A359143 -id:A359143 - cp's OEIS frontend

A361350 A variant of A359143 which includes the intermediate terms before digits are deleted (see Comments for precise definition).

11, 112, 1124, 11248, 1124816, 2486, 248620, 4860, 486018, 48601827, 4860182736, 486018273645, 8601827365, 860182736546, 86018273654656, 8601827365465667, 860182736546566780, 601273654656670, 60127365465667064, -1273545704, -127354570438, -12735457043849, -1273545704384962, -127354570438496270, 1273545743849627, 127354574384962777, 12735457438496277791, 273545743849627779

Offset: 0

N. J. A. Sloane, Mar 16 2023

This is essentially the same sequence as A359143 (so this too is a finite sequence), the difference being that it includes the terms before any digits are cancelled. Let S be the digit string of a(n), replacing a minus sign if present by 0.
Let T = S concatenated with the digit-sum of S.
If the leading digit of T is not present in the digit-sum of S, then a(n+1) = A359142(T), as in A359143.
If the leading digit of T is present in the digit-sum of S, then we add two new terms instead of one: a(n+1) = a(n) concatenated with the digit-sum of S, and a(n+2) = A359142(T), as in A359143.

			The digit strings for the initial terms are:
  11,
  112,
  1124,
  11248,
  1124816,
  2486,
  248620,
  4860,
  486018,
  48601827,
  4860182736,
  486018273645,
  8601827365,
  860182736546,
  86018273654656,
  8601827365465667,
  860182736546566780,
  601273654656670,
  60127365465667064,
  01273545704,
  0127354570438,
  012735457043849,
  01273545704384962,
  0127354570438496270,
  1273545743849627,
  127354574384962777,
  12735457438496277791,
  273545743849627779, ...
The sequence itself is obtained by replacing the leading zeros by minus signs.
For example, after the term 601273654656670, we first append its digit-sum 64, getting 60127365465667064. Since the leading digit 6 is present in 64, we cancel all the 6's, getting 01273545704. The corresponding term in the sequence is -1273545704.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Michael De Vlieger, Scatterplot of log_10(abs(a(n))), n = 1..10^3, showing negative terms in red.
Michael De Vlieger, Scatterplot of log_10(abs(a(n))), n = 1..10^4, showing negative terms in red.
Michael De Vlieger, Scatterplot of log_10(abs(a(n))), showing all terms, with negative terms in red.

Cf. A359142, A359143.

a[1] = {1, 1}; nn = 28;
Do[Which[ListQ[m], k = m; Clear[m],
      FreeQ[#3, #2], Set[k, #1~Join~#3],
      True, Set[k, #1~Join~#3];
      Set[m, 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 *)

More than the usual number of terms are shown in order to clarify the differences from A359143.

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 *)

A359142 Let s = sum of digits of n, let t = decimal concatenation of n and s, let u be obtained by deleting all copies of the leading digit of t from t, if this digit occurs in s. Then if u has only zero digits, a(n) = 0; if u has leading digit 0 but not all its digits are 0, delete all leading 0's from u and negate the result to get a(n); otherwise a(n) = u.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 112, 123, 134, 145, 156, 167, 178, 189, 90, 0, 213, 224, 235, 246, 257, 268, 279, 2810, 2911, 0, 314, 325, 336, 347, 358, 369, 3710, 3811, 3912, 0, 415, 426, 437, 448, 459, 4610, 4711, 4812, 4913, 0, 516, 527, 538, 549, 5510, 5611, 5712, 5813

Offset: 1

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

Since nonzero numbers may not have leading zeros, we indicate their presence by negating the number.
For use in A359142, we also define a(n) for improper decimal numbers n with leading zeros by following exactly the same steps. To get a(073), for example, we append the digit sum 10, and delete the 0's from 07310, getting a(073) = 731. To get a(074), we append 11, getting 07411, so a(074) = -7411.
The sequence of n such that a(n) < 0 begins (109, 1009, 1018, 1019, 1027, 1028, 1029, 1036, ...): see A359144. - M. F. Hasler, Feb 01 2023

			Examples:
  n.....s......t....u.....a(n)
  1.....1.....11....0......0
  2.....2.....22....0......0
  .....
  9.....9.....99....0......0
  10....1....101....0......0
  11....2....112..112....112
  12....3....123..123....123
  .....
  100...1...1001...00......0
  101...2...1012.1012...1012
  102...3...1023.1023...1023
  .....
  109..10..10910..090....-90
  .....

Michael S. Branicky, Table of n, a(n) for n = 1..10000
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]
Hans Havermann, Cycles in Eric Angelini's sum-and-erase, Personal blog "Glad Hobo Express", blogspot.com, Jul 28 2022.
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. A359143, A359144 (k such that a(k)<0).

A359142[n_]:= Module[{d=IntegerDigits[n],s,u},If[MemberQ[s=IntegerDigits[Total[d]],First[u=Join[d,s]]],u=DeleteCases[u,First[u]]];If[u=={}||First[u]==0,-1,1]FromDigits[u]];Array[A359142,100] (* Paolo Xausa, Oct 11 2023 *)

A359142(n) = { my(d=digits(n), s=digits(vecsum(d))); n>0 || d=concat(0,d); n=concat(d, s); setsearch(Set(s), d[1]) && n=select(c->c!=d[1], n); if(n && !n[1], -fromdigits(n), fromdigits(n)) }
apply(A359142, [0..99]) \\ M. F. Hasler, Feb 01 2023

def a(n):
    n = str(-n) if isinstance(n, int) and n < 0 else str(n)
    s = str(sum(map(int, n)))
    t = n + s
    u = t.replace(t[0], "") if t[0] in s else t
    return 0 if u == "" else (-int(u) if u[0] == "0" else int(u))
print([a(n) for n in range(1, 59)]) # Michael S. Branicky, Feb 01 2023

More terms from M. F. Hasler, Feb 01 2023

A372074 a(1) = 1; thereafter a(n+1) is obtained by appending the digit-sum of a(n) to a(n).

Original entry on oeis.org

1, 11, 112, 1124, 11248, 1124816, 112481623, 11248162328, 1124816232838, 112481623283849, 11248162328384962, 1124816232838496270, 112481623283849627077, 11248162328384962707791, 11248162328384962707791101, 11248162328384962707791101103, 11248162328384962707791101103107, 11248162328384962707791101103107115

Offset: 1

N. J. A. Sloane, Jun 16 2024

Paolo Xausa, Table of n, a(n) for n = 1..270

Cf. A372075.

Inspired by A359143.

s:=1; j1:=[s];
f:=proc(n) local t1,t2;
t1:=digsum(n);
t2:=length(t1); n*10^t2 + t1; end;
for n from 1 to 24 do s:=f(s); j1:=[op(j1),s]; od:
j1;

NestList[With[{s = DigitSum[#]}, #*10^IntegerLength[s] + s] &, 1, 20] (* Paolo Xausa, Jun 16 2024 *)

from itertools import islice
def A372074_gen(): # generator of terms
    yield (a:=1)
    while True: yield (a:=(s:=sum(map(int,str(a))))+a*10**len(str(s)))
A372074_list = list(islice(A372074_gen(),20)) # Chai Wah Wu, Jun 16 2024

cp's OEIS Frontend

A361350 A variant of A359143 which includes the intermediate terms before digits are deleted (see Comments for precise definition).

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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)).

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A372074 a(1) = 1; thereafter a(n+1) is obtained by appending the digit-sum of a(n) to a(n).

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