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There is a function FindSequenceFunction in Mathematica, that can identify a sequence of integers based on a few first elements. But what if I have a set of finite sequences sec[n] that grows with n? For example:

sec[0]={1}
sec[1]={1, 1}
sec[2]={1, 6, 1}
sec[3]={1, 15, 15, 1}
sec[4]={1, 28, 70, 28, 1}
sec[5]={1, 45, 210, 210, 45, 1}
sec[6]={1, 66, 495, 924, 495, 66, 1}
sec[7]={1, 91, 1001, 3003, 3003, 1001, 91, 1}
sec[8]={1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1}

Is there a way to use Mathematica to find the general expression to continue this sequence of sequences?

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    $\begingroup$ I flattened this to a single sequence. It is oeis.org/A086645. I haven't tried it in Mathematica yet, but flattening is probably a good first step. $\endgroup$ – mikado Oct 8 '16 at 17:11
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    $\begingroup$ More concisely: you have a (Riordan) triangle of integers that you want to identify. @mikado, that's when you're lucky and your triangle is listed in the OEIS. The question of how one might find a tentative formula for a given integer triangle is still an interesting one, tho. $\endgroup$ – J. M. is away Oct 8 '16 at 17:16
  • $\begingroup$ Wow, this finding is amazing, thank you! Yes, still wondering if there is a way to proceed in a general case. $\endgroup$ – Kagaratsch Oct 8 '16 at 23:42

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