[seqfan] Re: Pi in Pi (with a "divide sign")

M. F. Hasler oeis at hasler.fr
Thu Jul 17 13:13:46 CEST 2014


Hello Eric,
to beat this, it is sufficient to choose the start of an arbitrarily
long string of 99...99 or 100..00,
and it is most probable that there are infinitely many such.

For example, the string 99999 occurs at position 762. (This string
occurs 1956 times in the first 200M digits of Pi.
counting from the first digit after the decimal point: The 3. is not counted.)
The string 9999999 occurs at position 1722776. This string occurs 18
times in the first 200M digits of Pi.
The string 99999999 occurs at position 66780105. This string occurs 1
times in the first 200M digits of Pi.

So, if you cut after the 762 decimals and make the second cut (discard
the rest) after the 1524-th decimal digit, then the ratio equals Pi to
approx. 6 digits :
31415926..0721134/9999998372....  = pi + 5.747 x 10^(-7)

Nice summer vacation to all SeqFans,
Maximilian

On Thu, Jul 17, 2014 at 12:36 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>
> Hello SeqFans,
>
>
>
> Step 1 - I concatenate the first 173 consecutive digits of Pi:
>
> 3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938
>
>
>
> Step 2 - I insert the divide sign "/" somewhere to produce a fraction:
>
> 3141592653589793238462643383279502884197169399375105820974944592307816406286208/998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938
>
>
>
> Step 3 - The fraction = 3.1459...
>
>
>
> Can you beat that? (this is, get closer to Pi)?
>
>
>
> Best,
>
> É.
>
>
> _______________________________________________
>
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