Recaman Sequence (A005132 - - OEIS Link) is defined as follow;
seq(0) = 0; for n > 0, seq(n) = seq(n-1) - n if positive and not already in the sequence, otherwise seq(n) = seq(n-1) + n.
seq = 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9 ... index = 1, 2, 3 ,...
To avoid zero index, start indexing from 1. return the first n elements in Recaman Sequence
Related Challenges :
- Recaman Sequence - I
- Recaman Sequence - II
- Recaman Sequence - III
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Recaman sequence uses zero indexing, thus `seq(n) = seq(n-1) - (n-1)` or `seq(n) = seq(n-1) + (n-1)` (not the formula used in the description).
That is correct. I have modified the sequence to avoid zero indexing.
According to the definition here, seq(2) should be 2 because seq(n-1)+n when n=2, is 2. And so on...
Adiel, Thanks for your attention. I have changed the definition.
When i preinitialise y as follows:
y = zeros(1,x);
it doesn't pass, but when i just ommit preinitialisation - it works perfectly!
Hello, Mehmet OZC. I think the problem statement is still subject to misinterpretation. At the moment it looks like "index" means "n", which is inconsistent with the formula. To avoid this misunderstanding, you could insert one extra row to show explicitly the values of "n". ... Thus you would have: seq = 0, 1, 3, 6, ... . Then, n = 0, 1, 2, 3, ... . And finally, Index = 1, 2, 3, 4, ... . ____ Alternatively, you could omit any mention of indexing and just clarify with an _example_, such as: "The first four elements are [0, 1, 3, 6]."
David, you have already made a good explanation for a possible misinterpretation. I appreciate it. Thanks for your contribution. When I said indexing I meant something like linear indexing (https://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html)
good question :-)