Meh, it looks like the whole approach of trying to directly find the "magic" numbers is insoluble. So it seems we're left with trying to simplify/"elegantize" the "ugly" binary formula in some other way.
Anyway, I noticed that that "binary" formula can be written simply as a sum over subsets of an interval of naturals (note that binary = indicator functions), giving a neater, more compact form:
\( a_1 = r_{1,1} \)
\( a_n = \sum_{m=1}^{n-1} r_{n,m} a_m \).
Then,
\( a_n = r_{1, 1} \sum_{1 = m_1 < m_2 < \cdots < m_k = n\\2 \le k \le n}\ \prod_{j=2}^k r_{m_j,m_{j-1}}, n > 1 \).
I didn't try a form like this before since I didn't consider it "explicit" enough for my liking
(I'm a fan of sums over straight indices, and I was hoping the straight-index form would help, but alas, no dice.) But this lets us write the Schroder coefficients as
\( \chi_n =\ \sum_{1 = m_1 < m_2 < \cdots < m_k = n\\2 \le k \le n}\ \prod_{j=2}^k \frac{u^{m_j-1}}{1 - u_{m_j-1}} \frac{m_{j-1}!}{m_j!} \left{{m_j \atop m_{j-1}}\right}, n > 1 \).
We can eliminate the factorial quotient from the product since the product multiplies it to \( \frac{m_{k-1}! m_{k-2}! ... m_1!}{m_k! m_{k-1}! ... m_2!} = \frac{m_1!}{m_k!} = \frac{1}{n!} \), giving
\( \chi_n = \frac{1}{n!}\ \sum_{1 = m_1 < m_2 < \cdots < m_k = n\\2 \le k \le n}\ \prod_{j=2}^k \frac{u^{m_j-1}}{1 - u^{m_j-1}} \left{{m_j \atop m_{j-1}}\right}, n > 1 \).
A "simpler" formula for this would then mean one that sums over fewer terms and also has straight indices and no binary/subset/etc. stuff. This sums over \( 2^{n-2} \) terms. In theory, it should be possible to get it to \( \frac{n(n-1)}{2} \). How can we do this? The above is very general -- we can express Bernoulli numbers and what not in a similar form. But in that case, and in a lot of other cases, there exists a form with fewer terms, straight indices and no binary/subset/etc. stuff -- just some nested sums and products. Is that possible here as well? Is this formula already known somewhere?
Anyway, I noticed that that "binary" formula can be written simply as a sum over subsets of an interval of naturals (note that binary = indicator functions), giving a neater, more compact form:
\( a_1 = r_{1,1} \)
\( a_n = \sum_{m=1}^{n-1} r_{n,m} a_m \).
Then,
\( a_n = r_{1, 1} \sum_{1 = m_1 < m_2 < \cdots < m_k = n\\2 \le k \le n}\ \prod_{j=2}^k r_{m_j,m_{j-1}}, n > 1 \).
I didn't try a form like this before since I didn't consider it "explicit" enough for my liking
(I'm a fan of sums over straight indices, and I was hoping the straight-index form would help, but alas, no dice.) But this lets us write the Schroder coefficients as\( \chi_n =\ \sum_{1 = m_1 < m_2 < \cdots < m_k = n\\2 \le k \le n}\ \prod_{j=2}^k \frac{u^{m_j-1}}{1 - u_{m_j-1}} \frac{m_{j-1}!}{m_j!} \left{{m_j \atop m_{j-1}}\right}, n > 1 \).
We can eliminate the factorial quotient from the product since the product multiplies it to \( \frac{m_{k-1}! m_{k-2}! ... m_1!}{m_k! m_{k-1}! ... m_2!} = \frac{m_1!}{m_k!} = \frac{1}{n!} \), giving
\( \chi_n = \frac{1}{n!}\ \sum_{1 = m_1 < m_2 < \cdots < m_k = n\\2 \le k \le n}\ \prod_{j=2}^k \frac{u^{m_j-1}}{1 - u^{m_j-1}} \left{{m_j \atop m_{j-1}}\right}, n > 1 \).
A "simpler" formula for this would then mean one that sums over fewer terms and also has straight indices and no binary/subset/etc. stuff. This sums over \( 2^{n-2} \) terms. In theory, it should be possible to get it to \( \frac{n(n-1)}{2} \). How can we do this? The above is very general -- we can express Bernoulli numbers and what not in a similar form. But in that case, and in a lot of other cases, there exists a form with fewer terms, straight indices and no binary/subset/etc. stuff -- just some nested sums and products. Is that possible here as well? Is this formula already known somewhere?
