Auto comment: topic has been updated by TheSleepyDevil (previous revision, new revision, compare).

i had an idea where the answer is just summation ((n)/k) + ((n-1)/k) + ... till n is k-1 , but idk if that's true or not , or how to calculate it fast

Let j-i=q*k

j-i=x

x=k,2*k,3*k....,(n/k)*k

Let n/k=t (integer division)

n>=j>=x

Number of solutions is n-k+1+(n-2*k+1)+(n-3*k+1)...(n-t+1)=n*t+t-k*t*(t+1)/2