Determining values satisfying an inequality

Question

I have the following inequality:

$$\left\lceil \frac{\log((n-1)/6000)}{\log(3)} \right\rceil < \left\lceil \frac{\log((n-1)/3000)}{\log(3)} \right\rceil,$$ where $n$ is a positive integer, and I want to determine precisely for which values of $n$ it is verified.

I have no idea. Is it really possible?

Thanks.

Answer 1

We can write

$$\lceil \log_3\left(\frac{n-1}{3000}\right)\rceil=\lceil \log_3\left(\frac{n-1}{1000}\right)-1\rceil=\lceil \log_3\left(\frac{n-1}{1000}\right)\rceil-1$$

Similarly

$$\lceil \log_3\left(\frac{n-1}{6000}\right)\rceil=\lceil \log_3\left(\frac{n-1}{1000}\right)-\log_3(2)-1\rceil=\lceil \log_3\left(\frac{n-1}{1000}\right)-\log_3(2)\rceil-1$$

Thus the inequality is equivalent to

$$\lceil \log_3\left(\frac{n-1}{1000}\right)-\log_3(2)\rceil<\lceil \log_3\left(\frac{n-1}{1000}\right)\rceil.$$

Now $$\lceil \log_3\left(\frac{n-1}{1000}\right)-\log_3(2)\rceil=\lceil \log_3\left(\frac{n-1}{1000}\right)\rceil$$ if and only if:

$$\lceil \log_3\left(\frac{n-1}{1000}\right)\rceil-1< \log_3\left(\frac{n-1}{1000}\right)-\log_3(2)\leq\lceil \log_3\left(\frac{n-1}{1000}\right)\rceil.$$

The upper bound holds trivially. The other inequality is equivalent to

$$\lceil \log_3\left(\frac{n-1}{1000}\right)\rceil< \log_3\left(\frac{n-1}{1000}\right)+1-\log_3(2)\qquad (*)$$

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If you plot

$$\log_3\left(\frac{n-1}{1000}\right)+1-\log_3(2)-\lceil \log_3\left(\frac{n-1}{1000}\right)\rceil$$

you get a nice picture showing which value satisfy the inequality (where the plot is above zero, and where the inequality is an equality (where the plot is below zero).

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Now, one particular set of values for which the inequality $(*)$ holds (so that the original equality holds with equality) is where $\log_3\left(\frac{n-1}{1000}\right)$ is an integer, i.e. where there exists an integer $m$ such that $$3^m=\frac{n-1}{1000} \Leftrightarrow n=1000(3^m)+1$$

I am not sure where to go from here.

Answer 2

The following PARI/GP - program shows the intervals of numbers satisfying the inequality upto $667$

? for(k=-7,-2,s=exp(k*log(3))*3000+1;t=exp((k)*log(3))*6000+1;print(k," ",ceil(s
),"  ",truncate(t)))
-7 3  3
-6 6  9
-5 14  25
-4 39  75
-3 113  223
-2 335  667

Compare with the result obtained by the direct method :

? for(n=2,1000,s=ceil(log((n-1)/6000)/log(3)); t=ceil(log((n-1)/3000)/log(3)); i
f(s<t,print1(n," ")))
3 6 7 8 9 14 15 16 17 18 19 20 21 22 23 24 25 39 40 41 42 43 44 45 46 47 48 49 5
0 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 113
114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133
134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153
154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173
174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193
194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213
214 215 216 217 218 219 220 221 222 223 335 336 337 338 339 340 341 342 343 344
345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364
365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384
385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404
405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424
425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444
445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464
465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484
485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504
505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524
525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544
545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564
565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584
585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604
605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624
625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644
645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664
665 666 667
?

The results coincide, so the program should be correct.

I also calculated larger values, but I noticed that some values were not correcly calculated because $s$ and $t$ are (in theory) integers, so the ceil- and the truncate- command can give the wrong results. In this case, I do not know, which interval is correct.

But surely, someone can correct this issue. He/she is invited to edit my answer.