[seqfan] Re: Two make a palindrome

[Eric Angelini]

> picking 33 rather than 40 to follow 110.

... yes, my apologizes to all -- and
many thanks to you, Rob! It is fun 
and interesting to analyse your
results too!
Best,
É.
Propulsé d'un aPhone



Le 9 nov. 2013 à 17:10, "Rob Arthan" <rda at lemma-one.com> a écrit :

> Eric,
> 
> That's a fun sequence and an interesting conjecture. As you say, it is not easy to calculate by hand. To get a feel
> for the conjecture I wrote an ML program to do it. This is what I got for the first 200 values:
> 
>    [0, 11, 1, 10, 100, 12, 2, 20, 101, 22, 3, 13, 31, 103, 30, 110, 33, 4, 14, 41, 104, 40, 114, 24, 42, 112, 21, 102,
>     120, 201, 210, 1000, 105, 15, 5, 25, 52, 115, 35, 53, 113, 23, 32, 121, 26, 6, 16, 61, 106, 60, 116, 36, 63, 131,
>     34, 43, 134, 143, 314, 341, 413, 431, 1003, 111, 17, 7, 27, 72, 117, 37, 73, 137, 71, 107, 70, 170, 701, 710, 1007,
>     122, 18, 8, 28, 82, 118, 38, 83, 138, 81, 108, 80, 180, 801, 810, 1008, 128, 182, 218, 281, 812, 821, 1002, 123,
>     132, 213, 231, 312, 321, 1020, 124, 142, 214, 241, 412, 421, 1004, 133, 19, 9, 29, 92, 119, 39, 93, 139, 91, 109,
>     90, 190, 901, 910, 1009, 129, 192, 219, 291, 912, 921, 1029, 209, 290, 902, 920, 1012, 200, 44, 55, 66, 77, 88, 99,
>     141, 45, 54, 145, 51, 125, 152, 215, 251, 512, 521, 1005, 135, 153, 315, 351, 513, 531, 1030, 130, 301, 310, 1011,
>     140, 401, 410, 1014, 202, 50, 150, 501, 510, 1015, 151, 56, 65, 156, 165, 516, 561, 615, 651, 1006, 126, 62, 161,
>     46, 64, 146]
> 
> I think my program is right in picking 33 rather than 40 to follow 110.
> 
> Here are the first 200 values of the sequence giving the smallest missing integer at each stage of the calculation.
> 
>  [1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6,
>    6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8,
>    8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9,
>    9, 29, 39, 39, 39, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 44, 45,
>    45, 45, 45, 45, 45, 45, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46,
>    46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 47, 47, 47]
> 
> So if m(n) is the sequence of missing integers, m(3) = 2 say and m(198) = 47. 
> 
> My program is now in a loop printing out n, a(n) and m(n). The evidence currently supports your conjecture but m(n) is
> growing quite slowly:
> 
>   a(5846) = 589, m(5846) = 598
>   a(5847) = 598, m(5846) = 679
>   ...            
>   a(11539) = 1617, m(11539) = 679
>   a(11540) =  679 m(11540) = 697
> 
> So 697 persisted as the smallest missing integer for more than 5,000 stages. I will leave it running and report back if anything noteworthy occurs.
> 
> Regards,
> 
> Rob.
> 
> On 9 Nov 2013, at 10:07, Eric Angelini wrote:
> 
>> 
>> Hello SeqFans,
>> this seq P is so nice that I was almost
>> in tears all night ;-)
>> [and I'm convinced that P is a 
>> permutation of the non-negative
>> integers 0,1,2,3,4,5... +oo though 
>> it is far from being obvious (I might 
>> be overoptimistic)]:
>> 
>> P=0,11,1,10,100,12,2,20,101,22,3,13,31,103,30,110,40,...
>> 
>> The trick:
>> 
>> Take any pair of neighbouring terms; 
>> you can build a palindrome including all the digits of the pair.
>> 
>> Start with a(1)=0 then extend P
>> with the smallest unused integer
>> that fits the requirement.
>> 
>> 0 and 11 could produce 101, which
>> is a palindrome;
>> 11 and 1 produce 111
>> 1 and 10 produce 101
>> 10 and 100 produce 10001
>> 100 and 12 produce 10201
>> 12 and 2 produce 212
>> 2 and 20 produce 202
>> 20 and 101 produce 10201
>> 101 and 22 produce 12021
>> etc.
>> Extending P by hand is very
>> confusing -- but incredibly fun --
>> please excuse any potential mistake.
>> 
>> Best,
>> É.
>> 
>> 
>> 
>> 
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>> 
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> 
> 
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