Before I begin with the examples, I just noticed something mostly unrelated (maybe someone will find it interesting; perhaps Topol): looking through this big list of prime numbers http:// compoasso.free.fr/primelistweb/page/prime/liste_online_en.php, I'm not seeing a single one that equals any of the 3/6/9 numbers through Gematria number reduction. That's actually quite interesting. None of them.
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Let’s start with a = 6113, b = 13807
c = 84402191, d = 9187, e = 1222
(1,c) = (1222, 42191909, 4594) = {1222:42191909:9187:9186:1:84402191}
c = 9187, d = 95, e = 162
(1,c) = (162, 4499, 48) = {162:4499:95:94:1:9187}
c = 95, d = 9, e = 14
(1,c) = (14, 39, 5) = {14:39:9:8:1:95}
c = 9, d = 3, e = 0
e = 0 from a square, so we can calculate.
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Let's start with a = 11587, b = 14303
c = 165728861, d = 12873, e = 14732
(1,c) = (14732, 82851558, 6437) = {14732:82851558:12873:12872:1:165728861}
c = 12873, d = 113, e = 103
(1,c) = (104, 6324, 57) = {104:6324:113:112:1:12873}
c = 113, d = 10, e = 13
(1,c) = (13, 47, 5) = {13:47:10:9:1:113}
c = 10, d = 3, e = 1
This cell doesn't seem to exist since my spreadsheet calculated n to be 2.5 for the (1,c) cell (((a+b)/2)-d = ((1+10)/2)-3 = 2.5). So I guess this is where the recursive algorithm would end and we would find some number that allows us to calculate n-1 for c=113, allowing us to calculate n-1 for c=12873, allowing us to calculate n-1 for c=165728861. Probably.