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 = 3³ + 7³ + 1³, 8208 = 8⁴ + 2⁴ + 0⁴ + 8⁴ and 54748 = 5⁵ + 4⁵ + 7⁵ + 4⁵ + 8⁵. 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}, 3ⁿ = {3, 9, 27, 81}, 5ⁿ = {5, 25, 125}, 7ⁿ = {7, 49}, 11ⁿ = {11, 121}, 13ⁿ = {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 = 2⁵ 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 = 3³, 81 = 3⁴}. 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 = 2²×5 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 4² + 9² + 3² = 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 x³ = (845, …, 371), x⁴ = (493, …, 8208) and x⁵ = (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.