Given a positive integer $n$, let $p$ denote the largest prime $p$ less than $2^n$.

That is, $p$ equals

Max[Table[Prime[i], {i, 1, PrimePi[2^n]}]]

I want to form a sum by adding some terms for each $1\le m<2^n$.

Given the prime factorization $m=2^{a_2}\cdots p^{a_p}$, consider $k/m$ for primes $k$ with $a_k>0$.

For each $k$, form the sum $k/m+k/(2m)+k/(2^2m)+\cdots$ such that in reduced form the highest power of 2 in the denominator is $2^{n-2}$.

Specifically, if $k=2,$ take the sum

Sum[2/(2^i *m), {i,0,n-a_2-1}]

and if $k\not =2$, then take the sum

Sum[k/(2^i *m), {i,0, n-a_2-2}]

Then I want to add all of these new sums together (over the $k$'s with $a_k>0)$, divide the result by 2, and then take the minimum with $1/2^{n-1}$. This is the total contribution I want to assign to $m$, call it $Y_m$. What would be a code for $\sum_{1\le m<2^n}Y_m$?

Illustrative Example

If $n=5$ and $m=12=2^2*3$. Since $a_2,a_3>0$, I have $\color\red{k=2},\color\green{k=3}.$

For $k=2,$ I have $2/12=2/(2^23), $ so in the sum I include $\color\red{2/(2^23)+2/(2^33)+2/(2^43)}$

For $k=3,$ I have $3/12=3/(2^23)$, so in the sum I include $\color\green{3/2^2+3/2^33}$

Therefore, the contribution I assign to $m=12$ is $$Y_{12}=\min\bigg((1/2)[\color\red{(2/(2^23)+2/(2^33)+1/(2^43))}+\color\green{(3/(2^23)+3/(2^33))}],\frac{1}{2^{5-1}}\bigg).$$


1 Answer 1

Table[ms = Range[2, 2^n - 1];
  Total[MapThread[With[{Q = FirstCase[#, {2, q_} -> q, 0]},
    Min[#3, Total[Outer[Divide, #[[All, 1]]/#2, 2^Range[0, n - 2 - Q]],
     {1, 2}]/2 + If[OddQ[#2], 0, 2^(Q - n + 1)/#2]]] &,
       {FactorInteger[ms], ms, ConstantArray[2^(1 - n), 2^n - 2]}]], {n, 9}]


FindSequenceFunction[%, n] // FullSimplify


  • $\begingroup$ I don't see how you are taking the minimum with $1/2^{n-1}$ for each $m$. $\endgroup$ Jul 13, 2017 at 17:30
  • $\begingroup$ @TheSubstitute Min is called within MapThread as many times as the Length of ms $\endgroup$
    – Coolwater
    Jul 13, 2017 at 17:35
  • 1
    $\begingroup$ 2n-2 is how many values of $m$ to be summed over, excluding $m=1$ because 1 = "no factors" -> min(0, 2^(1-n)) -> 0. For other p it will be p^n - 2 $\endgroup$
    – Coolwater
    Jul 17, 2017 at 5:34

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