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Topic: Sequence with a twist (Read 364 times) |
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Margit
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Sequence with a twist
« on: Feb 3rd, 2006, 12:38am » |
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12, 122, 1224, 12248, 1224832, 122483224, 1224832246, 12248322464
What digit(s) never appear in this sequence ?
Prove it.
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towr
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Re: Sequence with a twist
« Reply #1 on: Feb 3rd, 2006, 1:00am » |
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I've at least identified the sequence
S1= '12'
Sn+1 = Sn.(Sn[n]*Sn[n+1])
where '.' is the concattenation operator, and the 'Sn[k]' extracts the kth digit from Sn.
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| « Last Edit: Feb 3rd, 2006, 1:02am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Grimbal
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Re: Sequence with a twist
« Reply #2 on: Feb 3rd, 2006, 8:58am » |
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Consider the set {1 2 3 4 6 8}
and their products:
1, 2, 3, 4, 6, 8,
2, 4, 6, 8,12,16,
3, 6, 9,12,18,24,
4, 8,12,16,24,32,
6,12,18,24,36,48,
8,16,24,32,48,64.
The multiples are all made from digits in the set, except for the 9 that appears for 3*3. We'll have to see that a 9 is not possible.
- 1 is always followed by 2, 6 or 8. Let's consider the first exception to the rule. A 1 followed by a digit different from 2,6,8 can only be the 1 as a result of 1*1. But for that, there must be a sequence 11 earlier, which would be an earlier exception to the rule.
- 33 can only happen as 3 followed by 3 or 32.
(a) 3 followed by 3 can happen only if a sequence 131 or 313 appears earlier. But there is no sequence 13.
(b) 3 can not be followed by 32 because 3 is 1*3 and 32 is 4*8, they have no common factor, so they can not be adjacent.
From (a) and (b) it follows that there is no 33 in the sequence.
- 9 can only be the result of 3*3, but there is no subsequence 33 in the sequence.
This shows that the digits 0,5,7,9 nowhere appear in the sequence. The others digits can be seen among the 10 first digits.
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| « Last Edit: Feb 3rd, 2006, 2:55pm by Icarus » |
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