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Re: Question on partitions?

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Subject: Re: Question on partitions?
From: danj4...@gmail.com (Dan joyce)
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by: Dan joyce - Mon, 19 Jun 2023 23:02 UTC

On Monday, June 19, 2023 at 10:36:53 AM UTC-4, V wrote:
> The best way to find Your answer is to ask god.
> On Monday, March 5, 2001 at 10:23:20 PM UTC+2, Dan Joyce wrote:
> > I am just getting into the concept of partitions and it looks very
> > interesting. So this could be a stupid question.
> > My question is, if n as the 1st integer to be partitioned then n->oo
> > for totals of ordered and unordered partitions would be---
> > starting with n=2
> > n = 2^(n-1) = Total number of ordered and unordered partitions
> > for any particular n. It appears true through the 7th integer.
> > Is this true, or do we have another problem similar to the Leo Moser's
> > spot problem where 2^n breaks down at 2^5.
> > If that were the case, it would have to break down somwhere after 2^6
> > in this instance.
> > This could be a very well known fact about partition counts of any
> > particular integer for ordered and unordered n count totals to =
> > 2^(n-1), but I have not found it yet.
> > Any comments appreciated.
> > Dan Joyce
Someone did answer it 22 some odd years ago but some smart a**
deleted it. You might know who posted this old thread is still around
and probably deleted the response to my original question.

Re: Question on partitions?

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Subject: Re: Question on partitions?
From: danj4...@gmail.com (Dan joyce)
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by: Dan joyce - Mon, 19 Jun 2023 23:04 UTC

On Monday, June 19, 2023 at 7:02:09 PM UTC-4, Dan joyce wrote:
> On Monday, June 19, 2023 at 10:36:53 AM UTC-4, V wrote:
> > The best way to find Your answer is to ask god.
> > On Monday, March 5, 2001 at 10:23:20 PM UTC+2, Dan Joyce wrote:
> > > I am just getting into the concept of partitions and it looks very
> > > interesting. So this could be a stupid question.
> > > My question is, if n as the 1st integer to be partitioned then n->oo
> > > for totals of ordered and unordered partitions would be---
> > > starting with n=2
> > > n = 2^(n-1) = Total number of ordered and unordered partitions
> > > for any particular n. It appears true through the 7th integer.
> > > Is this true, or do we have another problem similar to the Leo Moser's
> > > spot problem where 2^n breaks down at 2^5.
> > > If that were the case, it would have to break down somwhere after 2^6
> > > in this instance.
> > > This could be a very well known fact about partition counts of any
> > > particular integer for ordered and unordered n count totals to =
> > > 2^(n-1), but I have not found it yet.
> > > Any comments appreciated.
> > > Dan Joyce
> Someone did answer it 22 some odd years ago but some smart a**
> deleted it. You might know who posted this old thread is still around
> and probably deleted the response to my original question.

Some idiot is playing games here!

Subject	Author
Re: Question on partitions?	Dan joyce
Re: Question on partitions?	Dan joyce