rectangle tilings

[Tanya Khovanova]

On Thursday, February 15, 2007 at 13:42,
"Eric Angelini" <Eric.Angelini at kntv.be> wrote:
>Hello,
>- is there a simple formula giving the n-th term of A033307 ?
>(Champernowne's constant decimal expansion)

That depends on what you mean by "simple". Perhaps what you really
want to know is whether a(n) can be given in closed form. This can be
done, thereby allowing us to find a(n) without also having to find all
prior terms of the sequence:

Let "index" i = ceiling( W(log(10)/10^(1/9) (n - 1/9))/log(10) + 1/9 )
where W denotes the principal branch of the Lambert W function.

Then a(n) = mod(floor(10^(mod(n + (10^i - 10)/9, i) - i + 1)
ceiling((9n + 10^i - 1)/(9i) - 1)), 10).

Perhaps the above could be simplified significantly; I'm not sure.
(Does anyone have a neater way of giving a(n) in closed form?) The
corresponding Mathematica code is

i[n_] := Ceiling[FullSimplify[ProductLog[Log[10]/10^(1/9) (n - 1/9)]
/Log[10] + 1/9]];
a[n_] := Mod[Floor[10^(Mod[n + (10^i[n] - 10)/9, i[n]] - i[n] + 1)
Ceiling[(9n + 10^i[n] - 1)/(9i[n]) - 1]], 10];

If we want the equivalent of Champernowne's but in a different base b,
merely replace each instance of 9 and 10 above by (b-1) and b, resp.

BTW, what is the difference between A007376 and A033307? The latter
states that "A007376 is same sequence without initial 0." but A033307
does not have an initial 0 (despite its saying "Offset  0,2").

David W. Cantrell




David,  You are right - thanks for catching this

At one point I normalized the offsets for decimal expansions,
6,7,4,5,6,... with offset 2 

A number like 0.4563... would become the sequence 4,5,6,3,...
with offset 0 (and without any leading zero)

So A033307 lost its initial zero at that point!

I will merge the two entries, as A007376.

And add your formula!

Thanks

Neil



on second thoughts I am not going to merge A007376 and A033307

- there are just too many references to A033307 extant

But I will point out that, considered purely as sequences of digits,
they are identical

Neil



Counting 2 by 4 case you forgot the tiling with 2 squares.

---------- Original Message ----------------------------------

>Dear Joshua Zucker,
>
>Thank you for responding so quickly and clearly.
>
>Yes, I like your comment on   A001045  Jacobsthal sequence: a(n) =
>a(n-1) + 2a(n-2).
>"Number of ways to tile a 3 X (n-1) rectangle with 1 X 1 and 2 X 2
>square tiles. "
>
>I drew my 2xn rectangles long-way up (Maybe as nx3, ambiguous).  So
>I'll avoid confusing "horizontal" and vertical hereinbelow.
>

>
>2x4: we have the 5 tilings into dominoes, plus 5 with a square
>tetromino and 2 dominoes, with the square at either end, or in the
>middle and, if the square's at an end, the 2 dominoes either both
>horizontal or both vertical, so a(2x4) = 5+5 =10.
>



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