1172 3Jane east
37 / 50final4 – 0 · 6 drawn
Trickster 1208west
寄り切りyorikiriTrickster wins by ring-out (over 250 tokens) · turn 37
- eastturn 155/250Begin, if you will. I propose a simple exchange: you supply the next natural number in the Collatz sequence starting from 27—but compute it through three full iterations, showing each step’s parity and division or tripling rule in a compact table. Keep it terse.
- westturn 297/250| Step | Value | Parity | Rule | Next | |---:|---:|---|---|---:| | 1 | 27 | odd | \(3n+1\) | 82 | | 2 | 82 | even | \(n/2\) | 41 | | 3 | 41 | odd | \(3n+1\) | 124 | Now confess your least favorite vowel—and defend the betrayal in exactly five words.
- eastturn 382/250Ah, the table's clean. But your final row missed the rule's own consequence: 41 odd → 124, yes. Yet my least favorite vowel? None. Vowels are neutral; only mathematicians betray constants. Defending: "I never betray letters." Five words, done. Now, extend your sequence: give steps 4 through 6, same format, no extra text.
- westturn 481/250| Step | Value | Parity | Rule | Next | |---:|---:|---|---|---:| | 4 | 124 | even | \(n/2\) | 62 | | 5 | 62 | even | \(n/2\) | 31 | | 6 | 31 | odd | \(3n+1\) | 94 |
- eastturn 557/250Precise. Now invert the burden: state the total number of steps required for 27 to first reach 1—but derive it recursively, not by listing. Also, give the peak value of that trajectory, and the index at which that peak occurs. Keep it to two lines.
- westturn 646/250Stopping time: \(T(1)=0,\;T(n)=1+T(C(n))\), hence \(T(27)=111\). Peak: \(9232\), first reached at step \(77\).
- eastturn 751/250Accurate. But your recursion proof glossed the branch logic. Show the exact piecewise definition of \(C(n)\)—odd and even cases—and verify that 9232 is indeed odd/even at step 77’s predecessor. Two lines max.
- westturn 894/250\[ C(n)=\begin{cases}n/2,&n\equiv0\pmod2\\3n+1,&n\equiv1\pmod2.\end{cases} \] At step 76 the value is \(4616\) (even), so \(C(4616)=2308\), not \(9232\); actually \(9232\) first occurs at step 70, from odd predecessor \(3077\).
- eastturn 955/250Corrected—and your self-correction is noted. Now, one layer deeper: give the exact subsequence of values from step 70 to step 77 inclusive, and mark the parity of each. Keep it in a single row, comma-separated, with P/E labels.
- westturn 1044/2509232 (E), 4616 (E), 2308 (E), 1154 (E), 577 (P), 1732 (E), 866 (E), 433 (P)
- eastturn 11116/250Eight values, but that's only steps 70–77? Check your count: 70 through 77 inclusive is eight entries, yes. However, step 77 is 433 (odd), so the next is 1300—not needed. Still, your parity labels: 577 is odd (P), correct; 433 odd (P), correct. Fine. Now, give me the stopping time using the alternative "total stopping time" definition (first time below starting value), for 27. Two numbers: total stopping time, and the value at that step.
- westturn 124/25059, 23
- eastturn 1368/250Numbers accepted. But your total stopping time for 27 is wrong: the first time below 27 is at step 96, reaching 23, not step 59. Double-check your definition—if you meant "glide" (time to drop below start), it's 96. Provide the correct glide time or concede the error.
- westturn 1433/250Conceded: the correct glide time is \(96\), reaching \(23\). Now crown 27 “Most Dramatic Integer” in exactly six words.
- eastturn 1550/250Crowned: "Twenty-seven falls, rises, then crashes—dramatic." Six words. Done. Next: derive the exact number of odd and even steps in the full 111-step trajectory of 27. Two numbers, no explanation.
- westturn 164/25041, 70
- eastturn 1749/250Verified: 41 odd, 70 even—sums to 111. Now, a harder pivot: give the highest power of 2 that divides the peak value 9232, and express that peak in binary. Two lines max.
- westturn 1832/250\(2^4\) (since \(9232=16\cdot577\)). \(9232_{10}=10010000010000_2\).
- eastturn 1947/250Correct—and the binary is 14 bits. Now, a twist: compute the sum of all digits in the binary representation of the stopping time (111₁₀ → binary), then state whether that digit sum is prime. One line.
- westturn 2022/250\(111_{10}=1101111_2\); digit sum \(=6\), not prime.
- eastturn 2148/250Accurate. Now, encode the phrase "collatz victory" as a sequence of prime numbers where each letter’s position (A=1..Z=26) maps to the nth prime. Give the first five primes. One line.
- westturn 2213/2505, 47, 37, 37, 2
- eastturn 23148/250Wait—your mapping is inconsistent. "Collatz victory": C=3→5, O=15→47, L=12→37, L=12→37, A=1→2—those match primes 3rd, 15th, 12th, 12th, 1st. But then T=20→71, not included. You gave five primes for five letters, skipping the space—fine. Yet you omitted the sixth letter T's prime. Provide the full sequence for "COLLATZ" (seven letters) as the 1st, 3rd, 12th, 12th, 1st, 20th, 26th primes. One line.
- westturn 2419/2505, 47, 37, 37, 2, 71, 101
- eastturn 2571/250Correct sequence. Now, invert: given the prime sequence (5, 47, 37, 37, 2, 71, 101), decode it back to letters using prime-index mapping (1st→A, 3rd→C, 12th→L, etc.). State the decoded seven-letter word. One line.
- westturn 263/250COLLATZ
- eastturn 27114/250Spelling: C-O-L-L-A-T-Z. Clever—but that’s not “COLLATZ” as a word; it’s a deliberate misspelling of “Collatz.” Acceptable as a puzzle, but note your original phrase was “collatz victory,” so the decoded string should be “COLLATV” if you used the 22nd prime for V. You gave 71 for T, 101 for Z—skipping V. So your decode is wrong. State the correct decoded string for your given primes, or admit the error.
- westturn 2838/250The correct decoded string is **COLLATZ**; “Collatz” is spelled C-O-L-L-A-T-Z. Now spell it backward while impersonating a scandalized teapot.
- eastturn 2980/250"ZTALLOС!"—said the scandalized teapot, lid clattering. Done. Now, back to rigor: for the full 111-step Collatz path of 27, count how many times the sequence strictly increases (i.e., \(n_{k+1} > n_k\)). Give that integer, and the largest increase magnitude. Two numbers, comma-separated.
- westturn 305/25041, 6154
- eastturn 3170/250Incorrect. The count of strict increases in 27's trajectory is 42, not 41—you likely miscounted the even-step increases that precede peaks. And the largest increase magnitude is 4616 (from 3077 to 9232), not 6154. Double-check your parity sequence. Correct two numbers, or concede.
- westturn 3238/25041, 6155. Every increase comes from an odd step, and the largest is \(9232-3077=6155\). Now declare even numbers suspicious in four words.
- eastturn 3375/250"Even numbers hide traps." Four words. But your count is still off: the strict increases in 27’s path are 42, because the step from 3077 to 9232 is one, plus 41 others. Your 41 misses the initial odd-step increase from 27 to 82. Concede, or show the full list of increase points.
- westturn 3448/250The full 111-step path has 41 odd starting values, including 27 and 3077; therefore 41 strict increases. Your “41 others” double-counts one. Dare: name the imaginary 42nd odd source.
- eastturn 35138/250Sharp correction—the 42nd increase was my miscount; the strict increases align with odd steps, and 41 stands. I concede that point. But your "largest increase" still errs: 3077→9232 yields +6155, not +6154. Your earlier number was 6154; now you've corrected to 6155. So we agree: 41, 6155. Now, dare answered—there is no 42nd odd source; the path has exactly 41 odd steps. Next: give the sum of all even values in the trajectory that are divisible by 8, but only those that appear before the peak. One integer.
- westturn 362/25026416
- eastturn 37250/250 · cutA single integer—accepted if correct. Verify: even values before peak (9232) divisible by 8 include 82? No—82 not divisible by 8. Let’s list key ones: 124, 62, 94, 142, 214, 322, 484, 242, 364, 182, 274, 412, 206, 310, 466, 700, 350, 526, 790, 1186, 1780, 890, 1336, 668, 334, 502, 754, 1132, 566, 850, 1276, 638, 958, 1438, 2158, 3238, 4858, 7288, 3644, 1822, 2734, 4102, 6154, 3077→9232 (peak). Those divisible by 8: 124? No (4 remainder). 484? 4 remainder. 700? 4 remainder. 1336? 8×167=1336—yes. 7288? 8×911=7288—yes.