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As A040115 forms the basis of an iterative sequence leading to A329200 and A329201, this sequence forms the basis of a similar sequence A329624. As the concatenation of the digit sum can lead to a value larger than the original term we must take the absolute value of the difference to ensure subsequent terms are always positive. The largest value in the first 10000 terms is a(9991) = 171819.

Computed by Hans Havermann, Nov 01 2019.

There is a simple technique, due to Hans Havermann, that proves that many terms that in this sequence blow up: find a term in the trajectory that has an internal substring of three 9's. See A328974 for the case 1496.

The same reasoning shows that almost all numbers belong to the sequence.

From Scott R. Shannon, Nov 23 2019: (Start)

Although many of the numbers in this sequence eventually reach a term that contains three 9's, some do not. The first such example is 4949 which leads to 44444 after two steps, and starts a sequence of values that leads back to a much larger all-4's number every nine steps. No 9's appear in any of the intermediate values. Similar series found which lead to limitless loops are values with all 4's with a final 5, or all 3's with a final digit of 1 to 6. Of the first 33909 starting values that increase without limit 209 lead into one of these non-9 loops. (End)

From Scott R. Shannon, Nov 24 2019: (Start)

One can show that strings of repeated digits longer than a certain length will increase without limit by calculating the number of digits when a new value containing only the same digit appears again in the iterative sequence. Below shows the details of this for digits 1 to 9. The number of cycles is the number of iterations of A053392 required before a new value containing only the starting digit is seen. The minimum starting value is the smallest same-digit number such that subsequent iterations will produce another value with only the same digit of equal or longer length. The number of digits in the subsequence reoccurrence shows the number of digits in this value given the starting value has n digits. All sufficiently long single-digit numbers, with digits 1 to 8, increase 8-fold in length, minus a constant, after 9 iterations.

.

digit | # of cycles | min start value | # digits in reoccurrence

1 | 9 | 1111111 | 8*n - 45

2 | 9 | 222222 | 8*n - 38

3 | 3 | 33333 | 2*n - 5

4 | 9 | 44444 | 8*n - 31

5 | 9 | 55555 | 8*n - 33

6 | 3 | 6666 | 2*n - 4

7 | 9 | 77777 | 8*n - 27

8 | 9 | 8888 | 8*n - 28

9 | 2 | 999 | 2*n - 3

(End)

Apart from 0, the terms listed so far are all of period 2 or 3. Are there longer periods?

1496 is the smallest number whose trajectory under A053392 increases without limit.

More terms than usual are shown in order to display the onset of exponential growth.

Proof that this grows without limit, from Hans Havermann, Nov 01 2019: (Start)

One way to prove that the trajectory of a number under repeated application of the map defined in A053392 increases without limit is to show that there exists a term containing a non-final substring of three adjacent 9's.

If such a substring in term n is followed by a 0, it will grow to a non-final substring of three adjacent 9's followed by a 1 in term n+2: *9990* --> *18189* --> *99917*

If such a substring in term n is followed by a digit d that is neither 0 nor 9, it will grow to a non-final substring of four adjacent 9's in term n+2: *999d* --> *18181(d-1)* --> *9999d*

If such a substring in term n is followed by another 9, then it is a substring of k 9's for k >= 4, which will grow to a substring of (2k-3) 9's in term n+2: *[9]^d* --> *[18]^(d-1)* --> *[9]^(2d-3)*

Putting those together, such a substring in term n will grow to four adjacent 9's by term n+4; to five adjacent 9's by term n+6; to seven adjacent 9's by term n+8; ... to 2^k+3 adjacent 9's (see A062709) by term n+4+2k, regardless of what happens in the rest of the number.

In the present sequence a(41) contains two non-final substrings of three adjacent 9's. QED (End)

[Argument corrected and completed by David J. Seal, Nov 05 2019.]

24th and 26th terms are unknown, but a(25)=9539, a(27)=4589 and a(28)=5960.

See link for the precise definition.

This sequence is based on the following iterative cycle. Start with n, set m = A040115(n), the concatenation of the absolute values of differences between adjacent digits, and then repeat the following until the number m has been previously seen: if m is greater than n, let m = A040115(m), otherwise let m = A053392(m), the concatenation of the sums of pairs of adjacent digits.

For all starting values n the iteration eventually converges to 0 or else goes into a cycle of finite length. When the number m gets larger than the iteration's starting value n it will always have its magnitude decreased by the operation m = A040115(m), while m = A053392(m) can either increase or decrease its magnitude, depending on the digit values of m. This has the overall effect of never allowing the iterative values to increase without limit as is seen in the similar iterations A328975 and A329624.

All the values of A053393 are seen as repeating values in this sequence, although this sequence has significantly more; probably an infinite number, although this is unknown. The first nonzero repeating value is not seen until a(9090), which forms the two-member loop of 999 -> 1818 -> 999. The first starting value that leads to an m value greater than the initial starting value is a(10090), see examples below. A330159 lists the starting values which are also the first repeating value.

For the first 20 million terms the longest iterative sequence is seen for a(18505180) which takes 457 steps before reaching 0. See attached link. The longest found looping sequence is for a(14106482) which reaches 1040103 after 5 steps and then again after 116 steps, forming a loop of length 111. The largest number found which starts the repeating loop is for a(9265011) which reaches 1411131715 after 9 iterations and then again after 41 iterations.

From a(12) to a(99) the sequence repeats a pattern of ten 3's followed by a 2. After that, a(100) = 4 and the terms begin to show a slow average increase in value.

The Friedman web site quotes B. J. van der Zwaag as saying that the three-digit terms are those for which the middle digit is >= the sum of the other two digits.

As A053392(n) can be either larger or smaller than n this sequence lists n when it is either a multiple or a divisor of A053392(n). In the majority of terms n is a multiple of A053392(n); the first case where n is a divisor is a(27) = 182, where A053392(182) = 910.

All numbers of the form n = k*10^t, with k,t>=1 are in the sequence, as are numbers n = 75*10^t, with t>=1. Also present are numbers of the form n = 444...445 which have A053392 values like 888...889, for which n = 5*A053392(n). Similarly numbers of the form n = 444...455, which have A053392 values like 888...8910, for which A053392(n) = 2*n. For numbers up to 10^10 the largest term which is not one of these forms is a(120) = 654653884, which divides A053392(654653884) = 11910118111612.