Indexing n-th element with increasing offset.

Good day

I'm trying to find a whey how it would be possible to index a n-th element with increasing offset in a arrey of zeros (expectation - 1 0 1 0 0 1 0 0 0 1 0 0 0 0 1…). I found a whey, but its something like from first grade. Maybe someone would share some ideas, or give a hint of how it would be possible to reduce repeating typing…

x=1:1:100;
g=cumsum(x)<100;
g_idx=find(g);

X=zeros(size(x));
X(g_idx(1))=1;
X(cumsum(g_idx(1:2)))=1;
X(cumsum(g_idx(1:3)))=1;
X(cumsum(g_idx(1:4)))=1;
X(cumsum(g_idx(1:5)))=1;
X(cumsum(g_idx(1:6)))=1;
X(cumsum(g_idx(1:7)))=1;
X(cumsum(g_idx(1:8)))=1;
X(cumsum(g_idx(1:9)))=1;
X(cumsum(g_idx(1:10)))=1;
X(cumsum(g_idx(1:11)))=1;
X(cumsum(g_idx(1:12)))=1;
X(cumsum(g_idx(1:13)))=1

[Mario Malic]

Pascal's triangle, binomial theorem? You can check on Wiki, there's a formula for those numbers on it.https://wikimedia.org/api/rest_v1/media/math/render/svg/23050fcb53d6083d9e42043bebf2863fa9746043

[Andrew]

Thank You for Your guidance.

Answer 1

[Fangjun Jiang]

%%
a=1:13;
b=cumsum(a);
X=zeros(1,100);
X(b)=1