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I want to nest each element inside the previous at the mid point. How can I do this by indexing a very large list?

   {5/24,mid,19/24}/.mid->
   {7/24,mid,17/24}/.mid->
   {11/24,mid,13/24}

({5/24, {7/24, {11/24, mid, 13/24}, 17/24}, 19/24}) Yes, it's Goldbach that creates the lists. The sample is 5+19, 7+17 and 11+13 with the sum, 24, as the denominator. After the nest: we notice 5,7 on the left and 17,19 on the right. We put the sum of 17,19 divided by 2 (the mid point) into our solver to create the next chain link. I haven't shown any of that code, because the last time I did, I found out the ones who viewed were math phobic.

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    $\begingroup$ I don't understand the sequence 5, 7, 11,... in the numerators. Ditto for the upper bounds. $\endgroup$
    – Michael E2
    Apr 5 at 1:21
  • $\begingroup$ They are primes, since I am experimenting with the twin prime conjecture. $\endgroup$
    – Fred Daniel Kline
    Apr 5 at 1:30
  • $\begingroup$ What I meant was that you reference a very large list. Since the only lists in the question are of the form {a, mid, b}, it's unclear where the a's and the b's come from or how the large list is to be modified. $\endgroup$
    – Michael E2
    Apr 5 at 1:32
  • $\begingroup$ the triplet is just one of many. The a''s and b's are called p and q in my twin prime solver. Observation: only the last item in a twin, the others are twins with low prime on the left and the high prime on the rigjt---after the nesting. We rely on symmetry. $\endgroup$
    – Fred Daniel Kline
    Apr 5 at 1:52
  • $\begingroup$ I had to Reverse@list to get one working; otherwise it was tough choice. I did remove mid and the denominators, Thanks. $\endgroup$
    – Fred Daniel Kline
    Apr 5 at 11:27

2 Answers 2

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alist = {{5/24, 19/24}, {7/24, 17/24}, {11/24, 13/24}};

Fold[Riffle[#2, {#1}] &, Reverse@alist]

$$\left\{\frac{5}{24},\left\{\frac{7}{24},\left\{\frac{11}{24},\frac{13}{24}\right\},\frac{17}{24}\right\},\frac{19}{24}\right\}$$

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  • $\begingroup$ +1, I'll do some timings on large lists before accepting either answer. $\endgroup$
    – Fred Daniel Kline
    Apr 5 at 1:34
  • $\begingroup$ To see how it folds step by step: FoldList[Riffle[#2, {#1}] &, Reverse@alist] // Grid $\endgroup$
    – Syed
    Apr 5 at 1:36
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I'm assuming that mid is just a placeholder. So, you could do something like this:

list = {{11, 13}, {7, 17}, {5, 19}};
Fold[Insert[#2, #1, 2] &, list]

{5, {7, {11, 13}, 17}, 19}

(I've removed the denominators just for clarity.)

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4
  • $\begingroup$ +1, Can we put a list name on the right? $\endgroup$
    – Fred Daniel Kline
    Apr 5 at 1:25
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    $\begingroup$ yep. see my update. $\endgroup$
    – lericr
    Apr 5 at 1:26
  • $\begingroup$ hmm. looking at your comment, were you wanting to start with {5,7,11,13,17,19} and insert the brackets, or do you have the pairs in a list already like I assumed in my answer? $\endgroup$
    – lericr
    Apr 5 at 1:32
  • $\begingroup$ I start with the right-most twin prime and build the rest. I'm interested in building a chain of twin primes. $\endgroup$
    – Fred Daniel Kline
    Apr 5 at 1:40

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