[seqfan] Re: Question

[Bob Selcoe]

Hi Seqfans,

Alois Heinz raised the quite reasonable possibility with the two sequences 
proposed sequences that it may not be possible to determine if there are 
unique maximum and minimum numbers of partition elements in A000793.

It seems to me they are unique; but I can't prove it and certainly could be 
wrong.  And I don't have a way to test empirically to see if there's an 
obvious contradiction.

Can anyone say definitively one way or the other?  If I'm wrong, or if it's 
not clear, then I'll have to revise the proposed postings.

Thanks,
Bob S
(PS - Neil, I decided to post this question here anyway, so certainly no 
need to reply to my discussion post on A256445).

--------------------------------------------------
From: "Bob Selcoe" <rselcoe at entouchonline.net>
Sent: Sunday, March 29, 2015 2:16 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: Re: [seqfan] Re: Question

> Hi Allan, David and Seqfans,
>
> I proposed two sequences, A256443 and A256445 which are, respectively, 
> irregular triangles of the minimum and maximum number of partition 
> elements of A000793.  For example, A000793(11) = 30; so T256443(11,k) = 
> [5,6] and T256445(11,k) = [1,2,3,5].
>
> Hopefully this will be of some interest to the community.
>
> Best,
> Bob
>
> --------------------------------------------------
> From: "Allan Wechsler" <acwacw at gmail.com>
> Sent: Sunday, March 29, 2015 7:41 AM
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Question
>
>> Factoring 232792560 gives 2^4.3^2.5.7.11.13.17.19, which is definitely 
>> not
>> 23#. Factoring 223092870 does give all the primes through 23.
>>
>> I cannot prove quickly that 232792560 is the right value for A000793, but
>> it is definitely a lower bound, since 16+9+5+7+11+13+17+19 = 97 <= 100, 
>> so
>> a partition of 100 that achieves that value is [16,9,5,7,11,13,17,19,3]. 
>> So
>> Wilson's question is definitely answered in the negative, even if the 
>> given
>> value for A000793(100) is wrong.
>>
>> On Sun, Mar 29, 2015 at 1:03 AM, Bob Selcoe <rselcoe at entouchonline.net>
>> wrote:
>>
>>> Hi David and Seqfans,
>>>
>>> I would have thought so, but according to its b-file:  A000793(100) =
>>> 232792560.
>>>
>>> Since A007504(9) = 100 and prime(9) = 23, primorial(23) = A002110(9) =
>>> 223092870.
>>>
>>> So either there's an error in the b-file, or the answer is no.
>>>
>>> Cheers,
>>> Bob Selcoe
>>>
>>> --------------------------------------------------
>>> From: "David Wilson" <davidwwilson at comcast.net>
>>> Sent: Saturday, March 28, 2015 8:17 PM
>>> To: "'Sequence Fanatics Discussion list'" <seqfan at list.seqfan.eu>
>>> Subject: [seqfan] Question
>>>
>>>
>>>  A000793(A007504(n)) =? A002110(n)
>>>>
>>>>
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