Is there a smarter way to create a large vector, with repeated numbers in a diminishing quantity?

I need the vector to begin with 144 evenly spaced '2's, then 143 '3's, then 142 '4's, etc. I see the brute force method, but it doesn't seem very smart. Is there a way to get Matlab to create this vector for me?

thx

[Igor]

Valid code --

n=14;A1=(1:n)'*ones(1,n);A2=triu(A1)';A3=reshape(A2,1,n^2);A=A3(A3~=0),

[Sean de Wolski]

Any question that makes a golf challenge gets a vote from me!

Answer 1

[Sean de Wolski]

A one liner:

n = 144;
v = sort(nonzeros(triu(bsxfun(@times,(2:n+1).',ones(1,n)))));

or

v = sort(nonzeros(triu(toeplitz(2:n+1))));

Golf!

v=hankel(n+1:-1:2);
v=sort(v(~~v))

%33 characters

transpose it if you want a row vector.

Answer 2

[Friedrich]

I think the repmat command can help:

[repmat(2,144,1);repmat(3,143,1);repmat(4,142,1)]

[Igor]

no :) 144 times?

n=144;

A1=(1:n)'*ones(1,n);

A2=tril(A1);

A3=reshape(A2...);

find... all zero elements

[Friedrich]

Upps^^, than this way:

start_repeat = 144;

start_number = 2;

final_size = (start_repeat+1)*start_repeat / 2;

final_vec = zeros(final_size,1);

index = 1;

for i=start_repeat:-1:1

final_vec(index:index+i-1,1) = repmat(start_number+start_repeat-i,i,1);

index = index + i;

end

Answer 3

[Igor]

[1,2,2,3,3,3,4,4,4,4,...] similar task

n=1:10 trinv(n) = floor((1+sqrt(1+8*n))/2)

(<http://oeis.org/A002024>)

Answer 4

[Oleg Komarov]

Using rude for run length decoding:

n = 144;
val = 1:n;
len = val;

Out = rude(len,val); % rude on fex

Answer 5

[Andrei Bobrov]

n = 144;
vm = triu(repmat(n+1:-1:1,n+1,1));
vm(vm>0);

[Igor]

:) shortest :)

[Sean de Wolski]

Isn't right...