# Making an array using loop

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SWARNENDU PAL on 30 Dec 2020
Edited: Paul Hoffrichter on 1 Jan 2021
A=[2;3;4;5;12;13;14;15;16;17;24;25;26;27;28;29;36;37;38;39;40;41;48;49;50;51;52;53;60;61;62;63;64;65;72;73;74;75;76;77;84;85;86;87;88;89;96;97;98;99;100;101;108;109;110;111;112;113;120;121];
How to make an array like A using loop. Thank you.
Rik on 31 Dec 2020
The pattern is not obvious to me (nor, apperently, to Cris). It is also missing from the OEIS, which is generally an indication that the sequence is fairly obscure.

Paul Hoffrichter on 31 Dec 2020
Edited: Paul Hoffrichter on 1 Jan 2021
Change ORIGINAL_POST to false to get slightly different result.
clearvars; clc; % remove previous debug runs
lenA = 60; % assumes you know the length of A
A = zeros(lenA,1); % pre-allocate A
maxSeqLen = 6;
ORIGINAL_POST = true;
if ORIGINAL_POST
start = 2;
seqLength = 4;
else
% rng(123);
start = randi(maxSeqLen,1);
seqLength = randi(maxSeqLen);
end
nominalSequence = 1:maxSeqLen;
skip = 7;
% Initialize
A(1:seqLength) = start:(start + seqLength - 1);
startLoc = seqLength + 1;
k = seqLength + 1;
while startLoc + maxSeqLen - 1 <= lenA
start = A(startLoc - 1) + skip;
A(startLoc:startLoc + maxSeqLen - 1) = start: start + maxSeqLen - 1;
start = start + maxSeqLen - 1 + skip;
startLoc = startLoc + maxSeqLen;
end
if startLoc <= lenA
maxSeqLen = lenA - startLoc + 1;
A(startLoc:startLoc + maxSeqLen - 1) = start: start + maxSeqLen - 1;
% if you cannot pre-allocate A, and have to enter values one at a time,
% then you can add elements like this: A = [A; new-value];
end
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Paul Hoffrichter on 1 Jan 2021
clc - used to remove previous debug runs where printing was done.
clear all - good point. Edited to say clearvars.