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

[Hans Havermann]

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". 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". It is true that four adjacent 9's pop up here and there, sometimes prior three adjacent 9's showing up (in my list of 109 terms previously provided that I believe increase without limit, I count eleven such instances), but the showing up of four (or more) adjacent 9's does not invalidate the proof applied to (exactly) three adjacent 9's. I did not address the issue of (exactly) four adjacent 9's because my proof comment was applied to A328974 (the trajectory specifically of 1496) where the first occurrence of more than two adjacent 9's is three, not four. 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.

Replacing my "strictly internal substring" with "non-final substring" is likely an improvement.

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?