Find an integer $a$ such that $a\sqrt{2} $ has a given decimal part Suppose I want to find integers such that when I multiply it by an irrational number (e.g $\sqrt{2}$), its decimal part contains 0.14159... as starting digits. How do I find them? Is there a pattern for the digits of such numbers?
It is fairly easy to compute $\sqrt{2}$ to arbitrarily high precision.  See for example this page reporting a little more than one million digits.

If we are searching for a fixed initial string of length $n$, then of course if it can be located in $\sqrt{2}$ further along, that string can be brought to the front by multiplying by the appropriate power of ten.  One then has a (not necessarily minimum) solution for $a$.  In this case the string "14159" is found seven times within the first million digits.

Finding the minimum value of $a$ seems to be a difficult computational challenge.
We can show that there exists no integer such that $a \cdot \sqrt{2}$ has the full decimal part of $\pi$. Suppose there were such an integer $a$ such that $a \cdot \sqrt{2} = B + \pi -3$, where $B \in \mathbb{Z}$. Then, we would  also have that $a \cdot \sqrt{2} + 3 -B = \pi$. But then notice that the LHS is an algebraic number, whereas the RHS is a transcendental number. Contradiction.