I recently started using mathematica for simple tasks like simplifying etc.

Table of values

First of all let me say if this breaks any rules I apologize since I'm a novice in using this and I'll remove the question.

I want to use the table of values above to find a GENERAL FORMULA that represents the values.

What I want to achieve is a function in terms of '$V$' such that for any value of '$V$' I enter I will get the corresponding '$SS$' value.

In simple terms I'm looking for an equation such that when '$V$' is equal to '6' say, then I can input this into an equation which would probably look something like $D_{5}$ [Since it's the sum of the first 5 terms]= equation in terms of $V$ = $SS$ value.

But I need this equation to hold for an value of $V$ in the entire table.

The formula can also use the $E$ value I just thought it would be easier to study with respect to $V$.

I know there's the FindFormula operator but i'm not sure how to incoperate this with the table to solve such a problem.

  • 1
    $\begingroup$ The first step would be plotting the data. But with only 4 data points there's not much you can do for which you could be confident in future predictions. You might consider the following: v = {6, 9, 12, 15}; ss = {29, 84, 178, 320}; ListLogLogPlot[Transpose[{v, ss}]] followed by lm = LinearModelFit[Log[Transpose[{v, ss}]], x, x]. $\endgroup$
    – JimB
    Commented Sep 11, 2020 at 3:07
  • $\begingroup$ What is the meanings of rows and colums in your table? $\endgroup$ Commented Sep 11, 2020 at 4:31
  • $\begingroup$ @JimB If I were to calculate more $SS$ values(Which I can) Is it possible to get a better more suited general formula? And how would I do such. Thanks $\endgroup$ Commented Sep 11, 2020 at 17:00
  • $\begingroup$ @MustafaKösem The Rows and Columns correspond to a vertex and edge of a random graph in which I used a method to find the SS values. I am currently doing some research in different nanostar calculations. $\endgroup$ Commented Sep 11, 2020 at 17:00

1 Answer 1

f[n_]:=2*(3n+6-1)(n+2)+Total[9Table[Binomial[n,j]+n,{j,0,n}] ]


Also here is the hole Table:

Table[Flatten[Riffle[Table[{3n+6-1,3n+6-1},n+2],9Table[Binomial[n,j]+n,{j,0,n} ]]],{n,0,3}]           


  • $\begingroup$ Can you please elaborate on how you arrived at 2*(3n+6-1)(n+2)? This seems to be the crux of the formula. I'm inquiring because if i calculate different nanostar graphs my table rows and columns will change and the number of values etc. Is there a blueprint that I can use to change accordingly? Thank you. $\endgroup$ Commented Sep 11, 2020 at 17:05
  • 1
    $\begingroup$ That formula is the easy part... it just calculates the n-1 values for the initial values of V={6,9,12,15...} but it works for 0,1,2,3.... What values exactly do you need to know? The formula works for any n, and so does the table. $\endgroup$
    – ZaMoC
    Commented Sep 11, 2020 at 17:28
  • 1
    $\begingroup$ @JasonMoore the above formula generates the SS values that you want {29,84,178,320,528,838,1322,2124,3532,6122,11046....}. It doesn't really matter if you use {0,1,2,3,....) or 6+3n for n={0,1,2,3} (which is the V column) or 5+3n (which is the E column). The tricky part is to calculate the values {...,36,...,54,.....,54,,,,36} which come from Binomial[n,k]+n $\endgroup$
    – ZaMoC
    Commented Sep 11, 2020 at 18:16
  • $\begingroup$ Can you indulge me a bit further if possible. What changes would I have to make to this table [imgur.com/FodLUQF] to have the same type of outputs. Yes , the values are indeed different , however the $V$ and $E$ columns follow the same $n-1$ pattern as before but the above formula will not hold for this table. Can you please advise me on what tools specifically that I would have to master , or look at , to be able to manipulate the code to any such table? Thank you @J42161217 $\endgroup$ Commented Sep 15, 2020 at 20:26
  • $\begingroup$ [Imgur]imgur.com/gallery/ffNlsdQ Backup link of image incase it's taken down. $\endgroup$ Commented Sep 15, 2020 at 20:52

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