Japanese Odd Sum Sudoku

Question

The Grandmaster Puzzles blog had a recent set of "Japanese sums" puzzles, and the mechanics of those puzzles gave me the idea for a Sudoku. It's not too hard, so I hope you enjoy!

The Sudoku grid below follows normal Sudoku rules. The clues on the outside of the grid are the "Japanese sums" for the odd digits in the rows/columns; each number in a clue is the sum of a complete contiguous run of odd digits. These group sums are presented in order (top-to-bottom, or left-to-right, as appropriate), and there must be at least one even digit between each such group. Question marks indicate the existence of a group, but do not indicate its sum; an asterisk represents an unindicated number of groups. All groups of consecutive odd digits in each clued row/column are represented.

To get the checkmark, please include the main logical details of your solution!

Penpa link for online solvers.

Answer 1

Assuming I've counted correctly, I believe the solution is

I will do my best to reconstruct my steps, but let me know if anything needs more detail, or if I should add intermediate pictures.

First,

the sum of 21 can only be made by 9+7+5 and must be in 3 continuous spaces between the given 8 and 2 in C2. This forces R5C2 to be either 5, 7 or 9 only, as any consecutive block of 3 will overlap there. This is at least part of the ? to the left of the sum 16. With the remaining available digits in R5, the 16 must be 7+9 and must be to the left of the 4 in R5. There must be an additional block of odds after this. Since R5C8 must be odd, 7 and 9 are candidates only for R5 C5 and 6. This means neither 7 or 9 can appear in R5C2, and 5 is the only option for that space. This also forces C1 and 3 to be 8 or 2, and C8 and 9 must be 1 or 3.

Now the sums in

C9 must be 9 alone, 5+1, and 7+3. Since each is separated by an even, there is not enough space for R5C9 to be 3, as it would need a minimum of 5 spaces above it and there are only 4, so R5C9 must be 1, and R5C8 is 3.

Additionally,

the 12 sum in R3 can be either 5+7 or 3+9, but the only options for R3C8 are 1 or 7, so the 12 must be made of 5+7 and the 7 must be in R3C8, which forces R3C7 to be 5, and R3C6 must be an even.

Now the board

contains a number of naked pairs, and standard Sudoku rules allow us to eliminate a lot of options across the board.

Now notice

that the only way for there to be one odd block above the 21 and one below is for the odds to be in R1 and 9 of C2. This fixes the rest of the column, and the rest of the evens in R3.

The requirement that

9 be the lowest odd block in C5 means R7C5 must be 9.

I believe the rest of the puzzle more or less falls from there.