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• Hello, ... then ... Nice try, it looks as though you got the first frame correct with a couple of minor adjustments. I changed yours to ... * 10^(n-1) to
Message 1 of 5 , Mar 30, 2005
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Hello,

<mark.underwood@s...> wrote:

> The rows go from 2 to 29. There are 66 columns.
>
> If the number of a row (from 2 to 29) is represented by Ro (pun
> intended), and n is the iteration number from zero to infinity,
then
> the formula for the value of a row (to 66 digits) at the nth
> iteration is
>
>
> 11 * (Ro-2)!/Ro! * 10^(n mod 66)

Nice try, it looks as though you got the first frame correct with a
couple of minor adjustments. I changed yours to ... * 10^(n-1) to
match decimal point placement and because the mod 66 on n is not
necessary and not part of the formula (it actually would simply grow
to infinite size with more frames) and there's actually 61 digits per
row (I think), but these are minor adjustments.

I have added a couple of pages, one with the frames derived from your
formula

http://www.imathination.net/marksguess/marksguess.html

And one with the frames of the challenge animation

http://www.imathination.net/daframes/daframes.html

What I find fascinating and is perhaps a good hint, is that the
numbers to the left of the decimal for each "Ro" form a repeating
pattern of digits much like the repeating pattern to the right of
decimal point of a rational number, and the digits to the right of
the decimal point in the challenge animation cycle through different
values, one for each unique place digit of the repeating pattern on
the left of the decimal for each "Ro" entry. The tenth "Ro" in
particular is notably interesting.

Don't worry though, it doesn't take an awful lot to qualify for
acceptance to the Apartment of Mathematics.

Dick
• Hi Dick As you can tell I hurried - in fact I looked only at the first five rows of the first column, and then only the first row of the second column! How s
Message 1 of 5 , Mar 30, 2005
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Hi Dick

As you can tell I hurried - in fact I looked only at the first five
rows of the first column, and then only the first row of the second
column!

How's this then:

If sum(10^-n) means 10^-1 + 10^-2 + 10^-3 + ... + 10^-n, then the
formula is

(Ro-2)!/Ro! * 10^n * (1 + sum (10^-n))

So the third iteration of row = 29 is

27!/29! * 10^3 * (1 + 10^-1 + 10^-2 + 10^-3)

I checked a few more values this time. :) Has this to do with primes?

Mark

--- In primenumbers@yahoogroups.com, "Dick" <richard042@y...> wrote:
>
> Hello,
>
> --- In primenumbers@yahoogroups.com, "Mark Underwood"
> <mark.underwood@s...> wrote:
>
> > The rows go from 2 to 29. There are 66 columns.
> >
> > If the number of a row (from 2 to 29) is represented by Ro (pun
> > intended), and n is the iteration number from zero to infinity,
> then
> > the formula for the value of a row (to 66 digits) at the nth
> > iteration is
> >
> >
> > 11 * (Ro-2)!/Ro! * 10^(n mod 66)
>
> Nice try, it looks as though you got the first frame correct with a
> couple of minor adjustments. I changed yours to ... * 10^(n-1) to
> match decimal point placement and because the mod 66 on n is not
> necessary and not part of the formula (it actually would simply
grow
> to infinite size with more frames) and there's actually 61 digits
per
> row (I think), but these are minor adjustments.
>
> I have added a couple of pages, one with the frames derived from
your
> formula
>
> http://www.imathination.net/marksguess/marksguess.html
>
> And one with the frames of the challenge animation
>
> http://www.imathination.net/daframes/daframes.html
>
> What I find fascinating and is perhaps a good hint, is that the
> numbers to the left of the decimal for each "Ro" form a repeating
> pattern of digits much like the repeating pattern to the right of
> decimal point of a rational number, and the digits to the right of
> the decimal point in the challenge animation cycle through
different
> values, one for each unique place digit of the repeating pattern
on
> the left of the decimal for each "Ro" entry. The tenth "Ro" in
> particular is notably interesting.
>
> Don't worry though, it doesn't take an awful lot to qualify for
> acceptance to the Apartment of Mathematics.
>
> Dick
• ... Sorry to say I am running short of spare time in the next couple of days, but I sense you are closer. In fact, I am looking forward to checking because if
Message 1 of 5 , Mar 31, 2005
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<mark.underwood@s...> wrote:

> How's this then:
>
> If sum(10^-n) means 10^-1 + 10^-2 + 10^-3 + ... + 10^-n, then the
> formula is
>
> (Ro-2)!/Ro! * 10^n * (1 + sum (10^-n))
>
> So the third iteration of row = 29 is
>
> 27!/29! * 10^3 * (1 + 10^-1 + 10^-2 + 10^-3)

Sorry to say I am running short of spare time in the next couple of
days, but I sense you are closer. In fact, I am looking forward to
checking because if you have found a means to produce the original,
you have done so in a different way. My formula does not use the
factorial and I figured your match to the first frame was a
consequence of a triviality for n=2. If you've replicated mine with
yours, the challenge is truly back in my court.

My animation came out of an investigation of the convergence of
finite functions to Euler's gamma, which of course ultimately has to
do with the primes in general and via the Riemann hypothesis in
particular in this case (or so I am investigating/teaching myself).

I am working on a paper that is woefully incomplete at this stage and
I have recently asked someone to look at it privately. I will post
it here if they express no interest or otherwise advise me to go
ahead and post it, but I want to at least add a summary of the parts
I haven't had time to TeX up yet, and also, out of respect, give my
is dead on, nothing like getting it all in one place in neat and
clean form to show yourself what you got.

So, I will check your latest formula as soon as I can and with just a
little more patience, you will see the original work soon.

-Regards,

Dick Boland
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