Find $\sum_{k=1}^nk\lfloor k\varphi\rfloor$, where $\varphi$ is golden ratio

Question

I've got this far, which is nothing really. Assuming $S(n)=\sum_{k=1}^n\lfloor k\varphi\rfloor$, for which we have a recursive formula (see here: Solve summation $\sum_{i=1}^n \lfloor e\cdot i \rfloor $), we can write out sum as $$ 1\lfloor\varphi\rfloor+2\lfloor2\varphi\rfloor+\ldots+n\lfloor n\varphi\rfloor= $$ $$ =S(n)+(S(n)-S(1))+(S(n)-S(2))+\ldots+(S(n)-S(n-1))= $$ $$ =nS(n)-\sum_{k=1}^{n-1}S(k). $$ Computationally, this is obviously very slow. Anything faster? Thanks.