i want vector which is long size and each time size change but pattren is same in putting value

vector like this....i want in this sequence..this is for 1 to 8 row
      =[ 2 1 4 3 6 5 8 7   
         3 1 5 3 7 5 1 7 
         4 1 6 3 8 5 2 7 
         5 1 7 3 1 5 3 7
         6 1 8 3 2 5 4 7
         7 1 1 3 3 5 5 7
         8 1 2 3 4 5 6 7]

if size is increase than row change but sequence is same if 12 row than vector = [

   2  1  4  3  6  5  8   7  10  9  12  11
   3  1  5  3  7  5  9   7  11  9  1   11
   4  1  6  3  8  5  10  7  12  9  2   11
   5  1  7  3  9  5  11  7   1  9  3   11...............]

sequence is odd term remain same but even number move forward in sequence

[John D'Errico]

Huh? You are confused. Or, at least, confusing. Try again.

[Pratik Anandpara]

i have code which give only few entry

swapoddeven = reshape([2:2:12;1:2:12], 1, [])

[Jan]

This is a strange question. I do not have any idea about what you are searching for. What are the inputs, what is variable and what is fixed, and what do you want as output?

[Pratik Anandpara]

example : input is take digit from 1 to 8 output is vector in this pattern

   2 1 4 3 6 5 8 7 
   3 1 5 3 7 5 1 7
   4 1 6 3 8 5 2 7
   5 1 7 3 1 5 3 7
   6 1 8 3 2 5 4 7
   7 1 1 3 3 5 5 7
   8 1 2 3 4 5 6 7

here observe one pattern digit is reurn in that pattern 2nd-4th-6th-8th digit remain same and other numbers are in sequence..2nd-4th-6th-8th digit remian same at its place other digit move in sequence from 1 to 8... observe 1st digit-3rd-5th-7th digit in each row it rotate in sequence column wise....but 2nd-4th-6th-8th digit in each row is remain same... observe pls

Answer 1

[Roger Stafford]

Let your resulting matrix, M, have m rows and n columns.

   b = mod(0:n-1,2)==0;
   M = repmat((1:n).*b+(-1:n-2).*(~b),m,1);
   M = mod(M+bsxfun(@times,(0:m-1).',b),n)+1;