# Triangle T(n,k) = [x^(k*2^(n-1))] Product_{i=0..n-1} (1 + x^(2^i))^(2^(n-i-1))

Paul D. Hanna pauldhanna at juno.com
Sun Aug 19 21:19:44 CEST 2007

```Seqfans,
Here is a new triangle with an unexpected (conjectured) property
that I need help proving.

Consider A132318:  Symmetric triangle, read by rows, where
T(n,k) = [x^(k*2^(n-1))] Product_{i=0..n-1} (1 + x^(2^i))^(2^(n-i-1))
for n>0 with T(0,0)=1.

THE CONJECTURE:
Row sums equal 2^b(n) for n>=0, where
b = [0,1,2,5,12,27,58,121,248,...]
such that  b(n) = 2*b(n-1) + n-1 for n>0 with b(0)=0.

Notice that the rows consist of selected coefficients of a product
whose coefficients in x obviously sum to 2^(2^n - 1);
the unexpected observation is that the sum of those selected
coefficients is in turn an exact power of 2: 2^b(n), as described above.

Below I give the initial rows of the triangle and give examples.

Can anyone help prove the conjecture?
Thanks,
Paul
-------------------------------------------------------------

The triangle begins:
1;
1,1;
1,2,1;
1,15,15,1;
1,1024,2046,1024,1;
1,7048181,60060682,60060682,7048181,1;
1,469389728563470,72057594037927935,143176408618728932,72057594037927935,
469389728563470,1;

There are n*2^(n-1)+1 terms in
P(n) = Product_{i=0..n-1} (1 + x^(2^i))^(2^(n-i-1)) for n>0;
in this triangle, row n consists of coefficients of x^(k*2^(n-1)) in P(n)
as k=0..n.

Examples:
T(2,1) = [x^(1*2)] (1+x)^2*(1+x^2) = 2 ;
T(3,1) = [x^(1*4)] (1+x)^4*(1+x^2)^2*(1+x^4) = 15 ;
T(4,3) = [x^(3*8)] (1+x)^8*(1+x^2)^4*(1+x^4)^2*(1+x^8) = 1024 ;
T(5,3) = [x^(3*16)] (1+x)^16*(1+x^2)^8*(1+x^4)^4*(1+x^8)^2*(1+x^16) =
60060682.

(PARI)
{T(n,k)=if(n==0,1,polcoeff(prod(i=0,n-1,
(1+x^(2^i)+x*O(x^(k*2^(n-1))))^(2^(n-i-1))), k*2^(n-1)))}

END.

```