[seqfan] Re: Interesting set of palindromes
Dale Gerdemann
dale.gerdemann at googlemail.com
Sun Oct 30 10:53:22 CET 2011
Hello Seqfans
I wrote a while back:
> Here's an interesting set of palindromes that I uncovered. I wonder
> if there are other such sets
What I was referring to was a slight variation on Golden Ratio Base
(Bergman, 1957) in which each positive digit is balanced by a negative
digit.
With help from George Bergman, Martin Bunder and Robert Israel, I am
now able to answer my own question. There is apparently a sequence of
sequences, where each sequence is paired with a base, which satisfy my
requirements. Here are the first few series, labeled with the number
of required different digits, and as an example, the number 1000000
written in this base:
3. A001906 1/2 (3-Sqrt[5]) 11001010120010.1.01002101010011
4. A001353 1/2 (3+Sqrt[5]) 1311213030.0.0303121131
5. A004254 2-Sqrt[3] 32404322.3.22340423
6. A001109 2+Sqrt[3] 4210435.2.5340124
7. A004187 1/2 (5-Sqrt[21]) 1252601.1.1062521
8. A001090 1/2 (5+Sqrt[21]) 414355.2.553414
9. A018913 3-2 Sqrt[2] 202264.4.462202
10. A004189 3+2 Sqrt[2] 106153.0.351601
11. A004190 1/2 (7-3 Sqrt[5]) 651b1.0.1b156
12. A004191 1/2 (7+3 Sqrt[5]) 41c16.4.61c14
13. A078362 4-Sqrt[15] 2b023.c.320b2
14. A007655 4+Sqrt[15] 1d8e5.b.5e8d1
15. A078364 1/2 (9-Sqrt[77]) 15298.3.89251
16. A077412 1/2 (9+Sqrt[77]) g7g9.0.9g7g
17. A078366 5-2 Sqrt[6] d268.f.862d
18. A049660 5+2 Sqrt[6] 9cb4.d.4bc9
19. A078368 1/2 (11-3 Sqrt[13]) 7f75.c.57f7
Note here the occurrence of A049457.
The bit pattern is clearly more complex than Zeckendorf or Golden
Ratio Base. Nevertheless the pattern can be recognized by a
4-state automaton that starts in the middle and moves outward in the
direction of your choice.
Sfs0: highest two digits -> fs3, other digits-> fs1.
fs1: highest digit -> fs3, lowest digit -> s2, other -> fs1.
s2: highest digit -> fs3, lowest digit -> s2, other -> fs1.
fs3: second highest -> fs3, lowest digit -> s2, middle digits -> fs1.
All states except s2 are final. Sfs0 is initial.
Now a suggestion for the OEIS: Wouldn't it be nice if you could type in
A001906, A001353, A004254, ... and get this sequence of sequences?
Dale Gerdemann
On Mon, Oct 24, 2011 at 7:54 PM, Dale Gerdemann <
dale.gerdemann at googlemail.com> wrote:
> Thanks, Robert, for the comment.
>
> I had some email discussion with George Bergman a year or so ago about why
> the Golden Ratio representation is so symmetric. He explained it to me in
> terms of conjugates. But what he didn't explain was why the symmetry was
> imperfect. He has now written me a new email concerning the symmetric
> examples, and he has agreed that I can repost his email here:
>
> Dear Dale,
>> Yes, your b is the square of the Golden Section tau = (1 + sqrt 5)/2.
>> Looking back over my earlier e-mail to you, I see that the what made
>> the symmetry there imperfect was the fact that the conjugate of tau
>> was -tau^{-1}, so that when one conjugated an expression for
>> an integer, one got an expression with some terms -1 that had to
>> be dealt with. But the conjugate of tau^2 is the square of the
>> conjugate of tau, so the minus sign goes away and it is the inverse
>> of tau^2, and one gets perfect symmetry. Nice!
>> Yours, George Bergman
>
>
> In a second email, he writes:
>
> By the way, it is curious that one can get a representation for
>> every integer to base b using only digits 0, 1, 2, by repeatedly
>> adding 1, using the rule "030 -> 101" whenever a digit gets
>> above 1, since b satisfies b^2 - 3 b + 1 = 0, whence
>> b^{n+1} - 3 b^n + b^{n-1} = 0, whence 3b^n = b^{n+1} + b^{n-1}.
>> But there are nonunicities in these representations that one
>> doesn't discover directly in this way. In particular, it gives
>> 7 = 21.2, but one eventually discovers the representation you
>> used by continuing the above process up to 9 = 102.01, then noticing
>> that by subtracting 2, one has 7 = 100.01. (I agree that it
>> would be nicer to position the decimal point under the b^0
>> digit; but that would be hard to do in e-mail.) One can fix
>> this by throwing in a second rule, 02120 -> 10001. But one
>> would still need further rules; e.g., 0220 -> 1001. Probably
>> infinitely many.
>
>
>
> Dale Gerdemann
>
>
> On Mon, Oct 24, 2011 at 3:00 AM, <israel at math.ubc.ca> wrote:
>
>> Note that b^n + 1/b^n = L(2n) (where L(n) = A000032(n) are the Lucas
>> numbers). According to Zeckendorf, every integer can be represented
>> uniquely as a sum of non-adjacent Lucas numbers where L(0) and L(2) can't
>> both be used. Since L(2n+1) = L(2n+2) - L(2n), this representation can be
>> transformed to a representation (not necessarily the only one) by sums of
>> b^n + 1/b^n with coefficients in {-1,0,1}. Thus
>>
>> 1 = L(1) = L(2) - L(0) = b^1 - 2 b^0 + b^(-1)
>> 2 = L(0) = 2 b^0
>> 3 = L(2) = b^1 + b^(-1)
>> 4 = L(3) = L(4) - L(2) = b^2 - b^1 - b^(-1) + b^(-2)
>> 5 = L(1) + L(3) = L(4) - L(0) = b^2 - 2 b^0 + b^(-2)
>> 6 = L(0) + L(3) = L(4) - L(2) + L(0) = b^2 - b^1 + 2 b^0 - b^(-1) + b^(-2)
>> 7 = L(4) = b^2 + b^(-2)
>> 8 = L(1) + L(4) = L(4) + L(2) - L(0) = b^2 + b^1 - 2 b^0 + b^(-1) + b^(-2)
>>
>> etc
>>
>> Robert Israel israel at math.ubc.ca
>> Department of Mathematics http://www.math.ubc.ca/~israelUniversity of British Columbia Vancouver, BC, Canada
>>
>>
>> On Oct 19 2011, Dale Gerdemann wrote:
>>
>> Hello Seqfans,
>>>
>>> Here's an interesting set of palindromes that I uncovered. I wonder if
>>> there are other such sets.
>>>
>>>
>>> Let b=(3+sqrt(5))/2. And let a(n) be A001906(n), offset so that a(-1) =
>>> 0, a(0) = 1, a(1) = 3, etc
>>>
>>>
>>> 1 = 1 = a(0)
>>>
>>> 2 = 2 = 2a(0)
>>>
>>> 3 = b^1 + b^-1 = a(1) + a(-1)
>>>
>>> 4 = b^1 + 1 + b^-1 = a(1) + a(0) + a(-1)
>>>
>>> 5 = b^1 + 2 + b^-1 = etc (a-forms are palindromes just like b forms)
>>>
>>> 6 = 2b^1 + 2b^-1
>>>
>>> 7 = b^2 + b^-2
>>>
>>> 8 = b^2 + 1 + b^-2
>>>
>>> 9 = b^2 + 2 + b^-2
>>>
>>> 10 = b^2 + b^1 + b^-1 + b^-2
>>>
>>> 11 = b^2 + b^1 + 1 + b^-1 + b^-2
>>>
>>> 12 = b^2 + b^1 + 2 + b^-1 + b^-2
>>>
>>> 13 = b^2 + 2b^1 + 2b^-1 + b^-2
>>>
>>> 14 = 2b^2 + 2b^-2
>>>
>>> 15 = 2b^2 + 1 + 2b^-2
>>>
>>> 16 = 2b^2 + 2 + 2b^-2
>>>
>>> 17 = 2b^2 + b^1 + b^-1 + 2b^-2
>>>
>>> 18 = b^3 + b^-3
>>>
>>> 19 = b^3 + 1 + b^-3
>>>
>>> 20 = b^3 + 2 + b^-3
>>>
>>> 21 = b^3 + b^1 + b^-1 + b^-3
>>>
>>> 22 = b^3 + b^1 + 1 + b^-1 + b^-3
>>>
>>> 23 = b^3 + b^1 + 2 + b^-1 + b^-3
>>>
>>> 24 = b^3 + 2b^1 + 2b^-1 + b^-3
>>>
>>> 25 = b^3 + b^2 + b^-2 + b^-3
>>>
>>> 26 = b^3 + b^2 + 1 + b^-2 + b^-3
>>>
>>> 27 = b^3 + b^2 + 2 + b^-2 + b^-3
>>>
>>> 28 = b^3 + b^2 + b^1 + b^-1 + b^-2 + b^-3
>>>
>>> 29 = b^3 + b^2 + b^1 + 1 + b^-1 + b^-2 + b^-3
>>>
>>> 30 = b^3 + b^2 + b^1 + 2 + b^-1 + b^-2 + b^-3
>>>
>>> 31 = b^3 + b^2 + 2b^1 + 2b^-1 + b^-2 + b^-3
>>>
>>> 32 = b^3 + 2b^2 + 2b^-2 + b^-3
>>>
>>> 33 = b^3 + 2b^2 + 1 + 2b^-2 + b^-3
>>>
>>> 34 = b^3 + 2b^2 + 2 + 2b^-2 + b^-3
>>>
>>> 35 = b^3 + 2b^2 + b^1 + b^-1 + 2b^-2 + b^-3
>>>
>>> 36 = 2b^3 + 2b^-3
>>>
>>> 37 = 2b^3 + 1 + 2b^-3
>>>
>>> 38 = 2b^3 + 2 + 2b^-3
>>>
>>> 39 = 2b^3 + b^1 + b^-1 + 2b^-3
>>>
>>> 40 = 2b^3 + b^1 + 1 + b^-1 + 2b^-3
>>>
>>> 41 = 2b^3 + b^1 + 2 + b^-1 + 2b^-3
>>>
>>> 42 = 2b^3 + 2b^1 + 2b^-1 + 2b^-3
>>>
>>> 43 = 2b^3 + b^2 + b^-2 + 2b^-3
>>>
>>> 44 = 2b^3 + b^2 + 1 + b^-2 + 2b^-3
>>>
>>> 45 = 2b^3 + b^2 + 2 + b^-2 + 2b^-3
>>>
>>> 46 = 2b^3 + b^2 + b^1 + b^-1 + b^-2 + 2b^-3
>>>
>>> 47 = b^4 + b^-4
>>>
>>> 48 = b^4 + 1 + b^-4
>>>
>>> 49 = b^4 + 2 + b^-4
>>>
>>> 50 = b^4 + b^1 + b^-1 + b^-4
>>>
>>> 51 = b^4 + b^1 + 1 + b^-1 + b^-4
>>>
>>> 52 = b^4 + b^1 + 2 + b^-1 + b^-4
>>>
>>> 53 = b^4 + 2b^1 + 2b^-1 + b^-4
>>>
>>> 54 = b^4 + b^2 + b^-2 + b^-4
>>>
>>> 55 = b^4 + b^2 + 1 + b^-2 + b^-4
>>>
>>> 56 = b^4 + b^2 + 2 + b^-2 + b^-4
>>>
>>> 57 = b^4 + b^2 + b^1 + b^-1 + b^-2 + b^-4
>>>
>>> 58 = b^4 + b^2 + b^1 + 1 + b^-1 + b^-2 + b^-4
>>>
>>> 59 = b^4 + b^2 + b^1 + 2 + b^-1 + b^-2 + b^-4
>>>
>>> 60 = b^4 + b^2 + 2b^1 + 2b^-1 + b^-2 + b^-4
>>>
>>> 61 = b^4 + 2b^2 + 2b^-2 + b^-4
>>>
>>> 62 = b^4 + 2b^2 + 1 + 2b^-2 + b^-4
>>>
>>> 63 = b^4 + 2b^2 + 2 + 2b^-2 + b^-4
>>>
>>> 64 = b^4 + 2b^2 + b^1 + b^-1 + 2b^-2 + b^-4
>>>
>>> 65 = b^4 + b^3 + b^-3 + b^-4
>>>
>>> 66 = b^4 + b^3 + 1 + b^-3 + b^-4
>>>
>>> 67 = b^4 + b^3 + 2 + b^-3 + b^-4
>>>
>>> 68 = b^4 + b^3 + b^1 + b^-1 + b^-3 + b^-4
>>>
>>> 69 = b^4 + b^3 + b^1 + 1 + b^-1 + b^-3 + b^-4
>>>
>>> 70 = b^4 + b^3 + b^1 + 2 + b^-1 + b^-3 + b^-4
>>>
>>> 71 = b^4 + b^3 + 2b^1 + 2b^-1 + b^-3 + b^-4
>>>
>>> 72 = b^4 + b^3 + b^2 + b^-2 + b^-3 + b^-4
>>>
>>> 73 = b^4 + b^3 + b^2 + 1 + b^-2 + b^-3 + b^-4
>>>
>>> 74 = b^4 + b^3 + b^2 + 2 + b^-2 + b^-3 + b^-4
>>>
>>> 75 = b^4 + b^3 + b^2 + b^1 + b^-1 + b^-2 + b^-3 + b^-4
>>>
>>> 76 = b^4 + b^3 + b^2 + b^1 + 1 + b^-1 + b^-2 + b^-3 + b^-4
>>>
>>> 77 = b^4 + b^3 + b^2 + b^1 + 2 + b^-1 + b^-2 + b^-3 + b^-4
>>>
>>> 78 = b^4 + b^3 + b^2 + 2b^1 + 2b^-1 + b^-2 + b^-3 + b^-4
>>>
>>> 79 = b^4 + b^3 + 2b^2 + 2b^-2 + b^-3 + b^-4
>>>
>>> 80 = b^4 + b^3 + 2b^2 + 1 + 2b^-2 + b^-3 + b^-4
>>>
>>> 81 = b^4 + b^3 + 2b^2 + 2 + 2b^-2 + b^-3 + b^-4
>>>
>>> 82 = b^4 + b^3 + 2b^2 + b^1 + b^-1 + 2b^-2 + b^-3 + b^-4
>>>
>>> 83 = b^4 + 2b^3 + 2b^-3 + b^-4
>>>
>>> 84 = b^4 + 2b^3 + 1 + 2b^-3 + b^-4
>>>
>>> 85 = b^4 + 2b^3 + 2 + 2b^-3 + b^-4
>>>
>>> 86 = b^4 + 2b^3 + b^1 + b^-1 + 2b^-3 + b^-4
>>>
>>> 87 = b^4 + 2b^3 + b^1 + 1 + b^-1 + 2b^-3 + b^-4
>>>
>>> 88 = b^4 + 2b^3 + b^1 + 2 + b^-1 + 2b^-3 + b^-4
>>>
>>> 89 = b^4 + 2b^3 + 2b^1 + 2b^-1 + 2b^-3 + b^-4
>>>
>>> 90 = b^4 + 2b^3 + b^2 + b^-2 + 2b^-3 + b^-4
>>>
>>> 91 = b^4 + 2b^3 + b^2 + 1 + b^-2 + 2b^-3 + b^-4
>>>
>>> 92 = b^4 + 2b^3 + b^2 + 2 + b^-2 + 2b^-3 + b^-4
>>>
>>> 93 = b^4 + 2b^3 + b^2 + b^1 + b^-1 + b^-2 + 2b^-3 + b^-4
>>>
>>> 94 = 2b^4 + 2b^-4
>>>
>>> 95 = 2b^4 + 1 + 2b^-4
>>>
>>> 96 = 2b^4 + 2 + 2b^-4
>>>
>>> 97 = 2b^4 + b^1 + b^-1 + 2b^-4
>>>
>>> 98 = 2b^4 + b^1 + 1 + b^-1 + 2b^-4
>>>
>>> 99 = 2b^4 + b^1 + 2 + b^-1 + 2b^-4
>>>
>>> 100 = 2b^4 + 2b^1 + 2b^-1 + 2b^-4
>>>
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