Calculating sum of a ceil function involving golden ratio Let $\tau$ be a function on natural numbers defined as $\tau(n)=\lceil n\phi^2\rceil$, where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio.
Is there a way to calculate
$$\tau(1)+\tau(2)......+\tau(t)$$
for large values of $t$ ( $t$ is of the order $10^{16}$ ) ? 

The first thing that can be done is to rewrite $\tau(n)$ as
$$\lceil n\phi^2\rceil=\lceil n+n\phi\rceil=n+\lceil n\phi\rceil$$
The first half since $\phi^2=\phi+1$ and the second half because $n$ is an integer.  The first part of this new summation I'm sure you already know how to do, which leaves $\sum\lceil n\phi\rceil$.  From here, I attempted to use the best rational approximations of $\phi$.  For $\phi$, this is simply the ratio of consecutive Fibonacci numbers.  For example, $\frac32<\phi<\frac21$ and $\frac85<\phi<\frac53$.  More specifically, we want the best rational approximations which are greater than $\phi$, or $\frac{F_{2n+1}}{F_{2n}}$.
Now, let's say that $\frac ab$ is the best rational approximation greater than $\phi$ with denominator less than or equal to $N$.   It can be proven by contradiction that for $n\le N$, $\lceil n\phi\rceil=\lceil\frac{na}b\rceil$.  Now consider the sum $\lceil n(\frac ab)\rceil+\lceil(bk-n)\frac ab\rceil$.  If $b|n$, we are taking the ceilings of integers, so we can simply remove the ceilings to get a sum of $ak$.  Otherwise, both ceilings get rounded up, a combined total of $1$, yielding $ak+1$.
Let's say we want to calculate $\sum_{n=1}^{100}\lceil n\phi\rceil$.  We can use the approximation $\frac{F_{11}}{F_{10}}=\frac{89}{55}$.  So we have
$$\sum_{n=1}^{100}\lceil n\phi\rceil=\sum_{n=1}^{100}\lceil\frac{89n}{55}\rceil=\sum_{n=1}^9\lceil\frac{89n}{55}\rceil+\sum_{n=10}^{54}\left[\lceil\frac{89n}{55}\rceil+\lceil(110-n)\frac{89}{55}\rceil\right]+89=$$
$$\sum_{n=1}^9\lceil n\phi\rceil+45[2(89)+1]+89$$
One iteration has cut the sum drastically.  The next iteration with approximation $\frac{13}8$ will be less impressive before the final iteration with approximation $\frac53$ will complete the summation.
