490d — C492–C501: sequential research record
27 September 2026. Working continuation of C481. File52c’s latest supplied draft controls the rounded investigation. No canonical sources are changed. Second-order inversion remains deferred.
Each numbered entry records its question, declared inputs, exact results, finding, and reassessment. These are bounded research actions, not claims of independent discoveries. Parallel agents supplied source extraction and critical review; the main sequence was evaluated in order.
C492 — Resolve regular rounded-to-actual residual
Question. Is the regular +8 shift all rounding?
Source basis. File51a §§2.1,3.1–3.3; File18 §§4.1–4.2; File52c §3.14
Frozen inputs
{
"Actual_minus_Rounded_begetting_residuals": {
"Jared": 2,
"Methuselah": 2,
"Lamech": 2,
"Eber": -1,
"Reu": 2,
"Nahor": -1
},
"Rounded_MT": 4106,
"Rounded_LXX": 5486,
"selected_Shem_difference": 2
}
Results
{
"begetting_residual_sum": 6,
"MT_heads": [
4106,
4112,
4114
],
"LXX_heads": [
5486,
5492,
5494
]
}
Finding. No. Six row residuals sum+6; the selected standard versus strict Shem convention supplies the additional+2. The same decomposition applies to the supplied MT and LXX regular heads; it does not authorize upstream Gear transport.
Reassessment. Resolve the cumulative −2 separately before calling the combined shift a conservation law.
C493 — Cumulative residuals and completion selection
Question. Does cumulative rounding itself create the −2 change used in C481?
Source basis. File51a §§16.1–16.2; File22 §5.1
Frozen inputs
{
"MT_26_lifespans": [
930,
912,
905,
910,
895,
962,
365,
969,
777,
950,
600,
438,
433,
464,
239,
239,
230,
148,
205,
175,
180,
147,
137,
133,
137,
120
],
"terminal_anchor": 1406,
"actual_completion": 14004
}
Results
{
"actual_sum": 12600,
"rounded_sum": 12600,
"positive_rounding": 12,
"negative_rounding": -12,
"Moses_line_heads": [
14006,
14006
],
"completion_offset": -2
}
Finding. Cumulative rounding preserves12600 and the14006 Moses-line head exactly. The−2 is the selected lower completion endpoint14004, not a cumulative rounding error. The bridge uses both row residuals and explicit endpoint selection.
Reassessment. Combine the two separately explained shifts and test which endpoint alternatives preserve the anchor.
C494 — The exact residual conservation law
Question. Which declared endpoint changes preserve the weighted anchor?
Source basis. File51a §§3,16; File22 §§5.1,7.1; File52c §3.14
Frozen inputs
{
"rounded": [
14006,
4106
],
"actual_completion": [
14004,
4114
],
"strict_comparison": [
14006,
4112
],
"Year6_pair": [
14006,
4116
]
}
Results
{
"states": {
"rounded": {
"weighted_point": 12026,
"shift_from_12026": 0
},
"actual_completion": {
"weighted_point": 12026,
"shift_from_12026": 0
},
"strict_comparison": {
"weighted_point": "60136/5",
"shift_from_12026": "6/5"
},
"Year6_pair": {
"weighted_point": 12028,
"shift_from_12026": 2
}
},
"kernel": "4 delta_C + delta_R = 0",
"selected_vector": [
-2,
8
],
"general_integer_kernel": "t*(1,-4)"
}
Finding. The actual completion selection lies exactly in the kernel. Strict and Year6 alternatives give shifts6/5 and2. The conserved anchor is therefore a precise relation between selected states, not invariance under every actual/rounded convention.
Reassessment. Formalize the rounded path evaluator so convergence and nonconvergence can be explained by one rule.
C495 — Rounded chronology as path evaluation
Question. Can one rule explain both convergence and the LXX miss?
Source basis. File52c §3.3; C489
Frozen inputs
{
"anchor": 1406,
"paths": {
"MT_regular": [
1650,
1050
],
"MT_cumulative": [
9170,
3430
],
"LXX_regular": [
2250,
1830
]
}
}
Results
{
"rule": "V_a(path)=a+sum I(segment); chi(path)=sum(I(segment)-segment)",
"paths": {
"MT_regular": {
"raw_total": 2700,
"reversal_gain": 7920,
"evaluated_endpoint": 12026
},
"MT_cumulative": {
"raw_total": 12600,
"reversal_gain": -1980,
"evaluated_endpoint": 12026
},
"LXX_regular": {
"raw_total": 4080,
"reversal_gain": 4950,
"evaluated_endpoint": 10436
}
},
"composition": "chi(P followed by Q)=chi(P)+chi(Q), for an admitted concatenation"
}
Finding. Path gains are additive over retained segment lists. The MT regular gain7920 and cumulative gain−1980 offset their original9900 difference; LXX has different gains. Forgetting the partition discards necessary input.
Reassessment. Measure the exact effect of only the source-admitted Noah and Shem refinements.
C496 — Refinement defects
Question. When do documented refinements preserve the reversed path value?
Source basis. File52c §3.3; C489
Frozen inputs
{
"MT_regular_Noah": [
1650,
[
1050,
600
]
],
"MT_cumulative_Shem": [
9170,
[
8570,
600
]
],
"MT_cumulative_Noah": [
9170,
[
7620,
1550
]
],
"LXX_regular_Noah": [
2250,
[
1650,
600
]
]
}
Results
{
"defects": {
"MT_regular_Noah": 0,
"MT_cumulative_Shem": 990,
"MT_cumulative_Noah": 990,
"LXX_regular_Noah": 990
},
"definition": "delta_refinement=sum I(parts)-I(sum parts)"
}
Finding. The MT regular refinement has defect0; both cumulative refinements and the LXX regular refinement have defect990. Thus12026 versus13016 and10436 versus11426 are outcomes of the same path rule, not separate arbitrary anchors.
Reassessment. Test whether appending the documented Conquest-to-Nativity leg gives a uniform backbone continuation.
C497 — One shared tail links12026 and14726
Question. How does the documented1400 leg connect the two backbone anchors?
Source basis. File52c §§3.3–3.4
Frozen inputs
{
"Conquest": 1406,
"Nativity": 6,
"primary_reversed_sum": 10620
}
Results
{
"tail": 1400,
"reversed_tail": 4100,
"net_anchor_change": 2700,
"Conquest_backbone": 12026,
"Nativity_backbone": 14726,
"Nativity_span": 14720,
"E_span": 16000
}
Finding. Appending the shared1400 leg changes the reconstructed endpoint by4100−1400=2700. Both primary paths therefore continue together from12026 to14726. The backbone’s14720 span then expands to16000.
Reassessment. Identify the decimal gain law that explains the990 and2700 increments.
C498 — Decimal gains explain the path increments
Question. Which algebraic properties of decimal reversal generate the observed increments?
Source basis. File52c §1.2; C481; C495–C497
Frozen inputs
{
"three_digit_core": "10*(100a+10b+c)",
"two_digit_core": "100*(10a+b)",
"source_segments": [
1650,
1050,
9170,
3430,
1400
]
}
Results
{
"gains": {
"1650": 3960,
"1050": 3960,
"9170": -1980,
"3430": 0,
"1400": 2700
},
"three_digit_gain": "990*(c-a)",
"two_digit_gain": "900*(b-a)",
"general_invariant": "digit sum mod9; with10^k placeholders, gain divisible by9*10^k",
"not_23_preserving_example": {
"input": 230,
"inverse": 320,
"inverse_mod23": 21
}
}
Finding. The primary990 increments and the2700 tail increment are decimal-place effects. Reversal does not by itself preserve the23-lattice; its conjunction with the Key families still requires particular source inputs.
Reassessment. Determine the exact compatible scaling and translation rules for the rounded path module.
C499 — Where path evaluation is compatible with affine geometry
Question. Which operations commute with the path evaluator without losing segmentation?
Source basis. File52c §1.2; Strategy §§5C–5F; C495
Frozen inputs
{
"parts": [
1650,
1050
],
"anchor": 1406,
"test_translation": 215,
"decimal_scale": 10
}
Results
{
"translated_endpoint": 12241,
"scaled_endpoint": 120260,
"original_endpoint": 12026,
"whole_span_inverse": 7200,
"segmented_inverse_sum": 10620,
"general_rules": [
"all dates and anchor translated together: V shifts by same t",
"all durations and anchor scaled by10^k: V scales by10^k",
"I(s+t) need not equal I(s)+I(t)"
]
}
Finding. The module respects a change of origin and decimal rescaling when the entire declared object moves. It does not respect erasing segmentation. These are mathematical compatibility rules, not permission to translate source chronologies outside their admitted domains.
Reassessment. Connect the actual field to the three calendar Keys through their common measure.
C500 — The calendar family’s common measure
Question. Do the three Keys share a conserved quantity despite producing different year counts?
Source basis. Strategy §3.3; File52c §3.15; C481
Frozen inputs
{
"ratios": {
"E": "25/23",
"P": "70/69",
"J": "300/299"
},
"year_lengths": {
"E": 336,
"P": 360,
"J": 364
},
"source_seed": 12558
}
Results
{
"E": {
"calibration": "8400/23",
"seed_output": 13650,
"modeled_day_volume": 4586400
},
"P": {
"calibration": "8400/23",
"seed_output": 12740,
"modeled_day_volume": 4586400
},
"J": {
"calibration": "8400/23",
"seed_output": 12600,
"modeled_day_volume": 4586400
}
}
Finding. All three use K=8400/23 and give the same modeled day-volume4586400 for seed12558. The reusable rule is K_d(s)=(K/d)s. This explains common measurement, not every selected chronological placement.
Reassessment. Find the common integral input domain and apply it to the Jared–Noah–Flood field.
C501 — The three-Key integral domain
Question. Why does the middle member admit all three integral calendar outputs?
Source basis. File52c §3.15; C500
Frozen inputs
{
"Jared": 8372,
"Noah_G1": 8970,
"Flood_G2": 9568,
"Creation": 7912
}
Results
{
"common_input_lattice": 897,
"rows": {
"Jared": {
"E": 9100,
"P": "25480/3",
"J": 8400,
"all_integral": false
},
"Noah_G1": {
"E": 9750,
"P": 9100,
"J": 9000,
"all_integral": true
},
"Flood_G2": {
"E": 10400,
"P": "29120/3",
"J": 9600,
"all_integral": false
},
"Creation": {
"E": 8600,
"P": "24080/3",
"J": "103200/13",
"all_integral": false
}
},
"JNF_mod3_coefficients": [
2,
0,
1
]
}
Finding. The common integral span domain is897Z, since897=lcm(23,69,299). For598m, E and J are integral for all integer m; P additionally needs3|m. Of14,15,16 only Noah’s15 qualifies. This explains the three-Key middle role from the existing progression.
Reassessment. Use those outputs to derive the shared2926 endpoint and assess which placement assumptions remain.
Evidence
The accompanying evidence checkpoint contains the sequential journal, exact-arithmetic step scripts, source snapshots and SHA256 manifest. Latest step: C501. Assertions verify arithmetic or explicit scope decisions; they are not independent historical or statistical witnesses.