[seqfan] Re: Finite sequence?

jnthn stdhr jstdhr at gmail.com
Tue Sep 19 18:40:02 CEST 2017


Hans,

I have checked all numbers up to n=2000 for one or more repeated values of
T_m.  Other than those you have pointed out, I have found no more.  The
following output lists tuples of the form (n, repeated value of T_m):

[(5, 2), (6, 3), (7, 3), (8, 5), (10, 5), (11, 5), (11, 10), (13, 14), (13,
18), (14, 7), (14, 23), (15, 7), (19, 30), (22, 11), (23, 11), (26, 56),
(30, 15), (31, 15), (34, 297), (43, 5708), (44, 22), (45, 22), (46, 176),
(60, 30), (60, 1575), (61, 30), (68, 385), (84, 42), (85, 42), (112, 56),
(113, 56), (154, 77), (155, 77), (202, 101), (203, 101), (270, 135), (271,
135), (352, 176), (353, 176), (462, 231), (463, 231), (594, 297), (595,
297), (770, 385), (771, 385), (980, 490), (981, 490), (1254, 627), (1255,
627), (1584, 792), (1585, 792)]

I am currently using my 800MHz phone to run my program.  When I have a
chance to run it on my PC, I will check up to n=5000, which takes a couple
of hours.
On Sep 19, 2017 7:15 AM, "jnthn stdhr" <jstdhr at gmail.com> wrote:

> Hans,
>
> Thank you for pointing out the multiple pairs.  Later today I will modify
> my program to check for multiple pairs and report.
>
> Also, thank you for the Mathematica test.  I have coded an algorithm using
> Python (for its built-in big num capabilities) that produces the full
> triangle A008284(10,000) in about 3 minutes.  As a test of correctness I
> totaled row 10,000 and compared the result to the value of P(10,000) shown
> on the Wikipedia page for partitions and its begining and ending parts
> match.  I assume my method is novel, but who knows.  Any suggestions for
> how I should go about making this algorithm known?  Although it's rather
> simple, it's in a quick-and-dirty state and needs refinement, but I am
> willing to post it here for review.
>
> -Jonathan
> On Sep 19, 2017 5:49 AM, "Hans Havermann" <gladhobo at bell.net> wrote:
>
>> > For the values of m=8,13,19,26,34,43,46,68, none of which is equal to
>> 2P(n) or 2P(n)+1 for some integer n, there are repeated values greater than
>> 1 in T_m:
>>
>> Note that for m = 11, 14, and 60, there are two pairs of repeated values.
>> Perhaps only one of these pairs meets your 2P/2P+1 exclusion principle?
>>
>> > As an aside, would some mind timing how long it takes Mathematica to
>> produce the triangle A008284(10,000)?
>>
>> Blindly using the second (of three) Mathematica formulation in A008284,
>> I'd guess (based on a half-day run) that it would take me about three days.
>>
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>


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