A361350 -id:A361350 - cp's OEIS frontend

A359143 The sum-and-erase sequence starting at 11: a(0) = 11; for n>=1, let m = a(n-1), and if m < 0, change m to an improper decimal "number" by replacing the minus sign by a single leading zero; then a(n) = A359142(m).

11, 112, 1124, 11248, 2486, 4860, 486018, 48601827, 4860182736, 8601827365, 860182736546, 86018273654656, 8601827365465667, 601273654656670, -1273545704, -127354570438, -12735457043849, -1273545704384962, 1273545743849627, 127354574384962777, 273545743849627779

Offset: 0

N. J. A. Sloane, Jan 31 2023, based on suggestions from Eric Angelini and Hans Havermann

Although this entry was only created in January, 2023, the problem had already been extensively studied in 2022.
Comment from Michael S. Branicky, Jul 26 2022: (Start)
Starting at 11, this first reaches 0 at step 1399141.
The longest string encountered has length 222:
444444414144444454144444145454455155154545515454564756517555545657676664\
465677675961617616416561527541551562575592651853254255356658359962263264\
365667368971272273374676377982812823836853869892911922935952968991101010\
121016.
(End)
It is conjectured that every starting number will eventually enter a cycle or reach 0 (see A359142 for small examples).
The first nontrivial cycle has length 583792 and the smallest number in it is 3374 (see the "Cycles in ..." Havermann link).
There is no b-file, but instead there is an a-file from Hans Havermann giving the sequence in full in b-file format, from a(0) to a(1399141). Beware, this is a 106.5 MB file. - N. J. A. Sloane, Feb 01 2023
From Michael S. Branicky, Sep 06 2023: (Start)
There are additional cycles with lengths
- 20173, containing 34674044445,
- 46, containing 9982228989928229222222829202026260298265278295291026. (End)

Eric Angelini, Does this iteration end? (Sum and erase), Personal blog "Cinquante Signes", blogspot.com, Jul 26 2022.
Eric Angelini, Does this iteration end? (Sum and erase), Personal blog "Cinquante Signes", blogspot.com, Jul 26 2022. [Cached copy, pdf file, with permission]
Michael S. Branicky, Python program for the sum-and-erase sequence
Jean-Paul Delahaye, Des suites à la dynamique insaisissable, Pour la Science #549, July 2023, pp. 80-85. (Link requires a subscription.)
Hans Havermann, Table of n, a(n) for n = 0..1399141 [Beware, this is a 106.5 MB file.]
Hans Havermann, Cycles in Éric Angelini's sum-and-erase, Glad Hobo Express Blog, Jul 28 2022
Hans Havermann, A cycle of length 49 [Astonishing! - _N. J. A. Sloane_, Jan 31 2023]
N. J. A. Sloane, New Gilbreath Conjectures, Sum and Erase, Dissecting Polygons, and Other New Sequences, Doron Zeilberger's Exper. Math. Seminar, Rutgers, Sep 14 2023: Video, Slides, Updates. (Mentions this sequence.)

Cf. A359142, A361350, A361501.

a[1] = {1, 1}; nn = 21; Do[If[FreeQ[#3, #2], Set[k, #1~Join~#3], Set[k, #1~Join~#3]; Set[k, DeleteCases[#1~Join~#3, #2]]] & @@ {#, First[#], IntegerDigits@ Total[#]} &[a[n - 1]]; Set[a[n], k], {n, 2, nn}]; Array[(1 - 2 Boole[First[#] == 0])*FromDigits@ # &@ a[#] &, nn] (* Michael De Vlieger, Mar 16 2023 *)

A361501 A variant of A359143 in which all copies of a digit d are erased only when d is both the leading digit and the final digit of (a(n) concatenated with sum of digits of a(n)).

Original entry on oeis.org

11, 112, 1124, 11248, 1124816, 112481623, 11248162328, 1124816232838, 112481623283849, 11248162328384962, 1124816232838496270, 112481623283849627077, 2486232838496270779, 248623283849627077997, 248623283849627077997113, 248623283849627077997113118, 248623283849627077997113118128

Offset: 0

N. J. A. Sloane, Mar 17 2023

Since we cannot list nonzero numbers with leading digit 0, we use a minus sign to represent a leading zero.
To compute a(n+1), let m denote the decimal string formed from a(n) by replacing a minus sign (if present) by a leading 0.
Let k denote the concatenation of m and its digit-sum.
If the first and last digits of k are equal, delete all copies of that digit from k.
If k has any leading zeros, replace them with a minus sign.  The result is a(n+1).
A359143 eventually reaches 0, but we do not know if the present sequence will reach 0, enter a loop, or grow without limit towards +infinity or -infinity.
From Michael De Vlieger, Mar 17 2023: (Start)
First negative term is a(146).
Sequence continues beyond 2^30 terms. (End)
From Michael S. Branicky, Mar 21 2023: (Start)
The sequence enters a loop of period L = 224339586.
Specifically, a(35179968) = a(259519554) = 8863336630330333333663833080638368062852636350393323037363535737238.
In this loop, the term with the fewest digits is a(101772740) = 48623, and the term with the most digits is a(251014293), with 940 digits. (End)

			a(11) = 112481623283849627077, which has digit-sum 91.
So k = 11248162328384962707791 both begins and ends with 1.
Erasing all the 1's from k gives a(12) = 2486232838496270779.

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A359142, A359143, A361350.

a[1] = {1, 1}; nn = 17;
Do[If[And[#2 == Last[#3], n > 2],
       Set[k, DeleteCases[#1~Join~#3, #2]],
       Set[k, #1~Join~#3]] & @@
       {#, First[#], IntegerDigits@ Total[#]} &[a[n - 1]];
  Set[a[n], k], {n, 2, nn}];
Array[(1 - 2 Boole[First[#] == 0])*FromDigits[#] &@ a[#] &, nn] (* Michael De Vlieger, Mar 17 2023 *)

cp's OEIS Frontend

A359143 The sum-and-erase sequence starting at 11: a(0) = 11; for n>=1, let m = a(n-1), and if m < 0, change m to an improper decimal "number" by replacing the minus sign by a single leading zero; then a(n) = A359142(m).

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