# Some Sum-Of-Sum Congruences

Leroy Quet qq-quet at mindspring.com
Wed Mar 3 01:40:27 CET 2004

```I will just copy/paste my sci.math post.
I only post this message because I assume SOME of you may enjoy it.
I do not know if any of the {A(m,n)} sequences should be in the EIS.

But I will add that the below is a result of:

a(m+1) = sum{k=0 to m} binomial(m,k) (-1)^k a(k),

for all m, m>= 0.

Leroy

qqquet at mindspring.com (Leroy Quet) wrote in message
> qqquet at mindspring.com (Leroy Quet) wrote in message
> > Let, for each nonnegative integer k,
> >
> > a(6k) = a(6k +1) = 1;
> > a(6k +2) = a(6k +5) = 0;
> > a(6k +3) =a(6k +4) = -1.
> >
> > Let A(0,m) = a(m);
> >
> > and for all positive integers n, and for nonnegative integers m,
> >
> > A(n,m) = sum{k=0 to m} A(n-1,k);
> >
> >
> > Then, for q and r = any nonnegative integers:
> >
> >
> > m!*(A(q,m+1) -binomial(q+m,q-1))
> >
> > is congruent to
> >
> > m!*A(r,m) (-1)^m  (mod {m+q+r}).
> >
> >
> > So, more specifically, from the above we get:
> >
> > For ODD m,
> >
> > (m-1)!*A(r,m)
> >
> > is congruent to
> >
> > (r+m)!/(m r!)    (mod {m+2r+1}).
> >
> >
> >
> >
> > For EVEN m,
> >
> > (m-1)!*A(r,m)
> >
> > is congruent to
> >
> > (r+m-2)!/(m (r-2)!)  (mod {m+2r-2}).
> >
> >
> > (Someone might enjoy confirming the above congruences...)
> >
> > I wonder if any of these congruences have any interesting
> > number-theory implications...
> >
> > thanks,
> > Leroy
> >    Quet
>
> I must point out that
>
> A(n,m)  =
>
> sum{k>=0} (binomial(m+n-1-6k,n-1) +binomial(m+n-2-6k,n-1)
>           -binomial(m+n-4-6k,n-1) -binomial(m+n-5-6k,n-1)),
>
> where
>
> binomial(q,j) = q!/(j!(q-j)!)
> if q >=0, as before,
> but, here,
>  binomial(q,j) = 0
> for q < 0.
>
> thanks,
> Leroy Quet

I should probably rewrite the immediately previous sum, for number theory
purposes, as:

For n>= 2,

A(n,m) =

(1/(n-1)!)*
sum{k>=0} sum{j=1 to n-1} S(n-1,j)*(
[m+1-6k]^j +[m-6k]^j -[m-2-6k]^j -[m-3-6k]^j),

where S() is an unsigned Stirling number of the first kind,
and [x] =  maximum(0,x).

I should also probably give the first few terms of A(n,m):

A(0,m):  1, 1, 0, -1, -1,  0,  1,  1,  0, -1, -1, 0,...
A(1,m):  1, 2, 2,  1,  0,  0,  1,  2,  2,  1,  0, 0,...
A(2,m):  1, 3, 5,  6,  6,  6,  7,  9, 11, 12, 12, 12,...
A(3,m):  1, 4, 9, 15, 21, 27, 34, 43, 54, 66, 78, 90,...

Note that only {A(0,m)} and {A(1,m)} are periodic.

thanks,
Leroy Quet

```