Collection of Ideas

For a general partial fraction decomposition \[\frac{P(x)}{\prod_{i=1}^{m}q_i^{n_i}(x)} = \sum_{j=1}^m \sum_{i=\sum_{k=0}^{j-1}n_k}^{\sum_{k=1}^j n_k} \frac{A_i}{q_j^{i-\sum_{k=1}^{j-1}n_k}(x)}\]

where \(n_0\) is defined as \(1\).

Try \[F_{ij}^S= \sum _{k=j-i}^{j+i} S_k\]

Where \(S_k\) is the \(k^{th}\) element of some series starting from \(S_1\), and \(j>i\). Will result in a strictly upper triangular matrix. For example for the series \(S=1,2,3,4,5...\) we will have the matrix \[\begin{bmatrix} - & 6 & 9 & 12 \\ - & - & 15 & 20 \\ - & - & - & 28 \\ - & - & - & - \\ \end{bmatrix}\]

However, we may replace \(j-i\) and \(j+i\) with some other functions of \(i,j\) even the function itself in the right conditions. Try

\[F_{ij}^S= \sum _{k=(j-i)^2}^{j^2 -i^2} S_k\]

This results in

\[\begin{bmatrix} - & 3 & 22 & 69 & 156 \\ - & - & 10 & 60 & 174 \\ - & - & - & 21 & 114 \\ - & - & - & - & 36 \\ \end{bmatrix}\]

Of course we can now sum over the column and generate a new series \(N\), with the element \(1,2\) as \(N_1\). For example the new series generated by the first example is \(6, 24, 60, ...\) and the second example is \(3,32,150,480...\).

In general the new series \(N\) is generated by \[N= \sum_i F_{ij}^S= \sum_i \sum _{k=f(i,j)}^{g(i,j)} S_k\] where for all \(i,j\), \(g(i,j)>f(i,j)\) and of course \(f(i,j),g(i,j),i,j \in \mathbb{Z}\).

Of course the operation that forms a reduction of series \(S\) to \(F\) does not have to be a sum, it may be any operation which is appropriate. A question is then, can such a device be used to generate important series from other series, to porvide a direct mapping between say the numbers \(1,2,3,4,5...\) to the prime numbers \(2,3,5,7,11...\)?

To find such a mapping we know, if the column wer summed over, we are allowed a number of terms to form the final series element as the column adress minus one. In short, the first element of the series is one element from the matrix, the second the sum of two and so on. The ways to sum \(n\) elements into a number much greater than \(n\) is a problem, as there are many combinations, and perhaps only one is valid. However, any such direct mapping would have to be valid for small numbers as well as large, so it would be best to exhaust the possibilites for smaller elements and extrapolate the series upwards.

So a short program was written to analyse these sums further. Here are some sequences...

\[\begin{array}{|c|c|c|c|c|} \hline Operation & S &f(i,j) & g(i,j) & N \\ \hline + & \mathbb{Z}^+ & j-i & j+i & 6,24,60,120,210,336,504,720,990,1320,... \\ +,-,+,... & \mathbb{Z}^+ & j-i & j+i & 2,0,4,0,6,0,8,0,10,0,12,0,... \\ + & S_p^* & " & " & 1,2,3,5,7,11,13,17,19,23,29,35,41... \\ \hline \end{array}\]

The first sequence seems to contain many elementes which are in the factorial sequence \(1,1,6,24,120,720\) but extrapolating this stops around \(5040\).

The third sequence was chosen term by term to map the prime numbers well. However the sequence stops following on when higher order terms are not added. This is good, one can generate a sequence to compactify the prime numbers into a sequence of smaller integers. The sequence used to generate the digits seen was \(0,0,1,0,0,0,0,1,0,0,0,0,1,1,-1,-1,1,2,-2,0,1,0,1,1,0,0,0,0,0,0\), and was slowly built up to make it fit the desired result, there may be nothing meaningfull about it. However one can look for patterns still, if such a sequence was in anyway predictable it could be very benificial.

Upon extending the sequence it was made such that primes up to and including \(113\) could be found using the sequence \(S_p^*=0,0,1,0,0,0,0,1,0,0,0,0,1,1,-1,-1,1,2,-2,0,1,0,1,1,-2,0,1,2,-2,0,-1,0,1, \\ 1,0,0,1,0,-2,-1,2,1,0,-1,0,0,1,1,0,-2,1,1,2,-1,-2,0,0,0,0,-1,0,-1,1,1.\)

It appears that a term with a given index in the sequence will only affect the prediction of primes around that given index divided by \(2\) in the prime sequence. This provides reasonable flexibility and ease in predicting the next terms in the sequence when given the exisiting sequence of prime numbers. Now to look for patterns.

Adjusting the first term in this sequence seems to have no change suggesting it isn’t used by the function. adjusting a lower term in the starting sequence affects all terms beyond it in the new sequence thus making it incorrect very quickly. However, adjusting a higher term in the starting sequence doesn’t seem to affect the lower terms of the new sequence.

It is not know if it is necessary to have the \(1\) at the beginning of the sequence of primes. That is just how it was engineered in this circumstance.

Tp experiment with the effect of adding arbitrary 1’s and 0’s to the sequence one can start with a black sequence, \(S=0,0,0,0,0,0,0\) and add a \(1\) at a location. We denote the end sequence as \(N\).

\[\begin{array}{|c|c|c|} \hline i & S & N \\ \hline 1 & 0,0,0,0,0,0... & 0,0,0,0,0,0... \\ 2 & 1,0,0,0,0,0... & 0,0,0,0,0,0... \\ 3 & 0,1,0,0,0,0... & 1,1,1,1,1,1... \\ 4 & 0,0,1,0,0,0... & 1,2,2,2,2,2... \\ 5 & 0,1,1,0,0,0... & 2,3,3,3,3,3... \\ 6 & 0,0,0,1,0,0... & 1,2,3,3,3,3... \\ 7 & 0,0,0,0,1,0... & 0,2,3,4,4,4... \\ 8 & 0,0,1,0,1,0... & 1,4,5,6,6,6... \\ 9 & 0,0,0,0,0,1... & 0,1,3,4,5,5... \\ \hline \end{array}\]

We can see from the table that the sequences generated from adding a single term in \(S\) are summed as a result in \(N\) when multiple terms are added at once. The goal would be to write a sequence that doesn’t need negative values, this would be simpler in nature and perhaps more fundamental. It is strange how example \(7\) doesn’t progress through \(1\) when rising, also example \(9\) doesn’t progress through \(2\)

Thanks to Simon for suggesting/inspiring this idea. Attempting to write an input series to output the Fibonacci Series. Also, input the Fibonacci series and check the output. (coming soon)

When input the generated sequence is : \(4,18,62,188,536,1476,3984,10626,28138...\)

By changing \(f\) and \(g\) we change the transfomation that occurs on \(S\) to create \(N\). A closed form was found, let \(f(i,j)=i\) and \(g(i,j)=j\). The input sequence \(S_k=k\). Up to a few elements the out put series was \(N=3,11,26,50,85,...\). Wolfram-Alpha predicted this was a continuation on \[a_n=\frac{1}{6}(2n^3+9n^2+7n)\]

This is interesting. The transform represents a

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