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:

10_{2} → 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)

`10`_{13} → 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.