11/23/2007, 08:22 AM

This is indeed an interesting connection.

Now my quick 2 cents about it.

First, we want the explicit function of which the sums of the powers of the inverted fixed points are the coefficients. We compare Jay's beginnin with index 1 in the first row with Sloane's beginning at index 1 in the second row:

Obviously we have to move the lower row to the left which is the same as dividing Sloane's function by . We get then

and see that the sign is swapped for each uneven power, which can be achieved by using instead of .

So we get

with .

for . If we transform this further via we get

to prove.

Looks strange, perhaps I made an error somewhere.

Now my quick 2 cents about it.

First, we want the explicit function of which the sums of the powers of the inverted fixed points are the coefficients. We compare Jay's beginnin with index 1 in the first row with Sloane's beginning at index 1 in the second row:

Obviously we have to move the lower row to the left which is the same as dividing Sloane's function by . We get then

and see that the sign is swapped for each uneven power, which can be achieved by using instead of .

So we get

with .

for . If we transform this further via we get

to prove.

Looks strange, perhaps I made an error somewhere.