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There is a list with 19 elements a={0.500074`, 0.502676`, 0.510451`, 0.507502`, 0.52768`, 0.625164`, 0.935248`, 1.826161`, 3.845598`, 8.293859`, 18.014295`, 39.20736`, 85.3836`, 185.975252`, 405.095196`, 882.399269`, 1922.095421`, 4186.829292`, 9120.018288`}

The elements in the list are written according to a certain rule that I don't know. I would like to expand this list in both directions like here anew={... a, b, c, 0.500074`, 0.502676`, 0.510451`, 0.507502`, 0.52768`, 0.625164`, 0.935248`, 1.826161`, 3.845598`, 8.293859`, 18.014295`, 39.20736`, 85.3836`, 185.975252`, 405.095196`, 882.399269`, 1922.095421`, 4186.829292`, 9120.018288`, d, e, f ...} where ...a, b, c and d, e, f... are elements which fit the general rule.

How to find this rule and write down additional elements (or just find the most suitable elements)?

All elements should be a positive

ClearAll["Global`*"]
a={0.500074`, 0.502676`, 0.510451`, 0.507502`, 0.52768`, 0.625164`,    0.935248`, 1.826161`, 3.845598`, 8.293859`, 18.014295`, 39.20736`,    85.3836`, 185.975252`, 405.095196`, 882.399269`, 1922.095421`,    4186.829292`, 9120.018288`};
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1 Answer 1

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Look at the ratio of 2 consecutive elements:

a = {0.026284, 0.057253, 0.124713, 0.271658, 0.591744, 1.288974, 
   2.807725, 6.115966, 13.322188, 29.019243, 63.211573, 137.691494, 
   299.928425, 653.323290, 1423.110603, 3099.910594, 6752.423650};

Divide @@@ Partition[a, 2, 1]

{0.459085, 0.459078, 0.459081, 0.45908, 0.459081, 0.459081, 0.459081, \
0.459081, 0.459081, 0.459081, 0.459081, 0.459081, 0.459081, 0.459081, \
0.459081, 0.459081}

Therefore, each element is the predecessor times 1/0.459081= 2.17826. The continuation of 4 elelemnts to the left is, starting with a[1]]:

Reverse@NestList[0.4590811717341696` # &, a[[1]], 5]

{0.000535968, 0.00116748, 0.00254308, 0.0055395, 0.0120665, 0.026284}

and the continuation to the right starting with a[[-1]]:

NestList[1/0.4590811717341696` # &, a[[-1]], 5]

{6752.42, 14708.6, 32039.1, 69789.7, 152020., 331140.}
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  • $\begingroup$ Thanks! I'm, so sorry I gave a not very convenient sequence in the example. A better example is the following sequence: a = {0.500074`, 0.502676`, 0.510451`, 0.507502`, 0.52768`, 0.625164`, 0.935248`, 1.826161`, 3.845598`, 8.293859`, 18.014295`, 39.20736`, 85.3836`, 185.975252`, 405.095196`, 882.399269`, 1922.095421`, 4186.829292`, 9120.018288`} $\endgroup$
    – Mam Mam
    Commented May 5, 2023 at 14:01
  • $\begingroup$ Now if we use your construction Divide @@@ Partition[a, 2, 1] will be returned next list of coefficients: {0.994824, 0.984768, 1.00581, 0.961761, 0.844067, 0.668447, 0.512139,0.47487, 0.463668, 0.460404, 0.459462, 0.459191, 0.459113, 0.45909, 0.459084, 0.459082, 0.459081, 0.459081}, which are not the same. In the description of the task I have just changed of the list. I have many similar lists, each of which I need to correctly expand. $\endgroup$
    – Mam Mam
    Commented May 5, 2023 at 14:02
  • $\begingroup$ Is the creation rule for every list different? It is impossible to give a general answer if the rule changes from list to list.. It is better, if you explain your problem, so that a rule may be guessed. $\endgroup$
    – Daniel Huber
    Commented May 5, 2023 at 14:13
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    $\begingroup$ Interpolation fits a polynomial between succesive points and eventually extrapolates these plynomials at the beginning and end. Therefore "Interpolation" does not know about any rules. What you could try is to fit a polynomial to the first/last n points and extrapolate these and check how these behave. $\endgroup$
    – Daniel Huber
    Commented May 5, 2023 at 14:52
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    $\begingroup$ I tried it, but polynomials do not work well. But make a "ListLogPlot" that may tell you how to extrapolate. $\endgroup$
    – Daniel Huber
    Commented May 5, 2023 at 15:49

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