Problem 44338. Recaman Sequence - I

Created by Mehmet OZC
Appears in Cody5:Easy

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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 Feb 22, 2025

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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 :-)

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