[seqfan] Re: Interesting set of palindromes

Mon Oct 24 19:54:56 CEST 2011

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/~israel University
> 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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