Le $k≥1$ be positive integer. I am asking if it is possibe to find a closed formula for $$gcd(2^{2k+5}-3×2^{k+2}+1,2×(2^{k+2}-1)(2^{k+1}-1),2×(3×2^{k+1}-1)(2^{k+2}-1))$$ where $gcd$ is the greatest common divisor.
For $k=1$, we have
$$gcd(2²⁺⁵-3×2¹⁺²+1,2×(2¹⁺²-1)(2¹⁺¹-1),2×(3×2¹⁺¹-1)(2¹⁺²-1))= 7$$
Hint
Look at the value of the gcd for the first few values of $\ k\ .$ That suggests the following factorisation: \begin{align} 2^{2k+5}-3\cdot2^{k+2}+1&=2^{2k+5}-2^{k+3}-2^{k+2}+1\\ &=\big(\color{red}{2^{k+2}-1}\big)\big(2^{k+3}-1\big) \end{align} of the first number, and you already have $$ 2\big(\color{red}{2^{k+2}-1}\big)\big(2^{k+1}-1\big) $$ and $$ 2\big(3\cdot2^{k+1}-1\big)\big(\color{red}{2^{k+2}-1}\big) $$ for the other two numbers. The second of these minus $\ 3\ $ times the first is $$ 4\big(2^{k+2}-1\big)\ . $$