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(Might be a spoiler)

I was working on a famous problem, and I came up with a variation that can produce a sequence much harder to recognise than the original sequence.

Identify the problem which generates this sequence:

1, 1, 1, 2, 2, 2, 3, 4, 4, 5, 7, 8, 9, 12, 15, 17, 21, 27, 32, 38, 48, 59, 70, 86, 107, 129, 156, 193...

Hint:-

Fibonacci

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  • $\begingroup$ Welcome to Puzzling! (Take the Tour!) Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$
    – Rubio
    Commented Dec 19, 2017 at 12:57

1 Answer 1

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The answer is

a[i] = a[i-3]+a[i-4]

Because

Every number is the sum of its third and fourth predecessor.

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  • $\begingroup$ You just beat me to it. It's not too hard to figure out if you start at the end of the provided sequence of numbers. $\endgroup$ Commented Dec 11, 2017 at 12:02
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    $\begingroup$ This does not work for the first 2... $\endgroup$
    – Jason V
    Commented Dec 11, 2017 at 17:45
  • $\begingroup$ That was not what I expected. This kind of took away the "fun" of the question. Good observation though. I would accept a situational definition, and not a mathematical one, like how Fibonacci originally did with rabbit pairs. $\endgroup$ Commented Dec 12, 2017 at 13:18
  • $\begingroup$ Also, the solution should explain the first 5 terms (My solution requires a single parameter) $\endgroup$ Commented Dec 12, 2017 at 13:26

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