Preliminaries

Gauss sum formula

We will need a formula to compute, the sum of all natural numbers from \(1\) to an arbitrary natural number \(n\): $$T_n=1+2+3+4+\dots+n$$ This expression can also be written in the compact summation form: $$T_n=1+2+3+\dots+n=\sum_{k=1}^nk$$ \(T_n\) is called the triangular number for \(n\) and has a simple formula to compute, which is known as the Gauss sum formula: $$T_n=\frac{n(n+1)}{2}$$

Generalized Gauss sum formula

Using the Gauss sum formula from above, we can now also compute the sum of all natural numbers between an arbitrary natural number \(n\) and an arbitrary natural number \(m\), with \(n\lt m\): $$S_{n,m}=n+(n+1)+(n+2)+\dots+m$$ This expression can again be written in the compact summation form: $$S_{n,m}=n+(n+1)+(n+2)+\dots+m=\sum_{k=n}^mk$$ If we now add and subtract the triangular number \(T_{n-1}\) from \(S_{n,m}\), we can rearrange terms to find a formula for \(S_{n,m}\): $$S_{n,m}=S_{n,m}+\underbrace{T_{n-1}-T_{n-1}}_{0}=\underbrace{(n+(n+1)+(n+2)+\dots+m)}_{S_{n,m}}+\underbrace{(1+2+3+\dots+(n-1))}_{T_{n-1}}-\underbrace{(1+2+3+\dots+(n-1))}_{T_{n-1}}=\underbrace{\underbrace{(1+2+3+\dots+(n-1))}_{T_{n-1}}+\underbrace{(n+(n+1)+(n+2)+\dots+m)}_{S_{n,m}}}_{T_m}-\underbrace{(1+2+3+\dots+(n-1))}_{T_{n-1}}=T_m-T_{n-1}$$ Now we can simply plug in twice the Gauss sum formula from above: $$S_{n,m}=T_m-T_{n-1}=\frac{m(m+1)}{2}-\frac{n(n-1)}{2}$$

P3D Math

Relation of current token supply to current (untaxed) token price

Let \(s\) be the current token supply and \(p_s\) the (untaxed) token price in ETH given the token supply \(s\). The initial token price (price of the first token sold) is \(0.0000001\) ETH. The token price increases by \(0.00000001\) ETH per additional token. Therefore, we can compute the (untaxed) token price from the token supply: $$p_s = 0.0000001 + s\cdot 0.00000001$$ By rearranging terms, we can also compute the token supply given the current (untaxed) token price: $$0.0000001 + s\cdot 0.00000001 = p_s\Leftrightarrow s\cdot 0.00000001 = p_s - 0.0000001\Leftrightarrow s = \frac{p_s-0.0000001}{0.00000001}$$

Token supply needed for a token price of 1 ETH

Using the second formula derived right above, we can compute what the token supply would need to be, in order for the token price to be \(1\) ETH. We just set \(p_s\) equal to \(1\): $$s=\frac{p_s-0.0000001}{0.00000001}=\frac{1-0.0000001}{0.00000001}=99999990$$ So we need a token supply of \(99999990\) tokens, to have a token price of 1 ETH. You can verify this here.

Relation of current token supply to current amount of ETH in tokens

Let \(s\) be the current token supply, \(e\) the current amount of ETH in tokens and \(p_1, p_2, p_3,\ldots,p_s\) the prices paid for each token of the total supply (remember, the token price increases with every additional token). Now clearly, the current amount of ETH in tokens is equal to the sum of all token prices: $$e=p_1+p_2+p_3+\dots+p_s=\sum_{k=1}^sp_k$$ Now remember from above, that we could compute the token price \(p_s\) using the token supply \(s\) by: $$p_s = 0.0000001 + s\cdot 0.00000001$$ So we can expand the formula for \(e\) from above: $$e=\sum_{k=1}^sp_k=\sum_{k=1}^s(0.0000001+k\cdot 0.00000001)$$ Let us split this sum into two parts: $$e=\sum_{k=1}^s(0.0000001+k\cdot 0.00000001)=\underbrace{\sum_{k=1}^s0.0000001}_{s\cdot 0.0000001}+\underbrace{\sum_{k=1}^s(k\cdot 0.00000001)}_{0.00000001\cdot\sum_{k=1}^sk}=s\cdot 0.0000001+0.00000001\cdot\underbrace{\sum_{k=1}^sk}_{T_s}$$ The first summation just sums up \(0.0000001\) for a total of \(s\) times and can be replaced by a simple multiplication. In the second part, we can move the factor \(0.00000001\) out of the sum. Now observe that in the last expression, the summation is exactly the triangular number \(T_s\) finally giving us: $$e=s\cdot 0.0000001+0.00000001\cdot T_s=s\cdot 0.0000001+0.00000001\cdot\frac{s(s+1)}{2}$$

Amount of ETH in tokens needed for a token price of 1 ETH

We already know from above, that we need a token supply of \(99999990\) tokens, for the token price to be 1 ETH. Now we only need to set \(s\) to \(99999990\) in the formula derived right above, to get the amount of ETH needed for the token price to be 1 ETH: $$e=s\cdot 0.0000001+0.00000001\cdot\frac{s(s+1)}{2}=99999990\cdot 0.0000001+0.00000001\cdot\frac{99999990(99999990+1)}{2}=50000000.5$$ So we need almost exactly 50 million ETH in tokens, to have a token price of 1 ETH. You can verify this here.