So, we want the vector \( \\[15pt]
{[a_i]} \), from the matrix equation:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [a_i \right ]
=
\left [ \sum_{n=1}^{P(i)} \frac{a.ln(a)^{\sum_{j=1}^{i}c_{n,j}}}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}a_j^{c_{n,j}} \right ]} \)
where "r" is the row index of the first matrix at left, and "i" his column index.
Note that in the last equation, both r and i start counting from zero for the first row and column.
______________________________________
P(i) is the partition function
The first few values of the partition function are (starting with p(0)=1):
1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, … (sequence A000041 in OEIS; the link has valuable information about the partition function).
______________________________________
\( \\[15pt]
{c_{n,j}} \) is the number of repetitions of the integer j in the \( \\[15pt]
{n^{th}} \) partition of the number i
______________________________________
Solving the equation
If we do the substitution \( \\[25pt]
{a_i=\frac {b_i} {ln(a)} } \), we simplify the first equation to:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} \)
______________________________________
Special base.
This equation suggest a special number, which is m=1.7632228343518967102252017769517070804...
m is defined by \( \\[15pt]
{^2m=e} \)
For the base a=m, the equation gets simplified to:
\( \\[35pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
=
\left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} \)
But let's forget about m for now.
______________________________________
We are now very close to the solution. The only obstacle remaining is the product:
\( \\[15pt]
{ \frac{1}{ \prod_{j=1}^{i} c_{n,j}!} } \)
If we can do a substitution that get us rid of him, we have the solution:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} \)
At this point we only need to substitute \( \\[25pt]
{b_i=f^i} \), where f is arbitrary, to get:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [f^i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)} f^i \right ] \,=\, ln(^2a) . [P(i) . f^i]} \)
... and we get:
\( \\[15pt]
{ \left [f^i \right ]
= \left [ {{i} \choose {r}} \right ]^{-1} \cdot [ ln(^2a) . P(i) . f^i]} \)
The choice of f, very probably, determines the value for °a, and the branch of tetration.
{[a_i]} \), from the matrix equation:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [a_i \right ]
=
\left [ \sum_{n=1}^{P(i)} \frac{a.ln(a)^{\sum_{j=1}^{i}c_{n,j}}}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}a_j^{c_{n,j}} \right ]} \)
where "r" is the row index of the first matrix at left, and "i" his column index.
Note that in the last equation, both r and i start counting from zero for the first row and column.
______________________________________
P(i) is the partition function
The first few values of the partition function are (starting with p(0)=1):
1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, … (sequence A000041 in OEIS; the link has valuable information about the partition function).
______________________________________
\( \\[15pt]
{c_{n,j}} \) is the number of repetitions of the integer j in the \( \\[15pt]
{n^{th}} \) partition of the number i
______________________________________
Solving the equation
If we do the substitution \( \\[25pt]
{a_i=\frac {b_i} {ln(a)} } \), we simplify the first equation to:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} \)
______________________________________
Special base.
This equation suggest a special number, which is m=1.7632228343518967102252017769517070804...
m is defined by \( \\[15pt]
{^2m=e} \)
For the base a=m, the equation gets simplified to:
\( \\[35pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
=
\left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} \)
But let's forget about m for now.
______________________________________
We are now very close to the solution. The only obstacle remaining is the product:
\( \\[15pt]
{ \frac{1}{ \prod_{j=1}^{i} c_{n,j}!} } \)
If we can do a substitution that get us rid of him, we have the solution:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} \)
At this point we only need to substitute \( \\[25pt]
{b_i=f^i} \), where f is arbitrary, to get:
\( \\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [f^i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)} f^i \right ] \,=\, ln(^2a) . [P(i) . f^i]} \)
... and we get:
\( \\[15pt]
{ \left [f^i \right ]
= \left [ {{i} \choose {r}} \right ]^{-1} \cdot [ ln(^2a) . P(i) . f^i]} \)
The choice of f, very probably, determines the value for °a, and the branch of tetration.
(01/03/2016, 11:24 PM)marraco Wrote:
![[Image: lynxfBI.jpg?1]](http://i.imgur.com/lynxfBI.jpg?1)
I have the result, but I do not yet know how to get it.
