[seqfan] Re: Two make a palindrome
Rob Arthan
rda at lemma-one.com
Sat Nov 9 17:10:22 CET 2013
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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