After filling the grid, reading the columns as alphabet positions, ie 16 18 15 4 21 3 5 → PRODUCE and so on, gave the message “produce 38 hailstone numbers from 988 and fill grid”. The entries in the grid then had to be erased, apart from the circled cells, and refilled using the hailstone sequence starting 988, 494, 247, 742, 371 and so on.

This is one possible solution path.
As 1dn = N^{4} has two digits,
it must be 16 or 81, with N being 2 or 3.
As 8ac starts with the last digit of 1dn and must be greater than it
(coming later in the sequence of clues), 1dn can’t be 81,
so **N = 2**, 1dn = 16 and n is either 1 (n = N/2) or 4 (N = n/2).

For 8ac = 2P + PP to start with 6, we must have **P = 7**, 8ac = 63
and p is either 14 (P = p/2) or 22 (p = 3P + 1).
Now 4dn = NP = 14, which means that **p = 22**
(as no letter value is equal to a grid entry).

The clue order gives 2dn < 4dn and from the grid 2dn ends in 3, so 2dn = 13. As the sum of two adjacent hailstone numbers it’s either Z + Z/2 = 3Z/2 or Z + 3Z + 1 = 4Z + 1, for some integer value Z; 3Z/2 = 13 doesn’t have any integer solutions, so we have 4Z + 1 = 13, giving Z = 3, so R and r have the values 3 and 10 (3×3 + 1) in either order.

From the clue order, 14 < 10ac < 16, so 10ac = 15;
its clue gives R + T = 15, so T is either 5 or 12.
We know 25dn = 7t is less than 63, so t < 9.
T can’t be 5 because its only adjacent hailstone numbers
are 10 (preceding) and 16 (following),
so **T = 12**, **t = 6**, **R = 3** and **r = 10**,
giving 25dn = 42 and 28dn = 91.
From the clues, 26dn = n + 10 is less than 2dn = 13,
and 10 and 12 are already used as letter values (r and T),
so 26dn = 11 and **n = 1**.

From the clues, 5dn = 2m^{2} is less than 42, so m ≤ √(40/2) ≈ 4.47.
The values {1, 2, 3} are already assigned to {n, N, R}, so **m = 4**.
The hailstone number following 4 is 2, but that’s already assigned, so **M = 8** (preceding m) and 5dn = 32.

The clue ordering 28dn = 91 < 29dn < 27dn < 28ac means they all start with 9, which makes 28ac = A + 7 = 99 in the grid, so **A = 92**.
The value for a is then either 46 (following A) or 184 (preceding), but 46 is too small for 6dn = 343 + 8a to have four digits, so **a = 184** and 6dn = 1815.

The minimum value for 27dn = 4B is 94 (two clues on from 28dn = 91, with 92 already assigned); the only multiple of 4 in the range is 27dn = 96, with **B = 24**, which then makes 24ac = 68.
The hailstone number following 24 is 12, which is already assigned to T, so **b = 48** (preceding B), which makes 21ac = 580.

In the grid we have 1_1 for 2ac = 12u + 3, which is a multiple of 3, ie one of {111, 141, 171}, for which u is respectively {9, 11.5, 14}.
We can rule out the non-integer, and 14 is the value of 4dn, so **u = 9** and 2ac = 111.
The grid has 6_1 for 31ac = 24H + 9; the matching multiples of 3 are {621, 651, 681}, of which only 681 give an integer value for H, so **H = 28**, giving 31ac = 681.
The hailstone neighbours of 28 are {9, 14, 56}, but 9 = u and 14 = 4dn, so **h = 56**, giving 14ac = 112 and 19ac = 248.
U could be 18 (preceding u) or 28 (following), but 28 is now assigned to H, so **U = 18**, giving 7dn = 21.

The grid has 31_ for 5ac = 6g, so g is either 52 or 53, with hailstone neighbours of {17, 26, 104} or {106, 160} respectively.
As 30ac = 791 + G ends in 1, G must end in 0, so **G = 160** with **g = 53**, giving 5ac = 318 and 30ac = 951.

The grid now has 20_5 for 19dn = 53D + 160, so D ends in 5 and is between (2005 - 160)/53 ≈ 34.81 and (2095 - 160)/53 ≈ 36.51, ie **D = 35**, giving 19dn = 2015 and 17dn = 142.
For d, the hailstone neighbours of 35 are {70, 106}; 70 is too small to make 3dn = 15 + d a three-digit number, so **d = 106**, giving 3dn = 121, 20ac = 141 and 22ac = 213.

The grid has 8_1 for 9ac = 35C + 6, so 35C ends in 5, meaning C is odd.
29dn = 24 + c is between 91 and 96, with 92 already assigned, so c is one of {69, 70, 71}.
There are no odd neighbours of 69 or 71 (any odd Z is followed by 3Z + 1 (even) and could be preceded by 2Z (even) or (Z - 1)/3, not an integer in either case), so **c = 70**, giving 29dn = 94 and 11dn = 538.
The neighbours of 70 are {23, 35, 140}; as 35 is already assigned to D, **C = 23**, giving 9ac = 811.

The grid has _46 for 18dn = k^{2} + 46, which is between 318 and 538, so it’s either 346 or 446, with 346 - 46 = 300 isn’t a square number, so **k = 20**, giving 18dn = 446.
Its neighbours are {10, 40} but 10 = r, so **K = 40**, giving 12ac = 155

The grid has 1_5 for 17ac = 2S + 1, which is between 112 and 121, so 17ac = 115, with **S = 57**, also giving 23dn = 398.
The grid now has 518 for 13dn = 3s + 2, so **s = 172**.

The grid has 4_1 for 25ac = 24 + 9w, a multiple of 3 that is less than 18dn = 446, ie either 411 or 441.
The latter doesn’t give an integer value for w, so **w = 43**, giving 25ac = 411.
From the grid, 16dn = 6W + 35 ends in 5, so W must end in 0 or 5; the hailstone neighbours of 43 are {86, 130}, so **W = 130**, giving 16dn = 815.
Now we have 83_ for 16ac = 49X + 6; the only value that fits is **X = 17**, giving 16ac = 839.

The grid has 2_1 for 15dn = 219 + x, so x ends in 2.
The neighbours of X = 17 are {34, 52}, so **x = 52**, giving 15dn = 271.
The grid is now filled, as shown.

The preamble hints at decoding the grid downwards (thematically, like falling hailstones) rather than across. Looking at the first column, 16181542 can be split into alphabet positions (using A = 1, B = 2 etc) in a few ways. As 61, 81, 54 and 42 are not in the range 1 to 26, there must be splits between their digits, ie 16|18|15|4|2, with corresponding options AF/P, AH/R, AE/O, D, B/U (“borrowing” the 1 from the top of the second column), from which PRODU looks most promising. Extending this to all the columns, the message PRODUCE 38 HAILSTONE NUMBERS FROM 988 AND FILL GRID can be extracted.

The first 38 numbers in the hailstone sequence starting from 988 are: 988, 494, 247, 742, 371, 1114, 557, 1672, 836, 418, 209, 628, 314, 157, 472, 236, 118, 59, 178, 89, 268, 134, 67, 202, 101, 304, 152, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17. After clearing the contents of all but the circled cells, these can be fitted in to complete the final grid.

To start with, 6dn is 1_1_, so it can only be 1114, which leaves 19dn = 1672. There are then only single options for 16ac = 209, 17ac = 314, 30ac = 628 and 31ac = 988. 5ac and 13dn are both _18, so they must use {118, 418} in either order, which leaves 178 as the only option for 1_8 at 19ac, forcing 17dn = 371.

20ac and 18dn both have a central 4, so they must be {247, 742} in either order. For 15dn, there’s no number to fit _77, so it must be 472, making 14ac = 134, 20ac = 742 and 18dn = 247, and then 24ac = 76.

Ending in 8, 23dn can only be 268 (as 118 and 418 are accounted for), forcing 22ac = 152. Starting with 2, 16dn is either 202 or 236; for 21ac there’s no number to match 6_6, so 16dn = 202 and 21ac = 236. With 0 in the middle, 11dn is either 101 or 304; 10ac can’t be 13 (not in the list), so 11dn = 101 and 10ac = 11. Now 13dn can’t be 118 because there’s no number ending in 1 for 12ac, so 13dn = 418 and 5ac = 118. Only one number starting with 1 remains, so 9ac = 157.

Ending in 4, 12ac is either 304 or 494; as there are no numbers ending in 3 for 3dn, 12ac = 494 and 3dn = 304. Then 2ac has 3 in the middle, so 2ac = 836, and the last remaining three-digit number is 25ac = 557.

Of the two-digit numbers, there’s only one starting with 6, so 4dn = 67. That leaves only one ending in 7, so 7dn = 17. That leaves only one starting with 1, so 5dn = 19. Ending in 4, 29dn is either 34 or 44, but there’s no number ending in 3 for 28ac, so 29dn = 44, which forces 28ac = 34 and 28dn = 38.

The remaining numbers are {22, 29, 58, 59, 88, 89}. The 22 can’t go at 8ac because that would make 2dn 82, so 1dn = 22, forcing 8ac = 29 and 2dn = 89. Then 27dn can only be 59, forcing 25dn = 58 and then 26dn = 88, completing the final grid.