[seqfan] Re: Fibonacci concatenated (a list and some patterns)
Eric Angelini
Eric.Angelini at kntv.be
Thu Jan 28 11:42:36 CET 2016
Thank you, Jack and Robert, this
will spare us a lot of computer time!
Best,
É.
Catapulté de mon aPhone
> Le 27 janv. 2016 à 21:49, Jack Brennen <jfb at brennen.net> a écrit :
>
> Sorry, but there is no m for which the Fibonacci sequence
> starting with (8,m) contains the concatenation 8..m.
>
> After you eliminate the small values of m, you are left with
> a couple of rounded values for (8..m)/m which are allowable:
> 13, 21, 34, and 55, meaning that the concatenation 8..m
> must be either the 6th, 7th, 8th, or 9th term following
> m.
>
> Trying all possible values for the last four digits of m,
> only the following are plausible values of m:
>
> m ends with 0828 -> the 6th subsequent term ends with 0828.
> m ends with 0904 -> the 8th subsequent term ends with 0904.
> m ends with 3328 -> the 6th subsequent term ends with 3328.
> m ends with 5828 -> the 6th subsequent term ends with 5828.
> m ends with 8328 -> the 6th subsequent term ends with 8328.
>
> (All other values of m fail to match the last four digits
> at any of the 6th, 7th, 8th, or 9th terms.)
>
> Assume that the part of m excluding the last four digits
> is A. A = floor(m/10000).
>
> Then we have five possibilities to exclude:
>
> 8,10000*A+828, ..., 130000*A+10828
> 8,10000*A+904, ..., 340000*A+30904
> 8,10000*A+3328, ..., 130000*A+43328
> 8,10000*A+5828, ..., 130000*A+75828
> 8,10000*A+8328, ..., 130000*A+108328
>
> The differences between m and the candidate concatenation
> are respectively:
>
> 120000*A+10000
> 330000*A+30000
> 120000*A+40000
> 120000*A+70000
> 120000*A+100000
>
> One of those would need to be equal to 8 times a power
> of 10. That's impossible, since 8 times a power of 10
> is always 2 (mod 3), and none of those values are ever
> 2 (mod 3).
>
> - Jack
>
>
>
>> On 1/27/2016 7:29 AM, Eric Angelini wrote:
>> Hello SeqFans,
>> Jean-Marc Falcoz has computed around 50 terms of a nice seq.
>> Here is the idea.
>> Let "n" be the first term of a Fibonacci-like seq
>> and "m" the smallest integer such that the concatenation "nm"
>> is part of the said Fibo-like seq. The first 10 terms are:
>>
>> n m
>>
>> 1 4
>> 2 8
>> 3 23
>> 4 7
>> 5 71425
>> 6 1
>> 7 5
>> 8 0
>> 9 11
>> 10 0
>>
>> Explanation #1:
>> 14 is part of the Fibo-like 1,4,5,9,14
>> 28 is part of the Fibo-like 2,8,10,18,28
>> 323 is part of the Fibo-like 3,23,26,49,75,124,199,323
>> 47 is part of the Fibo-like 4,7,11,18,29,47
>> 571425 is part of ...
>> The last couple [5,71425] means that no other m < 71425
>> produces a concatenation "nm" with n that is part of its
>> own Fibo-like seq.
>>
>> Explanation #2:
>> The "0" that follows 8 doesn't mean that the concatenation
>> "80" is part of its own Fibo-like seq -- but means that
>> no "m" has been found < 10^7 [both Jean-Marc and me are
>> quite sure that _all_ possible couple "nm" will be part
>> at some point of its own Fibo-like seq. -- look at the
>> entry 56 of the hereunder list, for instance, where almost
>> one and a half million integers have been discarded before
>> the first "hit"].
>>
>> More comments after the list:
>>
>> n m
>>
>> 1 4
>> 2 8
>> 3 23
>> 4 7
>> 5 71425
>> 6 1
>> 7 5
>> 8 0
>> 9 11
>> 10 0
>> 11 0
>> 12 2
>> 13 0
>> 14 0
>> 15 445
>> 16 0
>> 17 0
>> 18 3
>> 19 0
>> 20 87
>> 21 623
>> 22 0
>> 23 0
>> 24 4
>> 25 0
>> 26 1802
>> 27 33
>> 28 0
>> 29 107
>> 30 5
>> 31 0
>> 32 0
>> 33 79
>> 34 0
>> 35 0
>> 36 6
>> 37 0
>> 38 0
>> 39 2703
>> 40 1063805
>> 41 0
>> 42 7
>> 43 0
>> 44 0
>> 45 805
>> 46 0
>> 47 0
>> 48 8
>> 49 0
>> 50 0
>> 51 0
>> 52 3604
>> 53 0
>> 54 9
>> 55 0
>> 56 1489327
>> 57 0
>> 58 214
>> 59 0
>> 60 0
>> 61 0
>> 62 0
>> 63 77
>> 64 1702088
>> ...
>>
>> Jean-Marc wanted to test the gaps between two "hits"
>> produced by the same "n" but a different "m". He found
>> quite a number of astonishing patterns. Here is what
>> he gets for n=1 to n=30:
>>
>> n <--> various m that produce a hit
>>
>> 1 <--> 4, 9, 49, 99, 499, 999, 4999, 9999, 14285, 49999,
>> 99999, 499999, 999999, 4999999, etc.
>>
>> 2 <--> 8, 98, 998, 9998, 28570, 99998, 999998, 9999998, etc.
>>
>> 3 <--> 23, 248, 2498, 42855, 249998, 2499998, etc.
>>
>> 4 <--> 7, 97, 997, 9997, 57140, 99997, 999997, 9999997, etc.
>>
>> 5 <--> 71425, ?
>>
>> 6 <--> 1, 46, 178, 496, 4996, 18178, 49996, 85710, 499996,
>> 1818178, 4999996, etc.
>>
>> 7 <--> 5, 95, 995, 9995, 99995, 999995, 9999995, etc.
>>
>> 8 <--> ?
>>
>> 9 <--> 11, 69, 161, 267, 744, 1661, 7494, 16661, 27267, 74994,
>> 166661, 749994, 1666661, 2727267, 7499994, etc.
>>
>> 10 <--> ?
>> 11 <--> ?
>>
>> 12 <--> 2, 92, 356, 992, 9992, 36356, 99992, 999992, 3636356,
>> 9999992, etc.
>>
>> 13 <--> ?
>> 14 <--> ?
>>
>> 15 <--> 445, 45445, 4545445, etc.
>>
>> 16 <--> ?
>> 17 <--> ?
>>
>> 18 <--> 3, 22, 322, 534, 3322, 33322, 54534, 333322, 3333322,
>> 5454534, etc.
>>
>> 19 <--> ?
>>
>> 20 <--> 87, 987, 9987, 99987, 999987, 9999987, etc.
>>
>> 21 <--> 623, 63623, 6363623, etc.
>>
>> 22 <--> ?
>> 23 <--> ?
>>
>> 24 <--> 4, 712, 72712, 7272712, etc.
>>
>> 23 <--> ?
>>
>> 26 <--> 1802, 181802, etc.
>>
>> 27 <--> 33, 483, 801, 4983, 49983, 81801, 499983, 4999983,
>> 8181801
>>
>> 28 <--> ?
>>
>> 29 <--> 107, 1232, 12482, 124982, 1249982, etc.
>>
>> 30 <--> 5, 890, 90890, 9090890, etc.
>>
>> ...
>>
>> Jean-Marc tells at 16:28 (Belgian and Swiss time) that the
>> search for a "m" producing a hit with "8" has reached 10^9
>> without success...
>>
>> Best,
>> É.
>>
>>
>>
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>>
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>
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