
Re: Tiling the plane with checkerboard patterns
Posted:
Jul 14, 2010 7:52 AM


> On Fri, Jul 09, 2010 at 06:00:32PM +0000, Avni Pllana > wrote: > > > > I think I found the right algorithm/formula for > a(n): > > > > %%%%%%%%%%%%%%%%%%%%%%%% > > > > for n=1:5 > > > > N=n^2; > > A=2; > > > > for i=1:N1 > > > > A=A+floor(nchoosek(N,i)/N)+mod(nchoosek(N,i),N); > > end > > > > a(n)=A; > > > > end > > > > %%%%%%%%%%%%%%%%%%%%%%%% > > > Avni, that's wonderful! It produces the exact > values for a(n), n=1,2,3,4,5. Do you have a proof > for the general case? > > Rouben > > Aside: Since the Matlab program above is limited > by its dependence on floating point arithmetic > to the range n=1,2,...,7, I recoded your algorithm > in Maple which performs integer arithmetic with no > limits. Here is the Maple's version of your formula: > > a := proc(n::posint) > local N, b; > N := n^2; > b := binomial(N,i); > return sum( iquo(b,N) + irem(b, N), i=0..N); > end proc: > > Then the command "seq(a(n), n=1..9);" produces > the values of a(n) for n=1,2,...,9: > > n a(n) > 1 2 > 2 7 > 3 64 > 4 4156 > 5 1342208 > 6 1908874521 > 7 11488774559744 > 8 288230376151712689 > 9 29850020237398251228192 >
Hi Rouben,
I tried to find further numerical evidence about my sum formula. A comparison of the number of equivalence classes using the Matlab program and using the sum formula is shown below:
n=1 N=1^2=1
Mat_1=[1 1] Sum_1=[1 1]
a(1)=2, Matlab a(1)=2, sum formula ________________________
n=2 N=2^2=4
Mat_2=[1 1 3 1 1] Sum_2=[1 1 3 1 1]
a(2)=7, Matlab a(2)=7, sum formula ________________________
n=3 N=3^2=9
Mat_3=[1 1 4 12 14 14 12 4 1 1] Sum_3=[1 1 4 12 14 14 12 4 1 1]
a(3)=64, Matlab a(3)=64, sum formula ________________________
n=4 N=4^2=16
Mat_4=[1 1 9 35 122 273 511 715 822 715 511 273 122 35 9 1 1] Sum_4=[1 1 15 35 125 273 508 715 810 715 508 273 125 35 15 1 1]
a(4)=4156, Matlab a(4)=4156, sum formula ________________________
n=5 N=5^2=25
Mat_5=[1 1 12 92 506 2130 x x ...] Sum_5=[1 1 12 92 506 2130 .......]
a(5)=1342208, Rouben a(5)=1342208, sum formula ________________________
n=6 N=6^2=36
Mat_6=[1 1 19 201 1649 x x ...] Sum_6=[1 1 35 210 1645 .......]
a(6)= ?, Matlab a(6)=1908874521, sum formula ________________________
Above results show that for n odd the sum formula gives exactly the number of equivalence classes and exact a(n).
For n even the sum formula does not give the exact number of equivalence classes, but it gives exact a(n), at least for n=2 and n=4.
These results resemble somehow the behavior of magic squares.
Best regards, Avni

