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

[David Seal]

Reading more of the sequences related to this problem, I noticed the following in the comments section of A328974:

"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 strictly internal substring of three adjacent 9's. 

 Such a substring in term n will grow to four adjacent 9's in
   term n+2; to six adjacent 9's in term n+4; to ten adjacent
   9's in term n+6; ... to 2^k+2 adjacent 9's (see A052548) in
   term n+2k, regardless of what happens in the rest of the
   number. 

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

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. The smallest such number and its first two successors are 19990 --> 1018189 --> 1199917.

It is fixable by replacing the second and third paragraphs by:

"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 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: *[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."

I am uncertain about the etiquette in such situations - should I just add the fixed proof to the comments, leaving the flawed proof as it is, or should I edit the flawed proof to correct it?

David