A231719 -id:A231719 - cp's OEIS frontend

A219666 The infinite trunk of factorial expansion beanstalk. The only infinite sequence such that a(n-1) = a(n) - sum of digits in factorial expansion of a(n).

0, 1, 2, 5, 7, 10, 12, 17, 23, 25, 28, 30, 35, 40, 46, 48, 52, 57, 63, 70, 74, 79, 85, 92, 97, 102, 109, 119, 121, 124, 126, 131, 136, 142, 144, 148, 153, 159, 166, 170, 175, 181, 188, 193, 198, 204, 213, 221, 228, 238, 240, 244, 249, 255, 262, 266, 271, 277

Offset: 0

Antti Karttunen, Nov 25 2012

a(n) tells in what number we end in n steps, when we start climbing up the infinite trunk of the "factorial beanstalk" from its root (zero).
There are many finite sequences such as 0,1,2,4; 0,1,2,5,6; etc. obeying the same condition (see A219659) and as the length increases, so (necessarily) does the similarity to this infinite sequence.
See A007623 for the factorial number system representation.

Antti Karttunen, Table of n, a(n) for n = 0..21622

Cf. A007623, A034968, A219651, A230411, A226061. For all n, A219652(a(n)) = n and A219653(n) <= a(n) <= A219655(n).
Characteristic function: Χ_A219666(n) = A230418(n+1)-A230418(n).
The first differences: A230406.
Other derived sequences: A230425-A230427, A230430, A230407-A230409, A219662 & A219663, A231723 & A231724, A230420, A230410, A231717, A231719.
Subsets: A230428 & A230429.
Analogous sequence for binary system: A179016, for Fibonacci number system: A219648.

nn = 10^3; m = 1; While[m! < Floor[6 nn/5], m++]; m; t = TakeWhile[Reverse@ NestWhileList[# - Total@ IntegerDigits[#, MixedRadix[Reverse@ Range[2, m]]] &, Floor[6 nn/5], # > 0 &], # <= nn &] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

;; Memoizing definec-macro from Antti Karttunen's IntSeq-library
(definec (A219666 n) (cond ((<= n 2) n) ((= (A226061 (A230411 n)) n) (- (A000142 (A230411 n)) 1)) (else (- (A219666 (+ n 1)) (A034968 (A219666 (+ n 1)))))))
;; Another variant, utilizing A230416 (which gives a more convenient way to compute large number of terms of this sequence):
(define (A219666 n) (A230416 (A230432 n)))
;; This function is for checking whether n belongs to this sequence:
(define (inA219666? n) (or (zero? n) (= 1 (- (A230418 (+ 1 n)) (A230418 n)))))

a(0) = 0, a(1) = 1, and for n>1, if A226061(A230411(n)) = n then a(n) = A230411(n)!-1, otherwise a(n) = a(n+1) - A034968(a(n+1)).

a(n) = A230416(A230432(n)).

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

Original entry on oeis.org

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

A226061 Partial sums of A219661.

Original entry on oeis.org

0, 1, 3, 8, 27, 110, 538, 3149, 21622, 172348, 1549896, 15401144, 168011252, 2003304293, 25928878272, 361788001015, 5411160126367, 86353882249911, 1464841397585335, 26323224850512719, 499551889319197565

Offset: 1

Antti Karttunen, May 28 2013

nonn

a(n) tells the position of (n!)-1 in A219666.

One less than A219665.
Cf. A219661, A219666.
Analogous sequence for binary system: A218600.
Cf. also A230410, A231719.

Accumulate@ Table[Length@ NestWhileList[# - Total@ IntegerDigits[#,
MixedRadix[Reverse@ Range[2, 120]]] &, (n + 1)! - 1, # > n! - 1 &] - 1, {n, 0, 8}] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

a(n) = a(n-1) + A219661(n-1) with a(1) = 0.
a(n) = A219652(n!-1).
a(n) = A219665(n) - 1.

Terms a(16) - a(21) computed from the new terms of A219661 by Antti Karttunen, Jun 27 2016

A230410 After a(0)=0, a(n) = A230415(A219666(n),A219666(n-1)).

Original entry on oeis.org

0, 1, 2, 2, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 1, 3, 1, 3, 1, 3, 3, 3, 1, 3, 4, 2, 2, 2, 4, 2, 2, 2, 2, 1, 3, 1, 3, 1, 3, 3, 3, 1, 3, 4, 2, 1, 3, 3, 3, 2, 4, 1, 3, 1, 3, 3, 3, 1, 3, 4, 2, 1, 2, 2, 2, 2, 3, 2, 2, 4, 3, 1, 3, 4, 2, 1, 2, 2, 2, 2, 3, 2, 1, 3, 2, 5, 2

Offset: 0

Antti Karttunen, Nov 10 2013

After zero, a(n) = number of positions where digits in the factorial base representations of successive nodes A219666(n-1) and A219666(n) in the infinite trunk of the factorial beanstalk differ from each other.

			a(8) = 1, because A219666(8)=23, whose factorial base representation (A007623(23)) is '321', and A219666(7)=17, whose factorial base representation (A007623(17)) is '221', and they differ just in one digit position.
a(9) = 3, because A219666(9)=25, '...01001' in factorial base, which differs from '...0321' in three digit positions.
Note that A226061(4)=8 (A226061(n) tells the position of (n!)-1 in A219666), and 1+2+3 = 6 happens to be both a triangular number (A000217) and a factorial number (A000142).
The next time 1 occurs in this sequence because of this coincidence is at x=A226061(16) (whose value is currently not known), as at that point A219666(x) = 16!-1 = 20922789887999, whose factorial base representation is (15,14,13,12,11,10,9,8,7,6,5,4,3,2,1), and A000217(15) = 120 = A000142(5), which means that A219666(x-1) = A219651(20922789887999) = 20922789887879, whose factorial base representation is (15,14,13,12,11,10,9,8,7,6,4,4,3,2,1), which differs only in one position from the previous.
Of course 1's occur in this sequence for other reasons as well.

Antti Karttunen, Table of n, a(n) for n = 0..3149

Cf. A230415, A230406, A231717, A231719, A232094. A230422 gives the positions of ones.

nn = 1200; m = 1; While[m! < nn, m++]; m; f[n_] := IntegerDigits[n, MixedRadix[Reverse@ Range[2, m]]]; Join[{0}, Function[w, Count[Subtract @@ Map[PadLeft[#, Max@ Map[Length, w]] &, w], k_ /; k != 0]]@ Map[f@ # &, {#1, #2}] & @@@ Partition[#, 2, 1] &@ TakeWhile[Reverse@ NestWhileList[# - Total@ f@ # &, nn, # > 0 &], # <= 500 &]] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

(define (A230410 n) (if (zero? n) n (A230415bi (A219666 n) (A219666 (- n 1))))) ;; Where bi-variate function A230415bi has been given in A230415.

a(0)=0, and for n>=1, a(n) = A230415(A219666(n),A219666(n-1)).

For all n, a(A226061(n+1)) = A232094(n).

A232096 a(n) = largest m such that m! divides 1+2+...+n; a(n) = A055881(A000217(n)).

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Nov 18 2013

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10080

A042963 gives the positions of ones and A014601 the positions of larger terms.

Cf. also A232097, A076733, A232094, A232095, A232098.

(define (A232096 n) (A055881 (A000217 n)))

a(n) = A055881(A000217(n)).

a(n) = A231719(A226061(n+1)). [Not a practical way to compute this sequence, but follows from the definitions]

A231717 After a(0)=0, a(n) = A231713(A219666(n),A219666(n-1)).

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 3, 1, 6, 3, 3, 3, 2, 1, 6, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 2, 3, 10, 3, 3, 3, 2, 1, 6, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 1, 3, 5, 5, 3, 10, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 1, 2, 4, 2, 4, 5, 3, 3, 9, 3, 1, 3, 6, 2, 1, 2, 4, 2, 4, 5, 3, 2, 4, 3, 10

Offset: 0

Antti Karttunen, Nov 12 2013

nonn

For all n, a(A226061(n+1)) = A232095(n). This works because at the positions given by each x=A226061(n+1), it holds that A219666(x) = (n+1)!-1, which has a factorial base representation (A007623) of (n,n-1,n-2,...,3,2,1) whose digit sum (A034968) is the n-th triangular number, A000217(n). This in turn is always a new record as at those points, in each significant digit position so far employed, a maximal digit value (for factorial number system) is used, and thus the preceding term, A219666(x-1) cannot have any larger digits in its factorial base representation, and so the differences between their digits (in matching positions) are all nonnegative.

Antti Karttunen, Table of n, a(n) for n = 0..3149

A231718 gives the positions of ones.

Cf. also A230410, A231719, A232095.

(define (A231717 n) (if (zero? n) n (A231713bi (A219666 n) (A219666 (- n 1)))))

a(0)=0, and for n>=1, a(n) = A231713(A219666(n),A219666(n-1)).

cp's OEIS Frontend

A219666 The infinite trunk of factorial expansion beanstalk. The only infinite sequence such that a(n-1) = a(n) - sum of digits in factorial expansion of a(n).

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

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A226061 Partial sums of A219661.

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A230410 After a(0)=0, a(n) = A230415(A219666(n),A219666(n-1)).

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A232096 a(n) = largest m such that m! divides 1+2+...+n; a(n) = A055881(A000217(n)).

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A231717 After a(0)=0, a(n) = A231713(A219666(n),A219666(n-1)).

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A231718 Positions of ones in A231717.

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