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I am working on a square matrix. Cell (1,1) starts at 1, and each following 2, then 3, .., working x to y order. Plus, each cell is computed after this, to Sum of Squares minus Sum of each cell. Sorry, a little unclear.

$ \sum\limits_{n = 1}^{n = x} {{x^2} - \sum\limits_{n = 1}^{n = x} x } $

Or course, the second Sum are the triangular numbers. (1) So to repeat. (2) A 10x10 matrix, built from consecutive integers. (3) Then, each cell, re-evaluated to the formula above.

Array[10 (#1 - 1) + #2 &, {10, 10}]
Partition[%, 10]
(* Out {{{1,2,3,4,5,6,7,8,9,10},{11,12,13,14,15,16,17,18,19,20},
{21,22,23,24,25,26,27,28,29,30},{31,32,33,34,35,36,37,38,39,40},
{41,42,43,44,45,46,47,48,49,50},{51,52,53,54,55,56,57,58,59,60},
{61,62,63,64,65,66,67,68,69,70},{71,72,73,74,75,76,77,78,79,80},
{81,82,83,84,85,86,87,88,89,90},{91,92,93,94,95,96,97,98,99,100}}}
 *)
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  • $\begingroup$ ... and your question is? I assume that you are asking how to build your matrix? Can you show the matrix you would like for a certain value of x or whichever parameter you use? $\endgroup$
    – MarcoB
    May 26, 2020 at 1:22
  • $\begingroup$ @MarcoB You're right. Consider the above a 10x10 matrix, not a 100 element vector. Values 1, 2, 3, 4, 5; become 0,2,8,20,40. ie. For 5; becomes 55-15=40. $\endgroup$
    – prog9910
    May 26, 2020 at 1:27

1 Answer 1

Reset to default
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ClearAll[matrix]
matrix[n_Integer] :=
 Partition[
   Accumulate[Range[n^2]^2 - Range[n^2]],
   n
 ]

matrix[10] // MatrixForm

picture from matrix form output

Note that you can directly obtain an explicit expression for each element:

Simplify[ Sum[n^2, {n, 1, n}] - Sum[n, {n, 1, n}] ]

(* Out: 1/3 n (n^2 -1) *)

Compare it with the definition I gave above to check that they are the same numbers:

Table[1/3 n (n^2 -1), {n, 1, 100}] == Flatten[matrix[10]]
(* Out: True *)

In this case, FindSequenceFunction also works:

FindSequenceFunction[Flatten@matrix[10], n] // FullSimplify
(* Out: 1/3 n (n^2 - 1) *)
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  • $\begingroup$ Oh, you can use Partition[ like that. Interesting. Everything is a function. Very good, that is what I needed. $\endgroup$
    – prog9910
    May 26, 2020 at 2:00
  • $\begingroup$ I'm thinking of Pascals Matrix too. Something new to study. mathematica.stackexchange.com/questions/100290/… $\endgroup$
    – prog9910
    May 26, 2020 at 2:06
  • $\begingroup$ Does FindSequenceFunction[ work on most anything, such as OEIS-related. $\endgroup$
    – prog9910
    May 26, 2020 at 2:34
  • $\begingroup$ @prog, it doesn't always work, but it is always useful to try. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    May 26, 2020 at 2:40

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