Stoneham number expansion

Question

How can we calculate the expansion of the a Stoneham number $\alpha_{10,3}$

I want to get the more digits as possible

Answer 1

In the Wikipedia article, $\alpha_{10,3}$ is defined as $$\sum_{k=1}^\infty\frac1{10^{3^k}3^k}$$ This converges very rapidly, so getting a lot of digits shouldn't be hard. The first question to answer is what is the truncation error caused by stopping after say $N$ terms. The error is $$\sum_{k=N+1}^\infty\frac1{10^{3^k}3^k}$$ To estimate this, set $$a_k=\frac1{10^{3^k}3^k}$$ Then $$\frac{a_{k+1}}{a_k}=\frac{10^{3^k}3^k}{10^{3^{k+1}}3^{k+1}}=\frac13\frac1{100^{2 \cdot 3^k}}<\frac1{300},$$ so that the truncation error is less than $$a_{N+1}(1+\frac1{300}+\frac1{300^2}+\cdots)=\frac{300}{299}a_{N+1}$$ That is, the truncation error is less than $\frac{300}{299}$ times the first omitted term. The terms decrease so fast that effectively, we don't have to worry about the truncation error. Choose $N$ so large that $a_{N+1}$ is negligible.

As for actually computing the digits, that depends on what kind of software you have available. You need some arbitrary-precision floating-point arithmetic package. A list can be found here.

Note that $\frac1{a_5}\approx 2.43\cdot10^{245}$ so calculating four terms will give you about $244$ decimal digits.