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Mar 20, 2019 at 16:32 vote accept chr
Mar 18, 2019 at 12:49 answer added Somos timeline score: 0
Mar 17, 2019 at 23:49 comment added bill s Use Solve[a[j] + a[j + 1] == a[j - 1] + a[j - 2], a[j]], which solves for a[j].
Mar 17, 2019 at 20:45 comment added J. M.'s missing motivation Use Indexed[].
Mar 17, 2019 at 20:40 history edited chr CC BY-SA 4.0
added 97 characters in body
Mar 17, 2019 at 20:39 comment added chr I don't want to do recursive stuff
Mar 17, 2019 at 19:28 history asked chr CC BY-SA 4.0