Talk:ResPlicate
Abstract tracking of 6 3 10 1 6 2 (2k) 1 pattern
Last line and the 6th last line are identical, confirming conjecture. (The last vertical ellipsis represents k-5 steps.)
6 3 10 1 6 2 (2k) 1 10 1 6 2 (2k) 1 10 1 6 2 (2k) 1 | 10 1 6 2 (2k) 1 10 1 6 2 (2k) 1 6 2 (2k) 1 10 1 | 6 2 (2k) 1 6 2 (2k) 1 6 2 (2k) 1 | 6 2 (2k) 1 10 1 6 2 (2k) 1 10 1 (2k) 1 | 6 2 (2k) 1 (2k) 1 6 2 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 | (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 (2k) 1 10 1 (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-12) 0's… 10 1 (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 | (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-11) 0's… (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-11) 0's… (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 | …2(k-11) 0's… (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 0 0 0 0 …2(k-11) 0's… (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 ⋮ (2k) 1 (2k) 1 10 1 (2k) 1 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 (2k) 1 10 1 (2k) 1 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-12) 0's… 10 1 (2k) 1 (2k) 1 0 0 0 0 (2k) 1 | (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-11) 0's… (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-11) 0's… (2k) 1 (2k) 1 0 0 0 0 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 | …2(k-11) 0's… (2k) 1 (2k) 1 0 0 0 0 (2k) 1 0 0 0 0 …2(k-11) 0's… (2k) 1 (2k) 1 0 0 0 0 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 ⋮ (2k) 1 (2k) 1 0 0 0 0 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 (2k) 1 0 0 0 0 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-12) 0's… 0 0 0 0 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-11) 0's… 0 0 (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-11) 0's… (2k) 1 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-11) 0's… 0 0 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-8) 0's… 0 0 (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-8) 0's… (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-8) 0's… (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-5) 0's… 6 2 (2k) 1 (2k) 1 6 2 | …2(k-4) 0's… …2(k-4) 0's… (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 ⋮ (2k) 1 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 (2k) 1 6 2 (2k) 1 (2k) 1 6 2 …2(k-5) 0's…
--Ørjan (talk) 15:09, 8 February 2014 (UTC)
6 3 10 1 6 2 n 1 for Odd N
Here's what the odds up to 500 do in the long run. If anyone sees a pattern, let me know.
Die: (3 7 9 15 17 19 23 31 33 35 37 43 47 49 53 55 61 65 67 71 73 77 79 81 85 87 89 91 93 101 103 111 115 125 127 129 133 137 145 147 149 159 165 167 169 171 173 175 177 181 185 189 191 195 199 201 203 209 213 215 217 221 223 227 229 231 235 237 239 241 245 247 253 255 259 261 263 265 267 273 277 279 281 283 285 287 289 295 297 299 301 303 307 309 311 319 321 329 331 337 339 341 343 347 349 351 355 357 359 361 363 365 367 371 373 377 379 389 391 393 397 399 401 403 411 413 415 417 421 423 425 427 429 437 439 445 451 455 457 465 469 471 477 479 481 483 491 493 495 497) Grow: (11 13 21 29 39 51 63 97 139 141 151 163 183 187 193 205 207 249 291 313 325 327 345 405 409 435 441 453 475 487 489) Twos: (1 5 25 27 41 45 57 59 69 83 95 99 105 107 109 113 117 119 121 123 131 135 143 153 155 157 161 179 197 211 219 233 251 257 269 271 275 293 305 317 323 335 353 369 375 381 383 385 387 395 407 419 431 433 443 447 449 461 463 467 473 485 499) Osc: (75 225 243 315 333 459)
--(this comment by Quintopia at 01:21, 14 February 2014 UTC; please sign your comments with ~~~~)
3 1 6 3 2 1 4 n
Proof of conjecture for n>=12:
3 1 6 3 2 | 1 4 n 1 4 n | 6 3 2 6 3 2 n n n n 2 n n n | n 0 2 n n n n 0 2 n n n n 0 n 0 2 n n n n 0 2 n n n n 0 …2n n's… …n+12 n's…
Testing shows it is also true for 0,1,3,6, and 11. The rest of 5,7,8,9, and 10 appear somewhat dubious. --Ørjan (talk) 23:55, 9 February 2014 (UTC)
Disproof for 5,7-10:
All of n=5,7,8,9 eventually give sequences containing the following subsequence:
…n-1 n's… 0 …n-1 n's… 0 …n-1 n's… 0 …n-1 n's…
Also, no number higher than n occurs otherwise in the sequence. From this we see that the previous command to run before this subsequence can delete at most n+1 integers from its front. In all the possible alignment cases, the subsequence reappears inside its produced result, thus preserving 0's indefinitely.
For n=10 a similar argument works for the subsequence
…8 10's… 0 0 …8 10's… 0 0 …8 10's… 0 0 …8 10's….