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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 228

Consider this equation.

Where Ps is the resulting Prime and n & p are integers.

If t=1, this equation reduces to the known prime (Refer OEIS) as follows:

Selecting the values of t and n for odd/prime Ps, yields the following example

Which is a prime of approximately 14,700 Digits.

By the way, does anyone know how to calculate how many digits this number is on Mathematica?

*Last edited by Stangerzv (2013-06-15 15:02:15)*

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**bobbym****Administrator**- From: Bumpkinland
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It has 14734 digits.

In Mathematica, this is simplest.

N[8^16315 - 7^16315 - 6^16315 - 5^16315 -4^16315 - 3^16315 - 2^16315, 20]

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**phrontister****Real Member**- From: The Land of Tomorrow
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And PrimeQ[8^16315 - 7^16315 - 6^16315 - 5^16315 - 4^16315 - 3^16315 - 2^16315] agrees that the result is prime.

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**Stangerzv****Member**- Registered: 2012-01-30
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Thanks bobbym and phrontister..alpertron won't work due to out of range.

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**bobbym****Administrator**- From: Bumpkinland
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Use this in Mathematica for a proof of the primality of a number:

Needs["PrimalityProving`"]

ProvablePrimeQ[

8^16315 - 7^16315 - 6^16315 - 5^16315 - 4^16315 - 3^16315 - 2^16315]

But it may take a long time!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**Stangerzv****Member**- Registered: 2012-01-30
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Thanks bobbym..I would run on my computer.

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**bobbym****Administrator**- From: Bumpkinland
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I turned it off after around 10 minutes. May take hours or days...

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**phrontister****Real Member**- From: The Land of Tomorrow
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bobbym wrote:

In Mathematica, this is simplest.

N[8^16315 - 7^16315 - 6^16315 - 5^16315 -4^16315 - 3^16315 - 2^16315, 20]

This one seems good too:

IntegerLength[8^16315 - 7^16315 - 6^16315 - 5^16315 - 4^16315 - 3^16315 - 2^16315]

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
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How about this one?

8^16315.- 7^16315 - 6^16315 - 5^16315 - 4^16315 - 3^16315 - 2^16315

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**phrontister****Real Member**- From: The Land of Tomorrow
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Yes...works too. What does the dot after the 5 do?

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**bobbym****Administrator**- From: Bumpkinland
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It tells M that number is done in floating point. This is much faster then exact arithmetic. Once M does one bit of an expression in floating point, it does the rest in it too.

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**phrontister****Real Member**- From: The Land of Tomorrow
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Oh, I see. Interesting option.

I recently installed a freeware Add-In into Excel that enables multi-precision floating point calcs up to 32760 significant digits. It was fine for what I was working on but it has some whiskers too, so I've disabled it.

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**bobbym****Administrator**- From: Bumpkinland
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That is interesting. I have been playing a little with geogebra's spreadsheet. If only they would interface geo with maxima a little better.

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**Stangerzv****Member**- Registered: 2012-01-30
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It seems that once the value of p becoming larger, you can simply taking log on the first term to get the digits. So, the number of digits=16315log8=14733.91.

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**bobbym****Administrator**- From: Bumpkinland
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Yep, that will will give an approximation ala Landau Notation.

But why settle for an approximation when you can get the exact answer?

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**Stangerzv****Member**- Registered: 2012-01-30
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I was not sure, until you gave me the exact value, which is more less the value by taking log on the first term.

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**bobbym****Administrator**- From: Bumpkinland
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Mathematically, it is correct to approximate it using just the first term when n is large.

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**phrontister****Real Member**- From: The Land of Tomorrow
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bobbym wrote:

Use this in Mathematica for a proof of the primality of a number:

Needs["PrimalityProving`"]

ProvablePrimeQ[

8^16315 - 7^16315 - 6^16315 - 5^16315 - 4^16315 - 3^16315 - 2^16315]But it may take a long time!

I didn't know that, so I looked it up and found this help file: PrimalityProving/tutorial/PrimalityProving

Near the end of the article it says "PrimeQ will be, in general, several orders of magnitude faster than ProvablePrimeQ." My PrimeQ test on this number took just under 6 minutes...but of course it doesn't give the mathematical proof (which I can't follow, anyway, having tested the function on a smaller prime and finding the proof too advanced for me).

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**bobbym****Administrator**- From: Bumpkinland
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Since the WZ algorithm they now use certificates to prove when an algorithm gave the right answer. I do not understand it either.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
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phrontister wrote:

IntegerLength[8^16315 - 7^16315 - 6^16315 - 5^16315 - 4^16315 - 3^16315 - 2^16315]

This form helped in another program of mine where I needed to plug the number of digits into another calculation, which I did via len=IntegerLength[a].

I guess that could also be done with other methods, but the one above seemed to be a simple option.

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**Stangerzv****Member**- Registered: 2012-01-30
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I have tested up to n=30,000 no prime so far. I would run up to n=200,000 and lets see whether there would be more prime or not.

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