[math-fun] a(n), a(n+1) and a(n+2) don't share any digit
Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.
Coincidentally, Éric's earlier posts made me think of this question, so I am ready with the answer: the only terms are 0,1,2,3,4,5,6,7,8. For all higher n, either n and n+1 will have the same first digit, or n+1 and n+2 will have the same first digit. On Thu, Oct 10, 2019 at 7:38 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.
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Hello Allan, once again, my question was vague :-(( Let me rephrase that properly: How many terms does the finite lexicographically earliest seq of distinct terms contain such that a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. My first terms are: S = 1,2,3,4,5,6,7,8,9,10,23,45,16,... We see that no three successive terms of S share any digit so far. Best, É. à+ É. Catapulté de mon aPhone
Le 10 oct. 2019 à 17:34, Allan Wechsler <acwacw@gmail.com> a écrit :
Coincidentally, Éric's earlier posts made me think of this question, so I am ready with the answer: the only terms are 0,1,2,3,4,5,6,7,8. For all higher n, either n and n+1 will have the same first digit, or n+1 and n+2 will have the same first digit.
On Thu, Oct 10, 2019 at 7:38 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.
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Oh, it was my mistake. I read it as "n, n+1 and n+2 have no common digits", which is absolutely not what you wrote. My apologies. I have no insight to offer into the problem you were actually posing. On Thu, Oct 10, 2019 at 1:05 PM Éric Angelini <eric.angelini@skynet.be> wrote:
Hello Allan, once again, my question was vague :-(( Let me rephrase that properly: How many terms does the finite lexicographically earliest seq of distinct terms contain such that a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0.
My first terms are:
S = 1,2,3,4,5,6,7,8,9,10,23,45,16,...
We see that no three successive terms of S share any digit so far. Best, É.
à+ É. Catapulté de mon aPhone
Le 10 oct. 2019 à 17:34, Allan Wechsler <acwacw@gmail.com> a écrit :
Coincidentally, Éric's earlier posts made me think of this question, so I am ready with the answer: the only terms are 0,1,2,3,4,5,6,7,8. For all higher n, either n and n+1 will have the same first digit, or n+1 and n+2 will have the same first digit.
On Thu, Oct 10, 2019 at 7:38 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.
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That is a lovely question, and finally after understanding it I think the sequence has 496 terms. The last three numbers are 971, 862, 3405 and the smallest number not used in the sequence is 143. Making the numbers distinct seems rather important, otherwise a loop may form, not that I checked it. For those desperate to know, the sequence I got is: 0 1 2 3 4 5 6 7 8 9 10 23 45 16 20 34 15 26 30 14 25 36 17 24 35 18 27 39 40 12 37 46 19 28 43 50 21 38 47 29 13 48 52 31 49 56 32 41 57 60 42 51 63 70 54 61 72 53 64 71 58 62 73 59 68 74 90 65 78 91 203 67 81 92 75 80 69 123 84 76 93 82 104 79 83 102 94 85 103 96 87 105 234 86 95 107 236 89 140 235 97 106 238 457 109 263 458 170 239 456 108 237 459 160 273 98 145 206 378 149 205 367 148 209 356 147 208 359 146 207 358 164 270 385 169 204 357 168 240 375 186 249 305 167 248 309 156 247 308 159 246 307 158 264 370 185 269 304 157 268 340 175 286 349 150 267 348 190 256 347 180 259 346 178 250 364 179 258 306 174 285 360 194 257 368 401 275 369 184 502 376 189 245 603 187 254 390 176 284 350 196 274 380 165 279 384 501 276 389 154 260 379 415 280 396 417 508 293 416 507 283 419 506 278 134 509 287 136 405 289 137 406 295 138 407 265 139 408 526 173 409 528 163 470 298 135 460 297 153 468 290 315 467 802 193 465 702 183 469 520 317 486 529 130 476 582 301 479 562 310 478 296 351 480 627 195 403 628 197 345 602 198 354 607 128 394 560 127 398 450 126 387 490 125 386 497 120 365 487 129 503 647 182 395 604 172 538 496 201 537 489 162 530 498 216 570 438 192 567 430 218 569 374 210 568 397 124 580 637 142 589 630 214 578 609 132 475 608 213 495 670 231 485 679 230 418 576 302 481 579 320 461 587 329 410 586 327 491 605 328 471 590 326 714 598 362 701 548 392 601 547 382 610 549 372 618 504 723 619 540 728 316 594 708 261 435 709 281 436 597 801 243 596 710 324 658 719 342 650 718 294 536 780 219 453 678 291 534 680 217 439 685 271 493 805 612 437 809 152 463 789 215 634 790 251 483 690 512 473 689 510 423 687 519 402 638 517 420 639 518 427 693 581 472 903 516 428 703 561 429 730 615 482 739 651 804 729 361 584 720 319 546 782 391 564 807 312 645 798 321 640 758 912 643 705 812 649 573 810 426 539 781 462 593 817 620 543 791 682 3045 917 826 3054 971 862 3405 Steve
On Oct 10, 2019, at 1:04 PM, Éric Angelini <eric.angelini@skynet.be> wrote:
Hello Allan, once again, my question was vague :-(( Let me rephrase that properly: How many terms does the finite lexicographically earliest seq of distinct terms contain such that a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0.
My first terms are:
S = 1,2,3,4,5,6,7,8,9,10,23,45,16,...
We see that no three successive terms of S share any digit so far. Best, É.
à+ É. Catapulté de mon aPhone
Le 10 oct. 2019 à 17:34, Allan Wechsler <acwacw@gmail.com> a écrit :
Coincidentally, Éric's earlier posts made me think of this question, so I am ready with the answer: the only terms are 0,1,2,3,4,5,6,7,8. For all higher n, either n and n+1 will have the same first digit, or n+1 and n+2 will have the same first digit.
On Thu, Oct 10, 2019 at 7:38 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.
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Pretty question! I confirm Steve's list. --Michael On Thu, Oct 10, 2019 at 2:03 PM Lucas, Stephen K - lucassk <lucassk@jmu.edu> wrote:
That is a lovely question, and finally after understanding it I think the sequence has 496 terms. The last three numbers are 971, 862, 3405 and the smallest number not used in the sequence is 143. Making the numbers distinct seems rather important, otherwise a loop may form, not that I checked it.
For those desperate to know, the sequence I got is: 0 1 2 3 4 5 6 7 8 9 10 23 45 16 20 34 15 26 30 14 25 36 17 24 35 18 27 39 40 12 37 46 19 28 43 50 21 38 47 29 13 48 52 31 49 56 32 41 57 60 42 51 63 70 54 61 72 53 64 71 58 62 73 59 68 74 90 65 78 91 203 67 81 92 75 80 69 123 84 76 93 82 104 79 83 102 94 85 103 96 87 105 234 86 95 107 236 89 140 235 97 106 238 457 109 263 458 170 239 456 108 237 459 160 273 98 145 206 378 149 205 367 148 209 356 147 208 359 146 207 358 164 270 385 169 204 357 168 240 375 186 249 305 167 248 309 156 247 308 159 246 307 158 264 370 185 269 304 157 268 340 175 286 349 150 267 348 190 256 347 180 259 346 178 250 364 179 258 306 174 285 360 194 257 368 401 275 369 184 502 376 189 245 603 187 254 390 176 284 350 196 274 380 165 279 384 501 276 389 154 260 379 415 280 396 417 508 293 416 507 283 419 506 278 134 509 287 136 405 289 137 406 295 138 407 265 139 408 526 173 409 528 163 470 298 135 460 297 153 468 290 315 467 802 193 465 702 183 469 520 317 486 529 130 476 582 301 479 562 310 478 296 351 480 627 195 403 628 197 345 602 198 354 607 128 394 560 127 398 450 126 387 490 125 386 497 120 365 487 129 503 647 182 395 604 172 538 496 201 537 489 162 530 498 216 570 438 192 567 430 218 569 374 210 568 397 124 580 637 142 589 630 214 578 609 132 475 608 213 495 670 231 485 679 230 418 576 302 481 579 320 461 587 329 410 586 327 491 605 328 471 590 326 714 598 362 701 548 392 601 547 382 610 549 372 618 504 723 619 540 728 316 594 708 261 435 709 281 436 597 801 243 596 710 324 658 719 342 650 718 294 536 780 219 453 678 291 534 680 217 439 685 271 493 805 612 437 809 152 463 789 215 634 790 251 483 690 512 473 689 510 423 687 519 402 638 517 420 639 518 427 693 581 472 903 516 428 703 561 429 730 615 482 739 651 804 729 361 584 720 319 546 782 391 564 807 312 645 798 321 640 758 912 643 705 812 649 573 810 426 539 781 462 593 817 620 543 791 682 3045 917 826 3054 971 862 3405
Steve
On Oct 10, 2019, at 1:04 PM, Éric Angelini <eric.angelini@skynet.be> wrote:
Hello Allan, once again, my question was vague :-(( Let me rephrase that properly: How many terms does the finite lexicographically earliest seq of distinct terms contain such that a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0.
My first terms are:
S = 1,2,3,4,5,6,7,8,9,10,23,45,16,...
We see that no three successive terms of S share any digit so far. Best, É.
à+ É. Catapulté de mon aPhone
Le 10 oct. 2019 à 17:34, Allan Wechsler <acwacw@gmail.com> a écrit :
Coincidentally, Éric's earlier posts made me think of this question, so I am ready with the answer: the only terms are 0,1,2,3,4,5,6,7,8. For all higher n, either n and n+1 will have the same first digit, or n+1 and n+2 will have the same first digit.
On Thu, Oct 10, 2019 at 7:38 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.
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participants (5)
-
Allan Wechsler -
Lucas, Stephen K - lucassk -
Michael Kleber -
Éric Angelini -
Éric Angelini