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.

  • $\begingroup$ It may be of interest to note that $S(n)$ as presented is sequence A054347 and has a reference therein that may lead to further information. $\endgroup$ – Leucippus Nov 22 '17 at 3:39

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