Problem 44338. Recaman Sequence - I

Created by Mehmet OZC

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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

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Last Solution submitted on May 13, 2020

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Stephen Cook on 16 Oct 2017

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).

Mehmet OZC on 16 Oct 2017

That is correct. I have modified the sequence to avoid zero indexing.

Adiel on 16 Oct 2017

According to the definition here, seq(2) should be 2 because seq(n-1)+n when n=2, is 2. And so on...

Mehmet OZC on 16 Oct 2017

Adiel, Thanks for your attention. I have changed the definition.

Svyatoslav Golousov on 23 Oct 2017

When i preinitialise y as follows:
y = zeros(1,x);
it doesn't pass, but when i just ommit preinitialisation - it works perfectly!

David Verrelli on 5 Nov 2017

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]."

Mehmet OZC on 6 Nov 2017

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)

Kodavati Mahendra on 25 Dec 2017

good question :-)

Solution Comments

Solution 1292583

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James on 17 Oct 2017

Any idea what the time limit for the Cody solutions is? I've got code that runs the test suite in about 2.3 seconds on my (six-year old) machine, but it's timing out repeatedly on Cody.

David Verrelli on 4 Nov 2017

Hi, James. The Cody time limit is reportedly around 30 seconds. [ https://www.mathworks.com/matlabcentral/cody/problems/44356 ] I just have confirmed a runtime of ~24 seconds (split across four tests) did not time out on Cody [Solution 1327589]. Although I would generally be aiming to get code complete within less than ~10 seconds in Cody. [Cf. Problem 44389.]

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Problem 44338. Recaman Sequence - I

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Problem 44338. Recaman Sequence - I

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