(not including $0^2 + b^2$ and only including odd numbers:)
Starting at $5 = 1^2 + 2^2$
then
$13 = 2^2 + 3^2$
$17 = 4^2 + 1^2$
$25 = 3^2 + 4^2$
$29 = 2^2 + 5^2$
What is the equation/s to find the
$\space n^{th}\space $
number in the sequence (eg: what number is the
$\space 397^{th}\space $
number or
$\space 100,578^{th}\space $ number in the sequence/s?) If someone were to have found this equation would that be new?
In this answer, I show it is unlikely that a single formula exists to meet your [global] criteria –– that it is only possible to meet your criteria on a set by set basis.
The numbers in the question are hypotenuse values of the first five [smallest primitive] Pythagorean triples in order of size. The following formula generate $\space sets\space$ of $\space C$-values from all triples where $\space GCD(A,B,C)=(2x-1)^2, x\in\mathbb{N},\space$ which includes all primitives and a few odd square multiples of primitives. These are the $\space values \space$ in the table below.
$$C=(2n-1)^2+2(2n-1)k+2k^2$$
$$\begin{array}{c|c|c|c|c|c|c|c} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 & \cdots\\ \hline Set_1 & 5 & 13& 25& 41& 61 & \underline{\mathit{85}} \\ \hline Set_2 & 17 & 29 & \mathbf{45} & \underline{\mathit{65}} & 89 & 117 \\ \hline Set_3 & 37 & 53 & 73 & 97 & \mathbf{125} & 157 \\ \hline Set_4 & \underline{\mathit{65}} & \underline{\mathit{85}} & 109 & 137 & 69 & 205 \\ \hline Set_5 &101 & 125 & \mathbf{153} & 185 & 221 & 261 \\ \hline \end{array} $$ The $\textbf{bold}$ numbers are from imprimitive triples and those $\underline{\textit {underlined italic}}$ are duplicates among primitive triples.
There will be an imprimitive any time $GCD\big(2(n-1),k\big)>1.$
There are $\space 2^{f-1} \space $ duplicates where $ \space f \space $ is then number of distinct prime factors of $\space C.$
Arranging the table entries in order of size, we can appreciate the difficulty of finding a pattern in the $\space (n,k) \space$ values needed to generate this order. If we enlarge the list to include more sets or higher $\space k$-values, the sequence changes, even in the middle of the list shown.
\begin{align*} F(1,1)&= (5) = 1^2 + 2^2\\ F(1,2)&= (13) = 3^2 + 2^2\\ F(2,1)&= (17) = 1^2 + 4^2\\ F(1,3)&= (25) = 3^2 + 4^2\\ F(2,2)&= (29) = 5^2 + 2^2\\ F(3,1)&= (37) = 1^2 + 6^2\\ F(1,4)&= (41) = 5^2 + 4^2\\ F(2,3)&= (45) = 3^2 + 6^2\\ F(3,2)&= (53) = 7^2 + 2^2\\ F(1,5)&= (61) = 5^2 + 6^2\\ F(2,4)&= (65) = 1^2 + 8^2 = 7^2 + 4^2\\ F(4,1)&= (65) = 1^2 + 8^2 = 7^2 + 4^2\\ F(3,3)&= (73) = 3^2 + 8^2\\ F(1,6)&= (85) = 7^2 +6^2 = 9^2 + 2^2\\ F(4,2)&= (85) = 7^2 +6^2 = 9^2 + 2^2\\ F(2,5)&= (89) = 5^2 + 8^2\\ F(3,4)&= (97) = 9^2 + 4^2\\ F(5,1)&= (101)= 1^2 + 10^2\\ F(4,3)&= (109)= 3^2 + 10^2\\ F(3,5)&= (125)= 5^2 + 10^2 = 11^2 + 2^2\\ F(5,2)&= (125)= 5^2 + 10^2 = 11^2 + 2^2\\ F(4,4)&= (137)= 11^2 + 4^2\\ F(5,3)&= (153)= 3^2 + 12^2\\ F(4,5)&= (169)= 5^2 + 12^2\\ F(5,4)&= (185)= 13^2 + 4^2 = 11^2 + 8^2\\ F(5,5)&= (221)= 5^2 + 14^2 = 11^2 + 10^2\\ F(5,6)&= (261)= 15^2 + 6^2\\ &\vdots\\ \end{align*} In the spreadsheet excerpt below (similar to Cantor's pairing ) we can see $\textbf{bolded}$ numbers which break with the "obvious" pattern.
Still, the $\space k^{th} \space$ triple in any set is predictable. For instance, the $\space 13^{th} \space $ number in $\space Set_{13}\space$ is $\space F(13,13)=1613= 13^2 + 38^2,\space$ and the $\space 5001^{st} \space $ number in $\space Set_{5000}\space$ is $\space F(5000,5001)=(250010001) = 5001^2 + 15000^2.$