Is there a method to find a math expression for a given pattern? I have this pattern and I am very curious to find out how can I generate it.

0 1 -1x 3 -4x 3x^2 5 -9x 9x^2 -5x^3 7 -16x 21.6x^2 -16x^3 7x^4 9 -25x 42.86x^2 -42.86x^3 25x^4 -9x^5 11 -36x 75x^2 -95.24x^3 75x^4 -36x^5 11x^6

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    $\begingroup$ The first column are the odd integers (save 0), and the second column are the negative perfect squares. As for the rest of them... $\endgroup$ – beanshadow May 30 '14 at 8:22
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    $\begingroup$ You really need to tell us more about the source of these numbers. In particular, are the decimal numbers rounded or exact? $\endgroup$ – Phira May 30 '14 at 8:33
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    $\begingroup$ The decimal numbers are exact. And if it helps, this is from a physics problem where each line of these coefficients are describing interference patterns of corresponding interaction order. $\endgroup$ – user2100362 May 30 '14 at 8:46
  • $\begingroup$ "The decimal numbers are exact" and "this is from a physics problem" don't go together well. - Are you sure tha $-75$ in the last line shouldn't be $75$? $\endgroup$ – Hagen von Eitzen May 30 '14 at 9:19
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    $\begingroup$ @heropup This sequence is the analytically calculated and simplified outcome of a physical problem, which is a particular interference pattern of up to 6 interacting particles. So there is an underlying math, but it doesn't directly tell anything about this sequence either. $\endgroup$ – user2100362 May 30 '14 at 12:39
  • The n-th row has n elements, starting with the first.
  • The signs alternate on each row, and each row starts with a $+$.
  • The first and last element of each row (starting with the second) host the odd numbers $2n-3$.
  • The second and next-to-last element on each row (starting with the third) host the perfect squares $(n-1)^2$. Notice that the differences of each two consecutive squares are exactly the odd numbers mentioned before.
  • The third and the next-to-next-to-last element on each row (starting with the fifth) host a number whose absolute value is very close to the arithmetic mean of the absolute values of the one to the side, and the one right above it, minus $4$.
  • As far as the element in the middle of the seventh row is concerned, I need to see more rows before determining a specific pattern.
  • Hope this helps !
  • $\begingroup$ Thanks, this is very helpful. But the main puzzle is really the decimal numbers. Writing the next rows gonna be an extremely lengthy calculations and I can't do it quickly. $\endgroup$ – user2100362 May 30 '14 at 11:45

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