorial base representation of n (see comments for the definition); number of drops in the n-th permutation of list A060117.

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A275734 Prime-factorization representations of "factorial base slope polynomials": a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).

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A275735 Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

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A275725 a(n) = A275723(A002110(1+A084558(n)), n); prime factorization encodings of cycle-polynomials computed for finite permutations listed in the order that is used in tables A060117 / A060118.

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A257511 Number of 1's in factorial base representation of n (A007623).

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