[seqfan] Re: Concatenate the sums of the neighboring digits

[David Seal]

Hans Havermann wrote:

> After quoting my proof in the comments section of a previous
> version of A328974, DS notes that "This proof is flawed, in
> that the assertion 'Such a substring in term n will grow to
> four adjacent 9's in term n+2' is not true if the substring
> is followed by 0 and not preceded by a fourth 9."
> 
> There were two ambiguities in my proof statements that might
> lead one down that road. The first ambiguity is that when I
> stated that "such a substring [of three adjacent 9's] in term
> n ..." I had not meant n to be the starting number. I would
> fix that by bracketing "n>1" after "term n". ...

I'll admit to not having noticed that ambiguity, but excluding the starting number isn't a good enough fix, because substrings of the form I describe can occur after the starting number. An example is 290900 --> 119990 --> 21018189 --> 31199917, in which a sequence of the type I describe arises after the first step.

However, on checking exactly what's required of multi-step predecessors for substrings of the form I describe, I find that they don't have >=2-step predecessors. Specifically:

* A predecessor of *n9990*, where n is not 9 and * is a wildcard representing any sequence of zero or more digits, must be of the form *(n+1)90900*, because working from right to left, the 0 must arise as 0+0, then the three 9s must arise as 9+0, 0+9 and 9+0, then the n must arise as (n+1)+9 = 1n.

* Predecessors of *m90900* with m non-zero do not exist because again working from right to left, each of the two 0s must arise as 0+0, then the 9 must arise as 9+0, then the 0 must arise as 1+9 = 10, but that would imply that the next digit to the left in *m90900* was 1, not the 9 that it actually is.

So your statement does turn out to be true, provided you amend it to exclude both the starting number and its first successor. It does however mean that the proof of the amended statement needs to include that backtracking through possible predecessors, making it somewhat more complicated.

> ... The second ambiguity is that when I stated "three adjacent
> 9's" I meant three adjacent 9's, as opposed to, say, four
> adjacent 9's. I would fix that by bracketing "exactly" before
> "three". ...

I did notice that ambiguity, but didn't regard it as worth saying anything about it, because the problem I described occurs with a particular sequence of exactly three 9s, and so is a flaw in your proof regardless of whether you meant "exactly three adjacent 9s" or "at least three adjacent 9s".

> ... I specifically started my proof with "One way to prove ..."
> because running into (exactly) four adjacent 9's is *another*
> way to prove, one that I did not feel necessary to explore for
> the trajectory of 1496.

Indeed, and showing that running into at least four adjacent 9s produces strictly growing numbers of adjacent 9s is easy, because *9999* --> *181818* --> *99999* (in which I'm still using * to denote any sequence of zero or more digits, whether or not they are further 9s). Exactly three adjacent 9s is the tricky case, because it only necessarily breaks even: *999* --> *1818* --> *999*. As a consequence, further analysis of the adjacent digits is needed to establish whether it actually grows.

Other repeated digits produce other ways to prove growth. At least six adjacent 3s or five adjacent 6s are enough to prove growth in three steps easily:

*333333* -->
*66666* --> *12121212* -->
*3333333* -->
*666666*

And exactly five adjacent 3s or four adjacent 6s produce a breaking-even loop rather than a growing loop:

*33333* -->
*6666* --> *121212* -->
*33333* -->
*6666*

Further analysis of the possibilities for the adjacent digits might produce criteria that cause it to start growing rather than just breaking even, as it does for *999*, but I haven't done that analysis.

And similarly, at least seven adjacent 1s, six adjacent 2s, five adjacent 4s, five adjacent 5s, four adjacent 7s or four adjacent 8s establish growth in nine steps easily:

*7777* --> *141414* -->
*55555* --> *10101010* -->
*1111111* -->
*222222* -->
*44444* -->
*8888* --> *161616* -->
*77777* --> *14141414* -->
*5555555* --> *101010101010* -->
*11111111111* -->
*2222222222* -->
*444444444* -->
*88888888*

But for these, reducing the number of adjacent copies of the digit produces a case where our knowledge of the successors shrinks rather than breaking even:

*111111* -->
*22222* -->
*4444* -->
*888* --> *1616* -->
*777* --> *1414* -->
*555* --> *1010* -->
*111* -->
*22* -->
*4* -->
**

covers the cases of six adjacent 1s, five adjacent 2s, four adjacent 4s, three adjacent 7s and three adjacent 8s, and:

*5555* --> *101010* -->
*11111* -->
*2222* -->
*444* -->
*88* --> *16* -->
*7* -->
**

finishes off by dealing with the case of four adjacent 5s.

I should add that while that completely analyses repeated single digits as possible ways of proving infinite growth, it still leaves open the question of using repeated longer blocks of digits - I can see roughly how I might go about investigating that, but haven't done so. Or indeed some sort of infinite growth not involving repeated blocks of digits, which is something I haven't even got as far on as seeing how to investigate it!
 
> DS: "If such a substring in term n is followed by another 9, then it
> is actually a substring of k 9's for k >= 4, which will grow to a
> substring of (2k-3) 9's in term n+2". Not (2k-2) 9's?

Not necessarily, e.g. *d99990* --> *1(d-1)1818189 --> *dd9999919* for d neither 0 nor 9 has a substring of exactly k=4 9s growing to a substring of exactly (2k-3)=5 9s in 2 steps. But a substring of k 9s for k >= 4 does grow to one of (2k-2) 9s in most cases, so I will clarify that I intended my statement to be interpreted in the "at least" way, not the "exactly" way. Basically, my view is that saying that a term contains a substring of three 9s means that it contains a substring "999" and doesn't say anything more about the term - i.e. it doesn't say anything about what it contains outside that substring. Unless of course extra words are added to indicate further meaning - for instance, saying that it contains a substring of exactly three 9s means that it contains a substring "999", the portion of it before that substring does not end with a 9 and the portion of it after that substring does not start with a 9, or saying that it contains an internal substring of exactly three 9s means additionally that neither of those portions of it before and after the "999" is empty.

David