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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)).

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

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OFFSET

0,1

COMMENTS

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)

LINKS

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

EXAMPLE

a(11) = 112481623283849627077, which has digit-sum 91.

So k = 11248162328384962707791 both begins and ends with 1.