The number of glyphs on side a of H can be considered to be 579 (580?), to be 648 (the sum of redmarked 627 and 21 in the table below), or to be 666 (= 648 + 18):

a1 50 50 burnt area could have contained the number of glyphs below 50 b1 *51 (?) 51
a2 58 108 108 b2 48 99
a3 52 160 160 b3 47 146
a4 56 216 216 b4 51 197
a5 59 275 275 b5 57 254
a6 64 339 *5 344 b6 54 308
a7 48 387 *3 395 b7 50 358
a8 46 433 *8 449 b8 *54 (?) 412
a9 40 473 *13 502 b9 65 477
a10 49 522 *18 569 b10 67 544
a11 36 558 *22 627 b11 53 597
a12 *21 (?) 579 *18 (?) 666 b12 *51 (?) 648
sum *666 (?) sum *648 (?)

It has been shown that the burnt area probably was there already when the creator carved his glyphs on the tablet (cfr at hahe), but that anyhow 648 (= 6 * 108) is a good alternative.

The visible glyphs on side a - including the vacant space for *Ha12-19 - sum up to the odd number 579, and maybe we should think '+ one more' as when we count to 472 glyphs for the text of G although there are only 471 glyphs on the tablet. 580 = 20 * 29.

However, the burnt area covers also a space at the end of line a12 where an estimated number of 18 glyphs can be imagined. 666 (= 6 * 111) is therefore also a possible alternative. But neither 579 nor 580 is divisible by 6.

On side b there is no burnt area. Yet, there are glyphs which are totally missing in line b1. There are 3 fully visible glyphs followed by 24 totally or partially invisible ones:

Hb1-1 Hb1-2 Hb1-3 Hb1-4 Hb1-5 Hb1-6
... ... ... ...
Hb1-7 Hb1-8 *Hb1-9 *Hb1-10 *Hb1-11 *Hb1-12
... ...
*Hb1-13 *Hb1-14 *Hb1-15 *Hb1-16 *Hb1-17 *Hb1-18
... ... ... ...
*Hb1-19 *Hb1-20 *Hb1-21 *Hb1-22 *Hb1-23 *Hb1-24
*Hb1-25 *Hb1-26 *Hb1-27
*Hb1-28 *Hb1-29 *Hb1-30 *Hb1-31 *Hb1-32

Then, suddenly, the glyphs are fully visible again. I guess those 24 more or less invisible glyphs are meant to be a sign. The redmarked 16 glyphs above are close to the 15 respectively 17 parallel ones in P and Q:

tablet number of glyphs
H ca 16
P 15
Q 17

The picture of glyph line Hb1 in Barthel does not in any way suggest 16 must be wrong. Therefore we should calculate with *32 + 19 (*Hb1-33--51) = *51 glyphs in line Hb1 (while remembering to think ± 1). But we have two more question marks to consider, viz. in lines Hb8 and Hb12. There is a very special absent glyph in line Hb8 and it was discussed at hakaturou. The parallel text in P has here an unusual sign which I have decided to name pito (navel):

...
Hb8-15 *Hb8-16 *Hb8-17 *Hb8-18 *Hb8-19 *Hb8-20 *Hb8-21
Pb9-32 Pb9-33 Pb9-34 Pb9-35 Pb9-36 Pb9-37

The creator of H has probably left a vacancy (clearly seen in the picture of the line in Barthel) for *Hb8-16. I have counted the vacant place among the *54 in the table above.

The last question mark on side b needs a more detailed description. The uncertainty comes at the very end of line Hb12:

Hb12-29 Hb12-30 Hb12-31 Hb12-32 (629)
Hb12-33 Hb12-34 Hb12-35 Hb12-36 (633)
Hb12-37 Hb12-38 Hb12-39 Hb12-40 (637)
Hb12-41 Hb12-42 Hb12-43 Hb12-44 Hb12-45 Hb12-46 (643)
... ...
Hb12-47 Hb12-48 Hb12-49 *Hb12-50 *Hb12-51 (648)

Maybe *Hb12-51 is a glyph which should be only imagined - like *Hb8-16. Or maybe we should accept that *Hb12-50 (where 12 * 50 = 600) is the last glyph of the text.

In the latter case we can still reach 648 by increasing 51 to 52 in line b1:

a1 50 50 burnt area could have contained the number of glyphs below 50 b1 *52 (?) 52
a2 58 108 108 b2 48 100
a3 52 160 160 b3 47 147
a4 56 216 216 b4 51 198
a5 59 275 275 b5 57 255
a6 64 339 *5 344 b6 54 309
a7 48 387 *3 395 b7 50 359
a8 46 433 *8 449 b8 *54 (?) 413
a9 40 473 *13 502 b9 65 478
a10 49 522 *18 569 b10 67 545
a11 36 558 *22 627 b11 53 598
a12 *21 (?) 579 *18 (?) 666 b12 *50 (?) 648
sum *666 (?) sum *648 (?)

And 52 (b1) + 48 (b2) = 100 seems to be equally good as 99.

413 = 14 * 29.5 instead of 412 for lines b1-b8 is better than 412. But then 648 - 413 = 235 suggests we should count also with *Hb12-51 in order to reach 236 (= 8 * 29.5). That decides it, I think, because 649 = 22 * 29.5 (and 22 suggests a full cycle because of the formula 22 / 7 = π).

The ordinal number (counted from Hb1-1) of the glyphs at the end of line Hb12 will then be as follows:

Hb12-29 Hb12-30 Hb12-31 Hb12-32 (630)
Hb12-33 Hb12-34 Hb12-35 Hb12-36 (634)
Hb12-37 Hb12-38 Hb12-39 Hb12-40 (638)
Hb12-41 Hb12-42 Hb12-43 Hb12-44 Hb12-45 Hb12-46 (644)
... ...
Hb12-47 Hb12-48 Hb12-49 *Hb12-50 *Hb12-51 (649)

In P, where all the glyphs are visible, the sums are 599 (side a) respectively 559 (side b). It could indicate a wish to end with 9 (here 99 respectively 59). Side a of G has 229 glyphs (excluding Gb8-30).

The last visible glyph in H is Hb12-49, where 49 once again appears (in addition to in 649). Possibly it means a 'square of 7' is finished. 12 * 49 = 588 (= 300 + 288) and 6 * 49 = 294 (the final day before 10 * 29.5 = a 'zero' day is reached).

I decide to change the tablet into:

a1 50 50 burnt area could have contained the number of glyphs below 50 b1 *52 (?) 52
a2 58 108 108 b2 48 100
a3 52 160 160 b3 47 147
a4 56 216 216 b4 51 198
a5 59 275 275 b5 57 255
a6 64 339 *5 344 b6 54 309
a7 48 387 *3 395 b7 50 359
a8 46 433 *8 449 b8 *54 (?) 413
a9 40 473 *13 502 b9 65 478
a10 49 522 *18 569 b10 67 545
a11 36 558 *22 627 b11 53 598
a12 *21 (?) 579 *18 (?) 666 b12 *50 (?) 648
sum *666 (?) sum *648 (?)

Considering the sum of the glyphs on side a and side b the following alternatives are possible:

side a side b sum
579 648 1227
580 648 1228
648 648 1296
666 648 1314

Mea ke in Hb6-29 should be an end glyph, and counting its ordinal number from Hb1-1 it becomes 284:

393
Ha10-30 Ha10-31 Ha10-32 Hb6-26 Hb6-27 Hb6-28 Hb6-29
400 / 2 = 200 days

284 can be read as 2 * 84 = 168 (or maybe as 28 'in a square'). Counting from Hb1-1 seems to result in relevant signs. Both signs (168 respectively 28 and 4) should indicate that light is ending.

Counting from Ha1-1 does also give more or less acceptable results:

side a Hb6-29 sum possible explanation
579 284 863 500 + 4
580 284 864 8 * 8 * 8
648 284 932 288
666 284 694 364 + 200

None of these alternatives for side a can be excluded.

If we should count with 649 glyphs on side b, then there will be 649 - 284 = 365 glyphs beyond Hb6-29. If we should count with only 50 glyphs in line Hb12 there will be 364 glyphs beyond Hb6-29.