Divisibility and Binomial Coefficients

[Stefan Steinerberger]

Dear Drew,

thanks for your computing time. Looking at the
solutions makes me wonder whether - if it is true
that there are infinitely many solutions and that
b-a <= 2 - half of the solutions (in an asymptotic
sense) satisfy b-a = 1 and half satisfy b-a = 2.
I am well aware that this might be unanswerable
and have no real reason to conjecture it, it just would
be nice if it were true. =)

I have not yet submitted the sequence to the OEIS;
given your work on it, would you like to submit it?

Thanks again,
Stefan



* Richard Mathar <mathar at strw.leidenuniv.nl> [May 16. 2008 22:22]:
> 

> ja> [...]
> ja> 
> ja> \\ not in OEIS:  partitions of 2*n into distinct even sqrfree parts
> ja> v=Vec( prod(n=1,N, 1+moebius(2*n)^2*x^(n) ) );
> ja> v2=Vec( prod(n=1,L2, (etaplus(x^(2*(2*n-1)^2))/etaplus(x^((2*n-1)^2)))^(-moebius(2*n-1))) );
> ja> (v-v2)==0 \\ OK up to 3000 terms
> ja> \\ ***** NOTE: apparently SAME AS partitions into distinct odd sqrfree parts
> ja> v=Vec( prod(n=1,N, 1+moebius(2*n-1)^2*x^(2*n-1) ) );
> ja> (v-v2)==0 \\ OK up to 3000 terms
> 
> I do not understand this. If 2*n is any even number, the count
> of partitions into distinct, odd and squarefree elements
> (distinct odd members of A005117, members of A056911) and the count of partitions
> into distinct, even and squarefree elements (distinct even
> member of A005117, distinct members of A039956) is not the same.

My fault: the lines should be:

for n == 4:
8 = 2 + 6
4 = 1 + 3





Here is a little trick that is handy in such cases.
We "disassemble" GFs by plugging in symbols:


----------- start pari/gp script ----------

N=11  \\ up to this order
x='x+O('x^(N+1));

1/prod(n=1,N, 1-sym(n)*x^n)  \\ OGF for all partitions

\\ same with symbols:
p1=prod(n=1,N, 1+sym(2*n)*moebius(2*n)^2*x^(n) )
\\ the other one with symbols:
p2=prod(n=1,N, 1+sym(2*n-1)*moebius(2*n-1)^2*x^(2*n-1) )

\\ now set all symbols to one to count:
for (n=1, 4*N, eval(Str(sym(n) "= 1")))

Ser(eval(truncate(p1)))
Ser(eval(truncate(p2)))

----------- end pari/gp script ----------



it will give (minor editing to make stuff readable):

? 1/prod(n=1,N,1-sym(n)*x^n)

? p1=prod(n=1,N,1+sym(2*n)*moebius(2*n)^2*x^(n))

? p2=prod(n=1,N,1+sym(2*n-1)*moebius(2*n-1)^2*x^(2*n-1))

? for(n=1,4*N,eval(Str(sym(n)"= 1")))
? Ser(eval(truncate(p1)))
? Ser(eval(truncate(p2)))


I find this technique very useful occasionally.






Dear Seqfans,  J. Lowell, who is not presently
a member of this list, asked me the following question:
Can someone help him?


>From jhbubby at mindspring.com Fri May 16 20:27:30 2008

There are 2 numbers, 332640 and 352800, that I want to check and see if they qualify for A134865.

The origin of 332640 is: I have to make sure some multiple of 11 is in the sequence. 2520*11=27720 can't be in this sequence because it is a multiple of 13860 with 72 divisors but not of 10080 (the smallest number with 72 divisors.) So we must find the LCM of 27720 and 10080, which is 110880. However, 110880 can't be in this sequence because it is a multiple of 18480 with 80 divisors but not of 15120 (the smallest number with 80 divisors.) So we must find the LCM of 110880 and 15120, which is 332640.

352800, on the other hand, is harder to think of given the probably true statement that 332640 qualifies. It became easier, however as follows:


49 = 7^2 with 3 divisors
LCM of 49 and 4 is 196 with 9 divisors
LCM of 196 and 36 is 1764 with 27 divisors
LCM of 1764 and 900 is 44100 with 81 divisors
44100 is a multiple of 30 but not of 24, so we must find the LCM of 44100 and 24 which is 88200 with 108 divisors
LCM of 88200 and 50400 (smallest number with 108 divisors) is 352800

Does either of these 2 numbers qualify??

(end)

Neil