Here's the math in the cited paper... I feel like they're making an error of 35960 as 36kilobits instead of, yeah, 15bits ? https://ieeexplore.ieee.org/documen

Here's the math in the cited paper... I feel like they're making an error of 35960 as 36kilobits instead of, yeah, 15bits ? https://ieeexplore.ieee.org/document/9438269 https://ieeexplore.ieee.org/document/9438269 If, in our previous example, the 4 wavelengths were to be selected from a palette of 32 different wavelengths, a single worfel location could store ~36 Kilobits of data. Thus, a 1 cm2 media with 10μ2 data locations (8μ2 worfels with 2μ spacing on all sides) = 1,000,000 worfels/cm2. For example, (32!/((32-4)! • 4!)) = 35,960 distinct states. (An analogous use of formula (1) is drawing a hand of 5 playing cards from a 52-card deck yields 2,598,960 distinct hands.) Applying the 35,960-state permutation table for k=4 (i.e., superimposing 4 wavelengths per worfel), and drawing from a palette, N, of 32 different wavelengths, yields 35,960,000,000 bits (≈35.9 gigabits) per cm2; or 35.9 x (6.42 cm2 per square inch) ≈ 230.4 gigabits/in2. And so for an example of a 4″x5″ media (20 in2), 20 x 230.4 ≈ 4.6 terabits per 4x5 inch media.