[seqfan] Re: Question

[Sean A. Irvine]

I have verified that A000793(100) = 232792560 by explicit computation
over all partitions of 100.

Sean.

On 30 March 2015 at 01:41, Allan Wechsler <acwacw at gmail.com> wrote:
> 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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