RSolve[{W[n]==W[⌊(n/2)⌋]+W[⌈(n/2)⌉]+n-1 , W[1]==0}, W[n], n]

Assume n = (2^k)and put A[k] = W[n].

W[n]==W[⌊(n/2)⌋]+W[⌈(n/2)⌉]+n-1 , W[1]==0

becomes

W[(2^k)]==W[((2^k)/2)]+W[((2^k)/2)]+(2^k)-1 , W[1]==0

W[n]==W[⌊(n/2)⌋]+W[⌈(n/2)⌉]+n-1 , W[1]==0

or

W[(2^k)]==W[(2^(k-1))]+W[(2^(k-1))]+(2^k)-1 , W[(2^0)]==0


or

W[(2^k)]==2W[(2^(k-1))]+(2^k)-1 , W[(2^0)]==0

or

A[k]==2A[k-1]+(2^k)-1 , A[0]==0

RSolve[{A[k]==2A[k-1]+(2^k)-1 , A[0]==0}, A[k], k]

So,

W[k] = n Log2[n] - n + 1 = (n-1) Log2[n] + 1

The above holds only for n = (2^k)