I want to obtain a partition of an Integer with an initial value and the value following it is smaller and the value following it is smaller than the previous value and no value repeats itself. within a limited number of terms For example Integer = 41 Intial Value = 11 number of terms = 8
let's say I am going to generate such partitions for number 42, how to do that?
I am not sure whether a super direct approach would exist, these problems with partitions and restrictions are usually complicated to compute since closed formulas are not easy to find and a lot relies on recurrences and bijections between different kind of partitions.
I guess the easiest way to attack what you say is via generating functions: since you don't only want terms to be different, but also have a condition on the total number of terms used, you will need two variables. One, say $x$, will count the actual contribution of the term, while $z$ will count how many terms this contributes. For example, a term $z^2x^4$ would be saying a contribution of $4$ with two terms, i.e. a $2 + 2$ or a $3 + 1$, although we don't know which contributed this term in the final product.
If your biggest term is $N$, then that is a $zx^N$. After that you allow any term to appear, but maybe it won't, so each possible term has a generating function $1 + zx^j$, for $1\le j \le N - 1$. So, the generating function for the partitions whose largest term is $N$, the remaining ones do not repeat and in which you count how much contributions you have is: $$zx^N(1 + zx^{N - 1})...(1 + zx).$$ To make an example, if $N = 5$ what you have to see is the product $$\begin{align*} &\;zx^5(1 + zx^4)(1 + zx^3)(1 + zx^2)(1 + zx)\\ &= zx^5(1 + zx + zx^2 + (z + z^2)x^3 + (z + z^2)x^4 + 2z^2x^5 + (z^2 + z^3)x^6 + (z^2 + z^3)x^7 + z^3x^8 + z^3x^9 + z^4x^{10})\\ &=zx^5 + z^2x^6 + z^2x^7 + (z^2 + z^3)x^8 + (z^2 + z^3)x^9 + 2z^3x^{10} + (z^3 + z^4)x^{11} + (z^3 + z^4)x^{12} + z^4x^{13} + z^4x^{14} + z^5x^{15} \end{align*} $$
For example, this would be telling us that for $10$ there are only $2$ ways to make this, each of them uses $2$ terms (that is the $z^2$), and sure enough these are $5 + 4 + 1$ and $5 + 3 + 2$, while for $9$ this would be saying that there is one way with $2$ terms and one with $3$ terms: $5 + 4$ and $5 + 3 + 1$.
These polynomials get complicated to manipulate as your parameters grow, nevertheless if you want to bound the number of terms, then you want to disregard any terms whose power of $z$ is bigger than some number. For example, if you did not want to count partitions with more than $3$ terms you take quotient modulo $z^4$ and the polynomial becomes: $$ zx^5 + z^2x^6 + z^2x^7 + (z^2 + z^3)x^8 + (z^2 + z^3)x^9 + 2z^3x^{10} + z^3x^{11} + z^3x^{12} $$ Finally, if what you want at this point is just to see how many there are of a given integer, without knowing how many terms they have since you already took care of your bound then you put $z = 1$ and you get: $$ x^5 + x^6 + x^7 + 2x^8 + 2x^9 + 2x^{10} + x^{11} + x^{12} $$ This, for example, would tell you: there are $2$ partitions of $10$ in which:
These would be $5 + 4 + 1$ and $5 + 3 + 2$. For eleven, there would be only one: $5 + 4 + 2$.
For $N = 42$ you can do this, tho you would probably need a computer that makes the calculations for you, and you need to do everything in the correct modulus from the start to make sure computations remain doable.
