Ha5-21	Ha5-22	Ha5-23	Ha5-24 (240)

We have used up 3 * 182 - 2 glyphs, i.e. 544 of the total 2 * 648 = 1296:

			540
Hb7-30	Hb7-31 (987)	Hb7-32		Ha5-17	Ha5-18 (234)	Ha5-19	Ha5-20
day 329			543 / 3 = 181				⅓

Possibly we should adjust the table into:

		540
Hb7-31 (987)	Hb7-32		Ha5-17	Ha5-18 (234)	Ha5-19	Ha5-20
546 / 3 = 182

Ha5-21 will be the 1st glyph of the rest = 1296 - 546 = 750 = 250 days. If there is another half year with 182 days in the glyph sequence from Ha5-21 up to and including Hb7-30, then another 546 glyphs will be used up, leaving only 204 (68 days).

If my reconstruction of the structure so far has been correct, then 204 glyphs should be somewhere among 750 glyphs. Only 3 'solstice honu' appear among these 750, and the distances between them are ca 200 glyphs:

	197		212
Ha10-35 (537)	65⅓	Hb2-35 (734)	70⅔	Hb6-45 (947)

I guess 4 of those 204 should be Ha5-21--Ha5-24. The searched for remaining glyphs are very close to what we can find between Ha10-35 and Hb2-35.

The position of honu at left in Ha10-35 indicates that another and new season has arrived. 537 / 3 = 179:

Counting from Ha5-21 (glyph number 237 counted from Ha1-1) Ha10-35 will be glyph number 301:

				295
Ha5-21 (237)	Ha5-22	Ha5-23	Ha5-24 (240)		Ha10-34 (536)	Ha10-35 (537)
300

It seems that sun has 100 days remaining at the beginning of the 2nd part of his cycle.

Ha10-30--31 do not fulfil a triplet if we will let honu in Ha10-35 be the first glyph in his triplet. But we have 4 similarly inconcruent ones at the beginning of the 300 glyphs, viz. Ha5-21--24. We can add 2 + 4 = 6 and divide by 3, and there we possibly have 2 intercalated days.

If so it means 300 glyphs are equal to 2 + 98 days. We have to reach 2 + 180 days, and we must find 180 - 98 = 82 more days, i.e. 3 * 82 = 246 glyphs. Hb7-32 is number 988, and 988 - 2 (Hb7-31--32) - 244 = 742:

Hb2-30 (729)	Hb2-31	Hb2-32	Hb2-33	Hb2-34
Hb2-35 (734)	Hb2-36	Hb2-37	Hb2-38	Hb2-39
Hb2-40	Hb2-41	Hb2-42	Hb2-43 (742)	Hb2-44

We can then consolidate the whole cycle:

		540
Hb7-31	Hb7-32 (988)		Ha5-17	Ha5-18 (234)	Ha5-19	Ha5-20
181				1

				295
Ha5-21	Ha5-22 (238)	Ha5-23	Ha5-24 (240)		Ha10-34 (536)
100

		195		7			242
Ha10-35 (537)	Ha10-36		Hb2-35 (734)		Hb2-43 (742)	Hb2-44		Hb7-30 (986)
66				3			81

181 + 1 = 182, the first half of 364. The other half will then be 100 + 81. And there are 66 + 3 days inserted. 182 + 181 + 69 = 432 = 1296 / 3. But presumably we should increase 81 to 82 by adding Hb2-42--44:

		195		5
Ha10-35 (537)	Ha10-36		Hb2-35 (734)		Hb2-41 (740)
66				2

			242
Hb2-42	Hb2-43 (742)	Hb2-44		Hb7-30 (986)
1			81