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Reference: A003230

A003230(n=1) =Number of simple squares of the Harter-Heighway dragon curve of degree (1+4) =4.


Example.jpg



a(n) = A003230

a(n) = expansion of 1/((1-x)·(1-2x)·(1-x-2x^3))


Explanation und supplements:


To obtain a simple representation for a(n) as an explicit formula, a number of constants first need to be defined. thrt stands for a third root and indicates the number:

thrt := (54+6·sqrt(87))^(1/3).

It is needed to specify the three solutions of 1-x-2·x^3 =0. The names of the three solutions are meant to be reminiscent of Root_Of.

The first Root_Of would be the real solution: Root_Of element R, denoted by ROR. One has

ROR:= thrt/6-1/thrt. -> {proof ROR}(see below) 

The second Root_Of would be the first complex solution: Root_Of element C, denoted by ROC. The third Root_Of is the second complex solution: conjugate(ROC).

A simple representation of ROC requires a further number, expediently designated RORext, as understood from the definition of ROR.

RORext:= thrt/6+1/thrt.

The only difference between ROR and RORext therefore is the minus sign. This allows an appropriate representation of ROC: One has

ROC:= (1/2)·[i·sqrt(3)·RORext-ROR],  where i is the imaginary unit. -> {proof ROC}(see below)

To be used finally with ROR and ROC are two constants for the purpose of keeping the representation of a(n) in check.

These would be AR and AC, whose appearance is symmetric in terms of R and C:

AR :=(2·ROR^2+ROR+2)/(2·ROR-3)    and
AC :=(2·ROC^2+ROC+2)/(2·ROC-3).

A003230 can now be written as

a(n):= (1/2)·[AR·ROR^(-(n+4))+2·Re{AC·ROC^(-(n+4))}+2^(n+4)+1].

Re{...} - as is customary - indicates the real component of the complex number in the curly brackets.

In more detail:

a(n) =   (1/2)· [              AR·   ROR^(-(n+4))
                  +            AC·   ROC^(-(n+4))
                  + conjugate( AC·   ROC^(-(n+4)) )
                  +             1· (1/2)^(-(n+4))
                  +             1·     1^(-(n+4))   ],

because Re{conjugate(AC·ROC^(-(n+4))} = Re{AC·ROC^(-(n+4))} and 2·Re(z) = z+conjugate(z) for all z element C.

As required, every solution of (1-x-2·x³)·(1-2x)·(1-x) =0 is accordingly accommodated in a term of the form AX(Root_Of)·Root_Of^(-(n+4)). The sum of these terms is a(n).



MAPLE allows the function a(n) thus defined to be understood and plotted for n element R too. It should furthermore be possible to test the function for accuracy.

To be defined for this purpose is a comparison function, i.e. that of the Taylor coefficients. (denoted subsequently by ac(n)).

For larger n, MAPLE outputs symbolic values even with simplify(a(n)), thus preventing a direct comparison ac(n)= a(n).

For this reason, the central part of the test program resorts to a use of ac(n)= round(evalf(a(n),max_digits)).

Except for the elapsed time at the end, this program provides no suggestion what a(n) = ac(n) might mean for all n element N.

The elapsed time at max_= 5000 can by all means be 4 min.

The equality of the terms from 0..max_ is no mathematical proof but a strong indication that, for all n element N: a(n) = ac(n).


summary:

p(x)   := 1-x-2·x^3
thrt   := (54+6·sqrt(87))^(1/3)
ROR    := thrt/6-1/thrt
RORext := thrt/6+1/thrt
ROC    := (1/2)·[i·sqrt(3)·RORext-ROR]



{proof ROR}: p(ROR) =0


    p(ROR)
=  1 -(thrt/6 -1/thrt) –2·(thrt/6 -1/thrt)^3
=  1 -(thrt/6 -1/thrt) –2·(thrt^3/6^3 –3·thrt/6^2 +3/(6·thrt) +1/thrt^3)
=  1 -thrt/6  +1/thrt  –2·thrt^3/6^3  +6·thrt/6^2 -6/(6·thrt) -2/thrt^3
=  1 -thrt/6  +1/thrt  –2·thrt^3/6^3     +thrt/6   -1/thrt    -2/thrt^3
=  1    	       –2·thrt^3/6^3                          -2/thrt^3
=  1 –2(54+6·sqrt(87))/6^3 -2/(54+6·sqrt(87))
=  1 –( 9 +sqrt(87))/(3·6) -(1/3)/(9+sqrt(87))
=  1 –( 9 +sqrt(87))/(3·6) -(1/3)·(9-sqrt(87))/(81-87)
=  1 –( 9 +sqrt(87))/(3·6) -(9-sqrt(87))/(3·6)
=  1 –( 9          )/(3·6) -(9         )/(3·6)
=  1 –( 1 )/(2) 	   -( 1 )/(2)
=  0
   q.e.d.

{proof ROC}: p(ROC) =0


   p(ROC)
=  1 -(1/2)·[i·sqrt(3)·RORext-ROR] -2·(1/2)^3·[i·sqrt(3)·RORext-ROR]^3
=  1 -(1/2)·[i·sqrt(3)·RORext-ROR] -(1/4)·[i·sqrt(3)·RORext-ROR]^3
=  1 -(1/2)·i·sqrt(3)·RORext +(1/2)·ROR -(1/4)·[i^3·sqrt(3)^3·RORext^3 -3·i^2·sqrt(3)^2·RORext^2·ROR +3·i·sqrt(3)·RORext·ROR^2 -ROR^3]
=  1 -(1/2)·i·sqrt(3)·RORext +(1/2)·ROR -(1/4)·[-i·sqrt(3)·3·RORext^3 +3·(3)·RORext^2·ROR +3·i·sqrt(3)·RORext·ROR^2 -ROR^3]
=  1 -(1/2)·i·sqrt(3)·RORext +(1/2)·ROR -(1/4)·[-i·3·sqrt(3)·RORext^3 +9·RORext^2·ROR +3·i·sqrt(3)·RORext·ROR^2 -ROR^3]
=  1 +(1/2)·ROR -(1/4)·[9·RORext^2·ROR -ROR^3] +i·{-(1/2)·sqrt(3)·RORext -(1/4)·[-3·sqrt(3)·RORext^3 +3·sqrt(3)·RORext·ROR^2]}
=  1 +(1/2)·ROR -(1/4)·ROR·[9·RORext^2 -ROR^2] +i·sqrt(3)·RORext·{-(1/2) -(1/4)·[-3·RORext^2 +3·ROR^2]}
=  1 +(1/2)·ROR -(1/4)·ROR·[9·RORext^2 -ROR^2] +i·sqrt(3)·RORext·{-(1/2) -(3/4)·[-RORext^2 +ROR^2]}
=  1 +(1/2)·(thrt/6 -1/thrt) -(1/4)·(thrt/6 -1/thrt)·[9·(thrt^2/36 +(1/3) +1/thrt^2) -(thrt^2/36 -(1/3) +1/thrt^2)]
  -(1/2)·i·sqrt(3)·RORext·{1 +(3/2)·[-(thrt^2/36 +1/3 +1/thrt^2) +(thrt^2/36 -1/3 +1/thrt^2)]}
=  1 +(1/2)·(thrt/6 -1/thrt) -(1/4)·(thrt/6 -1/thrt)·[8·thrt^2/36 +10/3 +8/thrt^2] -(1/2)·i·sqrt(3)· RORext·{1 +(3/2)·[-2/3]}
=  1 +(1/2)·(thrt/6 -1/thrt) -(1/2)·(thrt/6 -1/thrt)·[ 4·thrt^2/36 +5/3 +4/thrt^2]
=  1 +(1/2)·(thrt/6 -1/thrt) -(1/2)·[thrt^3/54 +5·thrt/18 +2/(3·thrt)   -4·thrt/36 -5/(3·thrt) -4/thrt^3]
=  1 +thrt/12    -1/(2·thrt) -(1/2)·thrt^3/54 -5·thrt/36 -1/(3·thrt)    +2·thrt/36 +5/(6·thrt) +2/thrt^3
=  1 +3·thrt/36  -3/(6·thrt) -(1/2)·thrt^3/54 -5·thrt/36 -2/(6·thrt)    +2·thrt/36 +5/(6·thrt) +2/thrt^3
=  1                         -(1/2)·(54+6·sqrt(87))/54                                         +2/(54+6·sqrt(87))
=  1 -(1/2) -sqrt(87)/18 +1/[3·(9+sqrt(87))]
=  1 -(1/2) -sqrt(87)/18 +((9-sqrt(87)/[3·(81-87)]
=  1 -(1/2) -sqrt(87)/18 -(9-sqrt(87)/18
=  1 -(1/2) -sqrt(87)/18 -(1/2) +sqrt(87)/18
=  0
    q.e.d.

I waive the proove: p(conjugate(ROC)) =0.


recursion:


a(n):= (1/2)·[AR·ROR^(-(n+4))+2·Re{AC·ROC^(-(n+4))}+2^(n+4)+1]. Abbr.: µ(z):= conjugate(z).

a(n  ):= (1/2)·[AR·ROR^(-(n+4))   +AC·ROC^(-(n+4))   +µ(AC·ROC^(-(n+4)))   +2^(n+4)   +1].
a(n+3):= (1/2)·[AR·ROR^(-(n+7))   +AC·ROC^(-(n+7))   +µ(AC·ROC^(-(n+7)))   +2^(n+7)   +1].
a(n+2):= (1/2)·[AR·ROR^(-(n+6))   +AC·ROC^(-(n+6))   +µ(AC·ROC^(-(n+6)))   +2^(n+6)   +1].


Proposition:
a(n+3):= a(n+2) +2·a(n) +2^(n+4)-1   *)
       = (1/2)·[AR·ROR^(-(n+6)) +AC·ROC^(-(n+6)) +µ(AC·ROC^(-(n+6))) +2^(n+6) +1]
              +[AR·ROR^(-(n+4)) +AC·ROC^(-(n+4)) +µ(AC·ROC^(-(n+4))) +2^(n+4) +1] +2^(n+4)-1
       = (1/2)·{  [AR·ROR^(-(n+6)) +AC·ROC^(-(n+6)) +µ(AC·ROC^(-(n+6))) +2^(n+6) +1]
               +2·[AR·ROR^(-(n+4)) +AC·ROC^(-(n+4)) +µ(AC·ROC^(-(n+4))) +2^(n+4) +1] }  +2^(n+4)-1
       = (1/2)·{   AR·[ROR^(-(n+6))+2·ROR^(-(n+4))] +AC·[ROC^(-(n+6))+2·ROC^(-(n+4))] +µ(AC·[ROC^(-(n+6))+2·ROC^(-(n+4))])
                  +2^(n+6)+1 +2^(n+5)+2 }  +2^(n+4)-1
       ---------------------------------------------------------------------------------------------------------------------
                      [ROX^(-(n+6))+2·ROX^(-(n+4))]
       = ROX^(-(n+7))·[ROX         +2·ROX^3       ] = ROX^(-(n+7))·[1], since [-1+ROX+2·ROX^3]=0 for ROX=ROR,ROC and µ(ROC).
       ---------------------------------------------------------------------------------------------------------------------
       =>
    *) = (1/2)·{  AR·[ROR^(-(n+7))] +AC·[ROC^(-(n+7))] +µ(AC·[ROC^(-(n+7))])  +2^(n+6)+1 +2^(n+5)+2  +2^(n+5)-2 }
       = (1/2)·{  AR·[ROR^(-(n+7))] +AC·[ROC^(-(n+7))] +µ(AC·[ROC^(-(n+7))])  +2^(n+7)+1 }
       = a(n+3).     q.e.d.