Thanks a lot, Kerry -- before going to bed, yesterday, I found also that the sequence had to start with 2, which is solution of: 1/a+1/b = (a+b)/10^0 I will submit this shortly to the OEIS. Best, É. __________________________________________________ http://angelink.be/?CeRaccourciFaitCinquanteSignes ----- Original Message ----- From: "Kerry Mitchell" <> To: "math-fun" <math-fun@mailman.xmission.com>; "math-fun" <math-fun@mailman.xmission.com>; <seqfan@ext.jussieu.fr> Cc: "Eric Angelini" <keynews.tv@skynet.be> Sent: Wednesday, March 09, 2005 7:32 PM Subject: Re: [math-fun] Sliding numbers
I ran an Excel spreadsheet and came up with these first 116 sliding numbers:
2, 7, 11, 20, 25, 29, 52, 65, 70, 101, 110, 133, 200, 205, 205, 250, 254, 290, 425, 502, 520, 641, 650, 700, 785, 925, 1001, 1010, 1100, 1258, 1330, 2000, 2005, 2050, 2050, 2225, 2500, 2504, 2540, 2900, 3157, 3445, 4025, 4250, 5002, 5020, 5200, 6266, 6325, 6410, 6500, 7000, 7850, 8125, 9250, 10001, 10010, 10100, 11000, 12508, 12580, 13300, 15689, 16265, 16625, 20000, 20005, 20050, 20500, 20500, 22025, 22250, 25000, 25004, 25040, 25400, 29000, 31282, 31570, 34450, 35125, 40025, 40250, 42500, 50002, 50020, 50200, 52000, 62516, 62660, 63250, 64100, 65000, 70000, 78253, 78500, 79405, 79625, 80125, 81250, 90925, 92500, 100001, 100010, 100100, 101000, 110000, 125008, 125080, 125800, 133000, 156314, 156890, 160625, 162650, 166250
They agree with what you posted and I included 2 because 1/1 + 1/1 = 2, and that shows that 20 is a multiple. I'm reasonably sure that no integer has been missed. I stopped at 116 because the next one should be 200,000 from 10^10 = 10^5 x 10^5 (I think), but I only checked exponents out to 9.
The powers of 10 corresponding to these numbers are:
0, 1, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 4, 3, 4, 4, 3, 4, 4, 3, 4, 4, 5, 5, 5, 5, 3, 4, 5, 4, 5, 6, 4, 5, 6, 6, 6, 4, 5, 6, 5, 6, 5, 6, 4, 5, 6, 5, 7, 6, 7, 7, 7, 6, 7, 4, 5, 6, 7, 5, 6, 7, 6, 7, 7, 8, 5, 6, 7, 8, 8, 8, 8, 5, 6, 7, 8, 6, 7, 8, 8, 6, 7, 8, 5, 6, 7, 8, 6, 7, 9, 8, 9, 9, 7, 9, 8, 9, 7, 8, 9, 9, 5, 6, 7, 8, 9, 6, 7, 8, 9, 7, 8, 8, 9, 9
and I think that the next one should be 10, but it was 6 in my truncated list.
As Eric pointed out, each sliding number leads to infinitely more, making this a fractal sequence. Here are the first several "primitive" elements, that is, those that aren't multiples of 10:
2, 7, 11, 25, 29, 52, 65, 101, 133, 205, 205, 254, 425, 502, 641, 785, 925, 1001, 1258, 2005, 2225, 2504, 3157, 3445, 4025, 5002, 6266, 6325, 8125, 10001, 12508, 15689, 16265, 16625, 20005, 22025, 25004, 31282, 35125, 40025, 50002, 62516, 78253, 79405, 79625, 80125, 90925, 100001, 125008, 156314, 160625
Perhaps it would be interesting to look at this problem with bases with more factors, like 6 or 12.
Kerry Mitchell
-------------- Original message from "Eric Angelini" <keynews.tv@skynet.be>: --------------
Hello SeqFan and Math-fun,
Let's start with an example of "sliding number":
1/4 + 1/25 = 0.29 --> 29 is a "sliding number" 1/8 + 1/125 = 0.133 --> 133 is a "sliding number" 1/2 + 1/5 = 0.7 --> 7 is a "sliding number" etc.
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