Registered: Jun 2005
Posts: 39

I think that this might make an integer that
looks like Pascal's Triangle if it's arranged
in rows of 30 numerals but I can't calculate
it at the moment to prove it.

(10**300)/((10**30)-(10**3)-1)

The ** means "to the power of".

I think the math looks Nice but even if it works
it's not really useful or efficient, just interesting.

I call this NKS for being (or at least looking) simple.

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Registered: Nov 2005
Posts: 34

Val, How did you think to look at this expression?

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Registered: Oct 2003
Posts: 20

The phenomenon described appears to be more general. In particular, one does not need to use the exponents 300, 30, 3. One very quick and crude way to see this is to evaluate the following expression

Manipulate[
Grid@Partition[
Module[{sd =
StringDrop[
ToString[N[((10^(100 i))/((10^(10 i)) - (10^i) - 1)), digits],
InputForm], -10], c}, c = Characters[sd];
PadLeft[c, pl + Length[c], "x"]], p], {{i, 3, "smith exponent"},
2, 7, 1, Appearance -> "Labeled"}, {{digits, 500,
"number of digits to compute"}, 100, 1000,
10}, {{p, 30, "row size"}, 20, 50, 1,
Appearance -> "Labeled"}, {{pl, 15, "pad left amount"}, 0, 30, 1,
Appearance -> "Labeled"}]

I have no earthly clue (a) why this works or (b) how to write a program that would break up the integers at the right location or (c) whether other combinations of integers produce similar results.

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Foundation Professor of Law
University of Houston Law Center
Houston, Texas 77204

Last edited by Seth J. Chandler on 08-20-2008 at 12:34 PM

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Registered: Oct 2003
Posts: 20

I'm afraid I found this very interesting. How can it be that the additive process by which Pascal's Triangle is generated can also be generated by the digit sequence of a division? And, if it were true, might it be a way of accelerating the computation of binomials?

The attached notebook contains code that examines the extent to which Smith's Conjecture is true. Preliminary findings: it works in base 10 so far as I can see. And this is true whether one uses 3 or other integers to generate the exponents. That is divisions of the form (10^(100 i))/((10^(10 i)) - (10^i) - 1)) work, at least for the values of i examined thus far and up to a about 1000 digits. This, of course, is NOT a proof.

I also thought I would see whether an analog to the proposition might also be true in other number bases, i.e. whether the digit sequence generated by (10^((b^2) i))/((10^((b^1) i)) - (10^((b^0)i) - 1)) matched Pascal's Triangle when transformed to base b. It appears to be "almost true," but in fact not true. The match breaks down after a while. But perhaps the formula I am using is not the right one.

Anyway, enjoy the exploration permitted by the attached notebook. I am sure the code and interface can be improved. Suggestions welcome.

Note: I am using regular expressions because (a) it is way faster than conventional Mathematica pattern matching using RepeatedNull[0] and (b) perhaps one should think of the relevant comparison as being between strings anyway.

Attachment: pascal triangle as integer.nb
This has been downloaded 366 time(s).

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Seth J. Chandler
Foundation Professor of Law
University of Houston Law Center
Houston, Texas 77204

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Registered: Oct 2003
Posts: 20

Turns out that Eric Rowland has begin work on this issue. See his materials titled "Numbers as Programs: The Power of Digit Sequences" at http://www.wolframscience.com/confe.../presentations/ and http://www.wolframscience.com/confe...ricRowland.html

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Seth J. Chandler
Foundation Professor of Law
University of Houston Law Center
Houston, Texas 77204

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Registered: Jun 2005
Posts: 39

re: Andrius and all:
On another occasion I rediscovered a power series fraction that generates all powers of x up to length n, and was told here that's what it was.

And since I stay off the beaten path in pursuit of
what I am searching for, and I thought I needed
Pascal's Triangle to solve something.
I had an idea to try to use powers of 2 and 11 to
generate it as needed. I thought that factorials
would be too slow an alternative solution for my
needs.

Powers of 11 go 121 1331 14641 and that's as
far as they go for the triangle because of carries. I searched for and found another fraction which
made room for the carries, and then another one, and then a pattern for sizes of rows and
numbers.Then I found numbers that looked
nice and I don't remember everything about
how it works but I'll guess that:
one term sizes the table,
one term sizes the rows,
one term sizes the numbers,
and the -1 is common to the pattern
of all my Pascal-generating power series
fractions, including for powers of 11.
Since it has since been shown that 3's aren't
necessary I don't remember why they are
there except maybe some intuition about
the size of the numbers in the table
generated by this fraction.

The -1 seems to act as like a two's
complement (negative) addition of 1
to something otherwise related to
powers of 10 which I can't visualize
at the moment. Idea: are the powers
of either 101 or 1001 also Pascal Triangles?
If so, the zeros make room for large numbers
and the generalized version makes Pascal
Triangles for the same reason powers of 11
does.

Failure in other bases would result because
one of these must be true:
-Only powers of 11 can make P.triangles.
-Only numbers that look like 11 can do it
So I think the number sizing terms might
probably be different than the terms that
make powers in other bases.
Notice HOW powers of 11 make P.triangles,
and think somehow the long division is
aligning the numbers in a similar way
and there is multiplication in the
blackboard method.
I'm going to guess that in hex, and
therefore also binary, 11h (17d) powers
may make a little more triangle like this:
1 11 121 1331 14641 15A51 16F61
If so, [01][01] = (11base256) powers too.
The point of this is high bases can hold
bigger numbers per digit and keep them
separate longer.
I do not recall but Odd-digits numbers NNNnn
with 1 at each end and zeros in between
might not align digits to make triangles.

．．．？？？Ｂｕｇ？？？２Ｂｃｏｎｔｉｎｕｅｄ

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Registered: Jun 2005
Posts: 39

...Continued:

I originally didn't think the fraction was
useful but now I think it would be very
useful if there is a way to spigot long
division efficiently either in general or in
this case. In this case there is probably
a way to predict the most recent relevant
blackboard state based on the position of the
desired row of the triangle, because there
is an obvious pattern in the numerator and
divisor for any case especially if the non-
integer version 1/((10**precision)-11)base(b)
is used. (Previously mentioned power series
fraction for powers of 11 or any other
number).Certainly there is more than one way
to spigot fractions, and long division can go on
forever without halting nor overflow while
generating a digit at a time and sending it
to an output device and forgetting about it.

This is useful because even though
Pascal's Triangle is CA Rule 90 with
carry and arbitrary precision, it seems
to me that it is otherwise picked with
factorials, which I do not know a fast
way to calculate if I need a long row.
I thought I saw similar patterns in
all each bits of the triangle to avoid
factorials with, when I first needed
to make large P's triangles.

Sorry if all that could be said in fewer words.

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