For each number in one of the sequences,
the next number was produced by taking the sum of
the nth powers of its digits,
with n = 3, 4 and 5 for the respective sequences.
Each sequence led to a number that generated itself
as the next number, thereby “ending” the sequence.
The three numbers were 371 = 3^{3} + 7^{3} + 1^{3},
8208 = 8^{4} + 2^{4} + 0^{4} + 8^{4}
and 54748 = 5^{5} + 4^{5} + 7^{5} + 4^{5} + 8^{5}.
These are known as Narcissistic numbers: n-digit numbers
that are equal to the sum of the nth powers of their digits.
There aren’t many of them:
there is just one with 10 digits (4679307774)
and there’s a 39-digit one (the largest possible).
The term was hidden in the last clues of the puzzle:
“N(A + r) − C”,
“YY + Y = I − S”,
“SYS = S(I − S)”,
“M + E = T − I + C”
and “N + U + m + S”.

This is one possible solving path.
It will be useful to prepare a list of
the possible prime powers for the clue letters.
The powers of 2 less than 200 are {1, 2, 4, 8, 16, 32, 64, 128},
3^{n} = {3, 9, 27, 81},
5^{n} = {5, 25, 125},
7^{n} = {7, 49},
11^{n} = {11, 121},
13^{n} = {13, 169}
and the rest are just the primes from 17 to 199.
Sorting them, we have {1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17,
19, 23, 25, 27, 29, 31, 32, 37, 41, 43, 47, 49, 53, 59, 61, 64,
67, 71, 73, 79, 81, 83, 89, 97, 101, 103, 107, 109, 113, 121,
125, 127, 128, 131, 137, 139, 149, 151, 157, 163, 167, 169,
173, 179, 181, 191, 193, 197, 199}.

For 25dn = Y(Y + 1) to have two digits we need Y ≥ 3.
Since 25dn says Y(Y + 1) = I − S
and 26dn gives YS = I − S,
we know Y(Y + 1) = YS, meaning Y + 1 = S.
That makes 26dn = Y(Y + 1)^{2}
and for that to have two digits we need Y ≤ 3.
Therefore **Y = 3**, **S = 4**,
**25dn = 12** and **26dn = 48**.
From 25dn we know I − S = 12, so **I = 16**.

For 1dn we now have U + 4(m + 1) to give a two-digit value,
so m < 24.
For 26ac we have O = 4_, ie one of {41, 43, 47, 49};
this is also 4 + UM, so UM must be odd, making U and M both odd.
For 8ac = 4 + Um/4 to be an integer, Um must be a multiple of 4;
but U is odd, so m is a multiple of 4, ie one of {8, 16}
(as 4 is already taken and 20 and 24 aren’t powers of a prime),
which makes Um a multiple of 8.
For 25ac we have 4Um = 1__, so Um is in the range [25, 49],
ie one of {32, 40, 48},
and U isn’t 1 because that would make m too big.
We can rule out 32 = 2^{5} because
it doesn’t have any odd factors for U;
for 48 the only odd factor is 3 but that’s already taken.
So Um = 40, **U = 5**, **m = 8**,
**25ac = 160**, **8ac = 14**,
**1dn = 41** and **17ac = 17**.

Now 26ac = 4 + 5M, which must end in 9 (M being odd),
so the grid entry is **26ac = 49 = O**
with **M = 9**.
This also completes **3dn = 18**.
Now 18ac = 4 + 16G/9 + A, and for that to be an integer
G must be a multiple of 9,
ie one of {27 = 3^{3}, 81 = 3^{4}}.
With 81, 16×81/9 = 144 is too big for the two-digit grid entry,
so **G = 27**.

From the first part of 27dn we have E + 9 = 9_ in the grid,
so E is in the range [81, 90], ie one of {81, 83, 89},
with 27dn being one of {90, 92, 98}.
From the second part of 30ac we have 144T + E + 4 = 2__ in the grid;
if T is 2 then the minimum of 288 + 81 + 4 = 373 is too big,
so **T = 1**.
For 31ac we now have 432N to fit one of {80__, 82__, 88__},
so N is between 8000/432 ≈ 18.5 and 8899/432 ≈ 20.6;
20 = 2^{2}×5 isn’t a prime power,
so **N = 19** and **31ac = 8208**,
with **27dn = 92** and **E = 83**;
this also completes **23ac = 472**,
**30ac = 231**, **13dn = 371**
and **29dn = 36**.
For 27dn we have C − 15 = 92, so **C = 107**,
which makes **19dn = 547**.

For 21dn, 432r − 1 = _63 in the grid;
3 for r would give a four-digit number, and 1 is taken,
so **r = 2**, **21dn = 863**
and **5dn = 514**.
Now 5ac = 87 + 8e = 5__ in the grid,
so e is ≥ (511 − 87)/8 = 53 and ≤ (599 − 87)/8 = 64,
ie one of {53, 59, 61, 64}.
That makes 22ac = e + 652 one of {705, 711, 713, 716},
but from the grid we know the middle digit is 1.
22dn is now filled as 701 = 16R − 4e − 79,
so R = (780 + 4e)/16;
the three possibilities for e give {63.5, 64, 64.75},
so **R = 64** with **e = 61**,
**22ac = 713**, **5ac = 575**,
**2dn = 845**, **14dn = 493**
and **22dn = 701**.
Now 18ac is filled as 95 for A + 52, so **A = 43**,
which makes **20ac = 98** and **24dn = 748**.
That completes 43 for 28ac = P − 4, so **P = 47**,
making **11ac = 134** and **6dn = 737**.
Also, we have 231 for 30ac = H + 62, so **H = 169**.

10ac = a − 4 is 8_ in the grid,
so a is in the range [84, 93] and the only available number is
**a = 89**, giving **10ac = 85**,
**4dn = 45** and **15dn = 882**.
For 9dn = o + 172 we have __1 in the grid, so o ends in 9.
For 2ac = 47o + 83W + 2 we have 8_14 in the grid;
47o ends in 3, so 83W ends in 9 and W ends in 3.
16ac = W + 15 has two digits, so W < 85,
restricting it to one of {13, 23, 53, 73}.
From the grid, 7dn = F + W − 80 = 54, so F + W = 134,
making F one of {121, 111, 81, 61},
from which we can eliminate 111 = 3×37 and 61 = e;
that leaves W as one of {13, 53}.
From 2ac we have o = (8_12 − 83W)/47.
If W is 13 then o is between (8012 − 83×13)/47 ≈ 147.5
and (8912 − 83×13)/47 ≈ 166.7,
for which the only available value ending in 9 is 149;
but 47×149 + 83×13 + 2 = 8084 doesn’t match 8_14 in the grid.
Therefore **W = 53**,
making **16ac = 68**, **F = 81**,
and o is between (8012 − 83×53)/47 ≈ 76.9
and (8912 − 83×53)/47 ≈ 96.02,
ie **o = 79**.
That completes the grid with **2ac = 8114**,
**12ac = 155**, **9dn = 251**
and **12dn = 169**.

In ascending order, the clue variables’ values are
{1, 2, 3, 4, 5, 8, 9, 16, 19, 27, 43, 47, 49, 53, 61, 64, 79, 81, 83, 89, 107, 169}
and the letters spell **TrY SUmMING A POWeR oF EaCH**,
which the preamble tells us relates to
the digits of each number in a sequence.
We’re also told that one of the sequences starts with 14dn = 493.
The sum of the squares of its digits is
4^{2} + 9^{2} + 3^{2} = 106, not a grid entry;
the sum of cubes is 64 + 729 + 27 = 820, also not a grid entry;
but the sum of fourth powers is 256 + 6561 + 81 = 6898,
which can be found as 16ac = 68 and 20ac = 98.
Applying the same sum-of-fourth-powers-of-digits rule to 6898
gives 16049 = 25ac & 26ac; then 16049 leads to 8114 = 2ac,
which leads to 4354 = 28ac & 7dn, then 1218 = 25dn & 3dn,
4114 = 1dn & 8ac, 514 = 5dn, 882 = 15dn, 8208 = 31ac,
and the next value is 8208 again,
so we’ve reached the end of that sequence:
(**493, 6898, 16049, 8114, 4354, 1218, 4114, 514, 882, 8208**, …).

The other sequences are presumably also sums of powers (other than 4) of digits, using the remaining 23 grid entries, namely {17, 36, 45, 48, 85, 92, 95, 134, 155, 169, 231, 251, 344, 371, 472, 547, 575, 701, 713, 737, 748, 845, 863}. Trying cubes of their digits gives, respectively, {344, 243, 189, 576, 637, 737, 854, 92, 251, 946, 36, 134, 155, 371, 415, 532, 593, 344, 371, 713, 919, 701, 755}, and eliminating the ones that aren’t grid entries leaves the pairs {17 → 344, 92 → 737, 134 → 92, 155 → 251, 231 → 36, 251 → 134, 344 → 155, 371 → 371, 701 → 344, 713 → 371, 737 → 713, 845 → 701}. These can be combined into the partial sequence (344, 155, 251, 134, 92, 737, 713, 371, …) with either (17) or (845, 701) at the front.

Leaving out the numbers in that partial sequence
and trying squares on the remaining grid entries gives
{17 → 50, 36 → 45, 45 → 41, 48 → 80,
85 → 89, 95 → 106, 169 → 118, 231 → 14,
472 → 69, 547 → 90, 575 → 99, 701 → 50,
748 → 129, 845 → 105, 863 → 109},
from which only the partial sequence (36, 45, 41) can be formed,
leaving most entries unused.
So, trying fifth powers gives
{17 → 16808, 36 → 8019, 45 → 4149, 48 → 33792,
85 → 35893, 95 → 62174, 169 → 66826, 231 → 276,
472 → 17863, 547 → 20956, 575 → 23057, 701 → 16808,
748 → 50599, 845 → 36917, 863 → 40787},
in which only 472 → 17863 can be split into grid entries,
namely 17 & 863.
Starting from there produces the sequence
(**472, 17|863, 575|95, 85|231, 36|169, 748|45, 547|48**, 547|48, …),
split into grid entries as shown
(54748 not being 54|748 because 54 is used in the 4th powers sequence).
That leaves only {701, 845} unused, so the cube-sums sequence can be completed as
(**845, 701, 344, 155, 251, 134, 92, 737, 713, 371**, …).

The three sequences are therefore x^{3} = (845, …, 371),
x^{4} = (493, …, 8208)
and x^{5} = (472, …, 54748),
so we can highlight the starting entries 2dn and 23ac.
Each sequence ends up repeating one number infinitely,
so the “last” terms in the sequences are the numbers
**371**, **8208** and **54748**.