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.

Answer 1

Let's denote $A(n) = \sum_{k=1}^n k \left\lfloor k\varphi \right\rfloor$.

Because $\varphi = \frac{1}{\varphi} + 1$, we have $\left\lfloor k\varphi \right\rfloor = k + \left\lfloor \frac{k}{\varphi} \right\rfloor$ and can write

$$ A(n) = \sum_{k=1}^n k \left\lfloor k\varphi \right\rfloor \\ = \frac{1}{6}n(n+1)(2n+1) + \sum_{k=1}^n k \left\lfloor \frac{k}{\varphi} \right\rfloor. $$

Let's write this last sum double sum and change its summation order:

$$\large \sum_{k=1}^n k \left\lfloor \frac{k}{\varphi} \right\rfloor = \sum_{k=1}^n \sum_{j=1}^{\left\lfloor \frac{k}{\varphi} \right\rfloor} k = \sum_{j=1}^{\left\lfloor \frac{n}{\varphi} \right\rfloor} \sum_{k= \left\lceil j\varphi \right\rceil }^{n} k $$

(Note: because $\varphi$ is irrational, $j\varphi$ is never an integer and the bounds go properly)

Now we can calculate the inner sum and it equals

$$ \large \sum_{k= \left\lceil j\varphi \right\rceil }^{n} k = \frac{1}{2}n(n+1) - \frac{1}{2} \lfloor j\varphi \rfloor\left( \lfloor j\varphi \rfloor + 1 \right) $$

Let's denote $m = \left\lfloor \frac{n}{\varphi} \right\rfloor$. So we get

$$ A(n) = \frac{1}{6}n(n+1)(2n+3m+1) - \frac{1}{2} \sum_{j=1}^m \lfloor j\varphi \rfloor\left( \lfloor j\varphi \rfloor + 1 \right) \\ = \frac{1}{6}n(n+1)(2n+3m+1) -\frac{1}{2}B(m) - \frac{1}{2} S(m) $$

where $S$ is as already defined (and known how to calculate) and $B(n) = \sum_{k=1}^n \left\lfloor k\varphi \right\rfloor^2$.

We get also a recursive formula for $B$ similarly. (Or see this link where it's done for $S$. For $B$ there's a bit more work and we have also the appearance of $A$ in the formula.) It is

$$ B(n) = \frac{1}{6}N(N+1)(2N+1) - B(J) - 2A(J) - \frac{1}{6}J(J+1)(2J+1) $$

where $N = \lfloor n\varphi \rfloor$ and $J = \left \lfloor \frac{N+1}{\varphi+1} \right\rfloor$.