(01/28/2011, 09:12 PM)mike3 Wrote: Then, the not-quite-symbolic but still "explicit" and "non-recursive" formula is
\( a_n = \sum_{j=1}^{2^{n-2}} \prod_{k=1}^{(\mathrm{number\ of\ 1\ bits\ in\ }2^{n-1} + 2j - 1)} r_{(\mathrm{position\ of\ }k\mathrm{th\ 1\ bit\ in\ }2^{n-1} + 2j - 1\mathrm{,\ with\ the\ LSB\ counted\ as\ position\ 1}),(\mathrm{if\ }k-1 > 0\mathrm{,\ position\ of\ }k-1\mathrm{th\ 1\ bit\ in\ }2j - 1\mathrm{,\ with\ the\ LSB\ counted\ as\ position\ 1,\ otherwise\ 1})} \).
I just came up with a symbolic form for this formula. It may not be the best (the "nth one bit's index" translations may not be the most "elegant" possible), but at least it gives the result.
\( a_n = \sum_{j=1}^{2^{n-2}} \prod_{k=1}^{\left(2j -\ \sum_{i=1}^{\lfloor\log_2(2j - 1)\rfloor} \lfloor\frac{2j - 1}{2^i}\rfloor\right)} r_{\left(1 +\ \sum_{i=0}^{\lfloor\log_2(2^{n-1} + 2j - 1)\rfloor}\mathbf{1}_{\{\nu \in \mathbb{N}_0: \nu < k\}}\left(\sum_{s=0}^{i} \frac{1 - (-1)^{\lfloor\frac{2^{n-1} + 2j - 1}{2^s}\rfloor}}{2}\right)\right),\left(1 +\ \sum_{i=0}^{\lfloor\log_2(2j - 1)\rfloor}\mathbf{1}_{\{\nu \in \mathbb{N}_0: \nu < k-1\}}\left(\sum_{s=0}^{i} \frac{1 - (-1)^{\lfloor\frac{2j - 1}{2^s}\rfloor}}{2}\right)\right)} \).
Letting \( r_{n, m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} \left{{n \atop m}\right} \), we now have \( \chi_n = a_n \), thus an explicit, non-recursive formula for the coefficients of the regular Schroder function of the decremented exponential.
However, as can be seen, this formula looks to be far too complicated and too general to be of any use in and of itself. However, it shows (or at least it will once the rigorous proof is complete!) that an explicit formula
exists. The big question now is, is there a simpler one? Note that this can be used to derive an explicit formula for the Lagrange inversion formula, thus we can plug all that together to get the fully-explicit, non-recursive formula for the coefficients of the regular inverse Schroder function of the decremented exponential, and so a fully-explicit, non-recursive formula for the coefficients of the Fourier series regular superfunctions of the exponential. But plugging together all those substitutions in this form is just a horrid nightmare. Yet if this monster formula exists, then it would seem likely that a simpler one does, too, considering how general this is.
I got this by plugging in the following formulas:
\( \mathrm{number\ of\ 1\ bits\ in\ }N = N -\ \sum_{i=1}^{\lfloor\log_2(N)\rfloor} \lfloor\frac{N}{2^i}\rfloor \)
\( \mathrm{position\ of\ }k\mathrm{th\ 1\ bit\ in\ }N\mathrm{,\ with\ the\ LSB\ counted\ as\ position\ 1} =\ \sum_{i=0}^{\lfloor\log_2(N)\rfloor} \mathbf{1}_{\{\nu \in \mathbb{N}_0 : \nu < k}}\left(\sum_{s=0}^{i} \frac{1 - (-1)^{\lfloor\frac{N}{2^s}\rfloor}}{2}\right) \).
Also, note that the sth bit of N, with s = 0 being the LSB, is \( \frac{1 - (-1)^{\lfloor\frac{N}{2^s}\rfloor}}{2} \).
