Inverse with floor

$= \displaystyle \int_{0}^{tan1} 0 + \displaystyle \int_{tan1}^{1729} 1$

$= 1729 - tan1$

$\begin{aligned}\displaystyle \int\limits_0^{1729} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x&=\int\limits_0^{\tan 1} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x + \int\limits_{\tan 1}^{1729} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x \\&=\int\limits_0^{\tan 1} 0 \mathrm{d}x + \int\limits_{\tan 1}^{1729} 1 \mathrm{d}x\\ &= 1729- \tan 1\\&\approx 1727.442592\end{aligned}\\\huge \boxed{\displaystyle\int\limits_0^{1729} \left \lfloor \tan^{-1} x\right \rfloor \mathrm{d}x = 1729 - \tan 1^c}$

$\displaystyle \int_{0}^{1729} \lfloor\tan^{-1}x\rfloor \mathrm{d}x=\int_{0}^{1} \lfloor\tan^{-1}x\rfloor \mathrm{d}x+\int_{1}^{2} \lfloor\tan^{-1}x\rfloor \mathrm{d}x +\int_{2}^{3} \lfloor\tan^{-1}x\rfloor \mathrm{d}x\\+\displaystyle\int_{3}^{4} \lfloor\tan^{-1}x\rfloor \mathrm{d}x+\ldots+\int_{1727}^{1728} \lfloor\tan^{-1}x\rfloor \mathrm{d}x+\int_{1728}^{1729} \lfloor\tan^{-1}x\rfloor \mathrm{d}x$

$\begin{aligned}\displaystyle&=\int_{0}^{1} 0 \mathrm{d}x+\int_{1}^{2} 0 \mathrm{d}x+\int_{2}^{3} 1 \mathrm{d}x+\int_{3}^{4} 1 \mathrm{d}x+\ldots+\int_{1927}^{1728} 1 \mathrm{d}x+\int_{1728}^{1729} 1 \mathrm{d}x\\ &=\displaystyle \int_{2}^{1729} 1 \mathrm{d}x=1729-2=\large\boxed{1727}\end{aligned}$

let x=tan (t) I=Integrate([t]sec^2(t)dt) from (0 to (tan^-1(1729)~1.57)) I=Integrate([t]sec^2(t)dt) from (0 to 1)) +Integrate([t]sec^2(t)dt) from (1 to 1.57)) the first part will vanish as [t]=0 in (0,1). and the second one = Integrate(sec^2(t)dt) from (1 to tan^-1(1729)))=1729- tan(1radians)=1727.(some fraction) thus [a]=1727

You have a thing with the 1729 number!