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User talk:Syed Iddi Hasan

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Hi,

I made the draft for the comment in OEIS A099009 as follows:

It appears that the comment made by Jens Kruse Andersen is missing one more family of terms (which starts with one or more digits "9" and ends with the digit "1"): 97508421, 9753086421, 9975084201, 975330866421, 997530864201, 999750842001, ...


What do you think ?

Regards, Alex (Alexander R. Povolotsky)

Hi!

What you have mentioned here is the sequence A214559. Please refer to my comment where I have already mentioned that A099009 is formed by the union of A214555 through A214559 and zero. So I guess I've already got it covered.

Regards, Iddi (Syed Iddi Hasan)

OK Iddi - thanks. I referred to A214559 and to your name in my A099009 comment.


Did you look for any dependencies in those parameters ?

I narrowed it down to four parameters. I ordered the digits from largest to smallest and smallest to largest, and by comparing them I was able to find the interdependent pairs of numbers. However, these four parameters seem to be independent of each other.

9(x1+1)//8(x2)//7(x3+1)//6(x2)//5(x3+1)//4(x2)//3(x4)//2(x2)//1(x3)//0//9(x2)//8(x3+1)//7(x2)//6(x4)//5(x2)//4(x3+1)//3(x2)//2(x3+1)//1(x2)//0(x1)//1 For what it is worth, adding digits, which are sharing the same "xi" parameter (where i=1,2,3,4), yields sums dividable by 9 (that is with the digital root being equal 9): i=1 9+0=9 i=2 8+6+4+2+9+7+5+3+1=45 i=3 7+5+1+8+4+2=27 i=4 3+6=9

Yes, that criteria will always hold. It originates from the fact that Kaprekar transformation always yields numbers divisible by 9 (or zero). As these are fixed points, the digits remain the same before and after the transformation, as they just get rearranged. And knowing that the difference between any two numbers with digital root 9 is also a number with digital root 9, we can state that the sum of the digits that are included in the subsequent number will always be a multiple of 9.

See http://mathoverflow.net/questions/200533/q-re-kaprekars-fixed-mapping-points

The comment by Jens Kruse Andersen are the sequences A214555, A214556, and a specific case of A214557 where x1=0. Refer to A214557 and A214558 which complete the whole set. Like I already said, A214555 through A214559 (and zero) makes up the whole sequence. I have no formal proof that there is any number which doesn't belong to any of these five sequences. However, I have verified (using programming) that there are certainly none until 10^100.

Thanks - I made clarification/correction in both places (my comment in A099009) and in my question http://mathoverflow.net/questions/200533/q-re-kaprekars-fixed-mapping-points

I believe you quoted A214558 instead of A214557 in your question in Math Overflow.

It is confusing that A214558 and A214557 overlap (contain the same terms, for example 864197532) - in my view the families of terms should be unique and overlapping should not happen unless one of them is subset of another - but which is set and which is subset - it is not clear to me ... Please advise ... and it would be helpful if you also would make clarifying comments re that in your both OEIS entries ... Also perhaps it would be helpful if you would mention in your sequences: A214555, A214556, A214557, A214558 that your work stems from the original comment of Andersen ?

The overlap occurs only for a few terms, but each has unique terms of its own. There might be a single expression to include both A214557 and A214558, but I can surely say that none of the two is a subset of the other. I had come up with the sequences independently while studying Kaprekar transformation, hence I had not mentioned Andersen in my work.

Historically Andersen's comment precedes your comments in A099009 and your A214555, A214556, A214557, A214558 OEIS entries for so long that it is really irrelevant whether you indeed came up with what you have or not ... - seriously ... :-) And, yes, you, in my opinion, should come up with the single expression to include both A214557 and A214558 and in the meanwhile to make some clarifying comments about the overlap. Cheers, Alex.

Point taken. I will try to combine the two of them. Hopefully it won't take too much time. :)

Thanks, much appreciated