[seqfan] Re: A033791

Fri Oct 17 01:31:35 CEST 2014

"    ...  while A033791 has a generating function Product(f(q^2^n)), where n 
goes from 0 to 5. "

This choice of "5" is an arbitrary truncation of  an intrinsically infinite 
product
t(z^2) . t(z^2^2) . t(z^2^3) . etc
The OEIS has a rule that arbitrary constants should be avoided, especially 
so if more general definition is available.

But, historically, I suppose the truncation to "5" was for calculation 
purposes  : q^2^5 = q^32 gives the sequence terms of A033791  up to  64 
terms.  Or, possibly its limitation to divisors of 32 had some 
lattice-theory foundation?

My advice is to submit a new sequence where the limitation to 5 factors is 
replaced,
and link the partitions comment to that one. Cross-referencing should then 
point out back to the
older (and by then obsolete?) A033791.

t(z^2) == Sum[(z^2)^(i (i+1)/2),{i, 0, inf]}]  ==   Sum[ z^(i (i + 1)), {i, 
0, inf]}]  ==  EllipticTheta[2, 0, z] / (2*z^(1/4))

Wouter.

-----Original Message----- 
From: David Newman
Sent: Friday, October 17, 2014 12:13 AM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A033791

Thank you Wouter.

In answer to your question:

"if I interpret your definition of carry-free frequencies correctly, then
the number of carry-free partitions of n equals A018819
"Binary partition function: number of partitions of n into powers of 2".
Can you confirm?"

Yes, I confirm this.

Here's how I see the situation:
If I let f(q)= 1+q+q^3+q^6+..., then my proposed sequence has generating
function  Product(f(q^2^n)), where n goes from 0 to infinity, while A033791
has generating function Product(f(q^2^n)), where n goes from 0 to 5.  I'm
still not sure if to propose this as a new sequence or as a comment to the
existing sequence since the two are identical for 60+ terms.

On Thu, Oct 16, 2014 at 8:59 AM, Wouter Meeussen <wouter.meeussen at telenet.be
> wrote:

> David,
>
> if I interpret your definition of carry-free frequencies correctly, then
> the number of carry-free partitions of n equals A018819
> "Binary partition function: number of partitions of n into powers of 2".
> Can you confirm?
>
> It is quite easy to extend both sequences to 128 terms, and they are
> perfectly equal.
> In the light of the above connection to A018819, the conjectured equality
> seems plausible if
> you consider the implementation of the theta_2 in terms of products of
> x^2^d.
>
> In Mathematica v. 10.0.1 : (*honni soit qui mal y pense *)
>
> carryfree[{li__Integer}]:=Block[{m,freq}, freq=Sort[Map[Last,Tally[{li}]
> ]];m=Max[freq];Max[Total[IntegerDigits[freq,2,1+Floor[Log[2,m]]]]]<2];
> your definition :
> Table[Count[IntegerPartitions[n,All, Table[ k (k+1) 2,{k,2+Floor[Sqrt[2n
> ]]}]],q_/;carryfree[q]],{n,128}]
> {1,1,2,1,2,3,3,2,3,4,3,6,5,5,7,6,6,7,6,9,10, ...
> ,237,221,220,246,225,242,262,243,249}
>
> same as A033791 :
>
> t[z_]:=Sum[z^(i (i+1)/2),{i,0,16}];
> CoefficientList[Series[Product[t[x^2^d],{d,0,10}],{x,0,128}],x]
>
> you just need to properly adapt the limits in the definition of t[z] (upto
> 16) and in the product t[x^2^d] (up to 10) over d .
>
> Nice find! Keep up the good work, David!
>
> Wouter.
>
>
> -----Original Message----- From: David Newman
> Sent: Thursday, October 16, 2014 3:16 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: A033791
>
> The sequence that I'm lookinh at is the number of partitions into summands
> which are triangular numbers and frequencies satisfying the "no binary
> carry" condition. A partition has the "no binary carry" condition if the
> sum of all its frequencies, when written in binary notation has no carry.
>
> On Wed, Oct 15, 2014 at 5:58 PM, Frank Adams-Watters <
> franktaw at netscape.net>
> wrote:
>
>  It would be easier to answer this question if you told us what your
>> proposed sequence is.
>>
>> Franklin T. Adams-Watters
>>
>>
>> -----Original Message-----
>> From: David Newman <davidsnewman at gmail.com>
>> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>> Sent: Wed, Oct 15, 2014 4:52 pm
>> Subject: [seqfan] Re: A033791
>>
>>
>> The sequence that I'm working with will agree with A033791 for n<64, but
>> will be different thereafter.  Should I propose it as new sequence or as 
>> a
>> comment on A033791.
>>
>> On Wed, Oct 15, 2014 at 12:32 PM, Andrew N W Hone <A.N.W.Hone at kent.ac.uk>
>> wrote:
>>
>>  Hi David,
>>
>>>
>>> I am not looking at the sequence just now but I guess theta_2 is a
>>>
>>>  theta
>>
>>  function in Jacobi's classical notation: see
>>>
>>> http://en.wikipedia.org/wiki/Theta_function
>>>
>>> under "auxiliary functions". In the combinatorial setting, counting
>>> partitions, you want to set the first argument z=0, which gives the
>>>
>>>  q-series
>>
>>
>>> \sum q^{(n+1/2)^2}
>>>
>>> where the sum is from n=-infinity to +infinity.
>>>
>>> A good classical reference is A Course of Modern Analysis by
>>>
>>>  Whittaker &
>>
>>  Watson. The first volume of Mumford's Tata Lectures on Theta is also
>>>
>>>  great,
>>
>>  and has a more modern point of view.
>>>
>>> All the best,
>>> Andy
>>>
>>> ________________________________________
>>> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of David
>>>
>>>  Newman [
>>
>>  davidsnewman at gmail.com]
>>> Sent: 15 October 2014 17:23
>>> To: Sequence Fanatics Discussion list
>>> Subject: [seqfan] A033791
>>>
>>> I have a partition type sequence which matches A033791 for the first
>>>
>>>  40
>>
>>  terms.  However, beyond the computational evidence I have no reason to
>>> think that the two are the same.  In fact, I'm guessing that they
>>>
>>>  differ
>>
>>  for n>63.  I am hampered in my efforts  in part because I don't know
>>>
>>>  which
>>
>>  function is meant by theta2.  Could someone give me the definition of
>>> theta2 or point me to a source for the definition?
>>>
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