I found this almost integer in studying the limit :
$$\lim_{x\to \infty}\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-x^3=-\infty$$
Well my goal was to find a constant such that :
$$\lim_{x\to \infty}\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-x^{\alpha}=constant$$
It appears that the constant in question have the following bound :
$$2.9999999990001<\alpha<2.999999999001$$
Found numerically obviously .
For the LHS the limit goes to infinity and with the RHS the limit goes to the opposite.
Now I'm not able to prove anything and curiously I didn't found anything on the websites .
So : Is the second limit viable ? Is this almost integer well know ? Can we hope to find a closed form ?
I do not want to play trouble fete !
When $x$ is large, if $$f(x)=\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)\Gamma\left(\sin\left(\frac{1}{x}\right)\right)$$ the expansion to $O\left(\frac{1}{x}\right)$ is $$f(x)=x^3-\gamma x^2+\frac{1}{12} \left(6 (\gamma -1)^2+\pi ^2\right) x+\left(-\frac{\zeta (3)}{3}-\frac{1}{12} \gamma \left(4+2 (\gamma -6) \gamma +\pi ^2\right)\right)$$ then no problem with the first statement.
Using $x=10^k$ and computing for $\alpha$ equal to the average of the two given bounds ($\alpha$ being a rational number), some results $$\left( \begin{array}{cc} k & f\left(10^k\right)-10^{\alpha k} \\ 4 & -5.77032\times 10^7 \\ 5 & -5.76056\times 10^9 \\ 6 & -5.63407\times 10^{11} \\ 7 & -4.16123\times 10^{13} \\ 8 & +1.26384\times 10^{16} \\ 9 & +2.01347\times 10^{19} \\ 10 & +2.29555\times 10^{22} \\ 11 & +2.53087\times 10^{25} \\ 12 & +2.76153\times 10^{28} \\ 13 & +2.99171\times 10^{31} \\ 14 & +3.22185\times 10^{34} \\ 15 & +3.45198\times 10^{37} \\ 16 & +3.68211\times 10^{40} \\ 17 & +3.91224\times 10^{43} \\ 18 & +4.14237\times 10^{46} \\ 19 & +4.37251\times 10^{49} \\ 20 & +4.60264\times 10^{52} \end{array} \right)$$
There is nothing strange in this table. As Newton iterates show below, it exists a $k$ such that, for this specific value of $\alpha$ $$f\left(10^k\right)=10^{\alpha k}$$
$$\left( \begin{array}{cc} n & k_n \\ 0 & 7.5000000 \\ 1 & 7.5224059 \\ 2 & 7.5229715 \\ 3 & 7.5229719 \end{array} \right)$$
In a second step, I computed $$g(k)=f(10^{20})-\big[10^{20}\big]^{3-10^{-k}}$$
$$\left( \begin{array}{cc} k & g(k) \\ 2 & 3.69043\times 10^{59} \\ 3 & 4.50074\times 10^{58} \\ 4 & 4.59458\times 10^{57} \\ 5 & 4.60411\times 10^{56} \\ 6 & 4.60506\times 10^{55} \\ 7 & 4.60517\times 10^{54} \\ 8 & 4.60500\times 10^{53} \\ 9 & 4.55947\times 10^{52} \\ 10 & 4.60517\times 10^{51} \\ 11 & 4.60517\times 10^{50} \\ 12 & 4.60505\times 10^{49} \\ 13 & 4.60644\times 10^{48} \\ 14 & 4.60109\times 10^{47} \\ 15 & 4.60287\times 10^{46} \end{array} \right)$$