Do you want a haunting literary image? You’ll find one of the strangest and strongest in Borges’ “La Biblioteca de Babel” (1941), which is narrated by a librarian in an infinite library. The librarian anticipates the end of his life:

Muerto, no faltarán manos piadosas que me tiren por la baranda; mi sepultura será el aire insondable; mi cuerpo se hundirá largamente y se corromperá y disolverá en el viento engenerado por la caída, que es infinita. — “La Biblioteca de Babel”

When I am dead, compassionate hands will throw me over the railing; my tomb will be the unfathomable air, my body will sink for ages, and will decay and dissolve in the wind engendered by my fall, which shall be infinite. — “The Library of Babel” (translation by Andrew Hurley)

The infinite fall is the haunting image. Falling is powerful; falling for ever is more powerful still. But it can’t happen in reality: soon or later a fall has to end. Objects crash to earth or splash into the ocean. Of course, you could call being in orbit a kind of infinite fall, but it doesn’t have the same power.

However, there’s more kinds of falling than one and I think the arithmophile Borges would have liked one of the other kinds a lot. Numbers can fall — you sum their digits, take the sum from the original number, and repeat. That is, n = n – digsum(n). Here are some examples:

10 → 9 → 0
100 → 99 → 81 → 72 → 63 → 54 → 45 → 36 → 27 → 18 → 9 → 0
1000 → 999 → 972 → 954 → 936 → 918 → 900 → 891 → 873 → 855 → 837 → 819 → 801 → 792 → 774 → 756 → 738 → 720 → 711 → 702 → 693 → 675 → 657 → 639 → 621 → 612 → 603 → 594 → 576 → 558 → 540 → 531 → 522 → 513 → 504 → 495 → 477 → 459 → 441 → 432 → 423 → 414 → 405 → 396 → 378 → 360 → 351 → 342 → 333 → 324 → 315 → 306 → 297 → 279 → 261 → 252 → 243 → 234 → 225 → 216 → 207 → 198 → 180 → 171 → 162 → 153 → 144 → 135 → 126 → 117 → 108 → 99 → 81 → 72 → 63 → 54 → 45 → 36 → 27 → 18 → 9 → 0

The details are different in other bases, like 2 or 16, but the destination is the same. The number falls to zero and the fall stops, because digsum(0) = 0:

10₂ → 1 → 0 (n=2)
100 → 11 → 1 → 0 (n=4)
1000 → 111 → 100 → 11 → 1 → 0 (n=8)
10000 → 1111 → 1011 → 1000 → 111 → 100 → 11 → 1 → 0 (n=16)
100000 → 11111 → 11010 → 10111 → 10011 → 10000 → 1111 → 1011 → 1000 → 111 → 100 → 11 → 1 → 0 (n=32)
1000000 → 111111 → 111001 → 110101 → 110001 → 101110 → 101010 → 100111 → 100011 → 100000 → 11111 → 11010 → 10111 → 10011 → 10000 → 1111 → 1011 → 1000 → 111 → 100 → 11 → 1 → 0 (n=64)

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10₁₃ → C → 0 (n=13)
100 → CC → B1 → A2 → 93 → 84 → 75 → 66 → 57 → 48 → 39 → 2A → 1B → C → 0 (n=169)
1000 → CCC → CA2 → C84 → C66 → C48 → C2A → C0C → BC1 → BA3 → B85 → B67 → B49 → B2B → B10 → B01 → AC2 → AA4 → A86 → A68 → A4A → A2C → A11 → A02 → 9C3 → 9A5 → 987 → 969 → 94B → 930 → 921 → 912 → 903 → 8C4 → 8A6 → 888 → 86A → 84C → 831 → 822 → 813 → 804 → 7C5 → 7A7 → 789 → 76B → 750 → 741 → 732 → 723 → 714 → 705 → 6C6 → 6A8 → 68A → 66C → 651 → 642 → 633 → 624 → 615 → 606 → 5C7 → 5A9 → 58B → 570 → 561 → 552 → 543 → 534 → 525 → 516 → 507 → 4C8 → 4AA → 48C → 471 → 462 → 453 → 444 → 435 → 426 → 417 → 408 → 3C9 → 3AB → 390 → 381 → 372 → 363 → 354 → 345 → 336 → 327 → 318 → 309 → 2CA → 2AC → 291 → 282 → 273 → 264 → 255 → 246 → 237 → 228 → 219 → 20A → 1CB → 1B0 → 1A1 → 192 → 183 → 174 → 165 → 156 → 147 → 138 → 129 → 11A → 10B → CC → B1 → A2 → 93 → 84 → 75 → 66 → 57 → 48 → 39 → 2A → 1B → C → 0 (n=2197)

But the fall to 0 made me think of another kind of number-fall. What if you count the 0s in a number, take that count away from the original number, and repeat? You could call this a z-fall (pronounced zee-fall). But unlike free-fall, z-fall doesn’t last long:

10 → 9
100 → 98
1000 → 997
10000 → 9996

And the number always comes to rest far above the ground, as it were. In a fall using digsum(n), the number descends to 0. In a fall using zerocount(n), the number never even reaches 1. At least, never in any base higher than 2. But in base-2, you get this:

10 → 1 (n=2)
100 → 10 → 1 (n=4)
1000 → 101 → 100 → 10 → 1 (n=8)
10000 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=16)
100000 → 11011 → 11010 → 11000 → 10101 → 10011 → 10001 → 1110 → 1101 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=32)
1000000 → 111010 → 111000 → 110101 → 110011 → 110001 → 101110 → 101100 → 101001 → 100110 → 100011 → 100000 → 11011 → 11010 → 11000 → 10101 → 10011 → 10001 → 1110 → 1101 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=64)

When I saw that, I had a wonderful vision of how even the biggest numbers in base 2 could z-fall all the way to 1. Almost all binary numbers contain 0, after all. So the z-falls would get longer and longer, paying tribute to la caída infinita, the infinite fall, of the librarian in Borges’ Library of Babel. Alas, binary numbers don’t behave like that. The highest number in base 2 that z-falls to 1 is this:

1010001 → 1001101 → 1001010 → 1000110 → 1000010 → 111101 → 111100 → 111010 → 111000 → 110101 → 110011 → 110001 → 101110 → 101100 → 101001 → 100110 → 100011 → 100000 → 11011 → 11010 → 11000 → 10101 → 10011 → 10001 → 1110 → 1101 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=81)

Above that, binary numbers land on what you might call a shelf:

1010010=82 → 1001110=78 → 1001011=75 → 1001000=72 → 1000011=67 → 111111=63 (n=82)

If binary numbers are an infinite tall mountain, 1 is at the foot of the mountain. 111111 = 63 is like a shelf a little way above the foot. But I conjecture that arbitrarily large binary numbers will z-fall to 63. For example, no matter how large the power of 2, I conjecture that it will z-fall to 63:

10 → 1 : 2 → 1 (count of steps=2)
100 ... → 1 : 4 ... → 1 (c=3)
1000 ... → 1 : 8 ... → 1 (c=5)
10000 ... → 1 : 16 ... → 1 (c=8)
100000 ... → 1 : 32 ... → 1 (c=16)
1000000 ... → 1 : 64 ... → 1 (c=27)
10000000 ... → 111111 : 128 ... → 63 (c=21)
100000000 ... → 111111 : 256 ... → 63 (c=60)
1000000000 ... → 111111 : 512 ... → 63 (c=130)
10000000000 ... → 111111 : 1024 ... → 63 (c=253)
100000000000 ... → 111111 : 2048 ... → 63 (c=473)
1000000000000 ... → 111111 : 4096 ... → 63 (c=869)
10000000000000 ... → 111111 : 8192 ... → 63 (c=1586)
100000000000000 ... → 111111 : 16384 ... → 63 (c=2899)
1000000000000000 ... → 111111 : 32768 ... → 63 (c=5327)
10000000000000000 ... → 111111 : 65536 ... → 63 (c=9851)
100000000000000000 ... → 111111 : 131072 ... → 63 (c=18340)
1000000000000000000 ... → 111111 : 262144 ... → 63 (c=34331)
10000000000000000000 ... → 111111 : 524288 ... → 63 (c=64559)
100000000000000000000 ... → 111111 : 1048576 ... → 63 (c=121831)
1000000000000000000000 ... → 111111 : 2097152 ... → 63 (c=230573)
10000000000000000000000 ... → 111111 : 4194304 ... → 63 (c=437435)
100000000000000000000000 ... → 111111 : 8388608 ... → 63 (c=831722)
1000000000000000000000000 ... → 111111 : 16777216 ... → 63 (c=1584701)
10000000000000000000000000 ... → 111111 : 33554432 ... → 63 (c=3025405)
100000000000000000000000000 ... → 111111 : 67108864 ... → 63 (c=5787008)
1000000000000000000000000000 ... → 111111 : 134217728 ... → 63 (c=11089958)
10000000000000000000000000000 ... → 111111 : 268435456 ... → 63 (c=21290279)
100000000000000000000000000000 ... → 111111 : 536870912 ... → 63 (c=40942711)
1000000000000000000000000000000 ... → 111111 : 1073741824 ... → 63 (c=78864154)

So the z-falls get longer and longer. But z-falling to 63 doesn’t have the power of z-falling to 1.