View Single Post 2020-08-05, 08:40 #392 garambois   Oct 2011 25×11 Posts I think I am now finally able to formulate a general conjecture that encompasses the two little conjectures stated in posts #384 and #387. General conjecture : s(n) = sigma(n)-n If p = s(3^i) is a prime, then we have : - s(3^(2i)) = m * p and s(s(3^(2i))) = m * r, where r is any integer. - s(3^(2i * k)) = m * p * u and s(s(3^(2i * k))) = m * t, with u and t any integers but p and m which remain the same whatever k integer k>=1 for a given i. I hope I have made this conjecture clear in English ! For a better understanding, here are some numerical examples below (I looked up all the i<=500 such that p = s(3^i) is a prime number) : Code: i = 3 p = 13 m = 28 = 2^2 * 7 i = 7 p = 1093 m = 2188 = 2^2 * 547 i = 13 p = 797161 m = 1594324 = 2^2 * 398581 i = 71 p = 3754733257489862401973357979128773 m = 7509466514979724803946715958257548 = 2^2 * 853 * 2131 * 82219 * 3099719989 * 4052490063499 i = 103 p = 6957596529882152968992225251835887181478451547013 m = 13915193059764305937984450503671774362956903094028 = 2^2 * 619 * 3661040653 * 1535090713229126909942383374434289901 If we take for example i=71, we have : i=71 p=3754733257489862401973357979128773 m=7509466514979724803946715958257548=2^2*853*2131*82219*309919989*4052490063499 So, we can say that for all sequences that begin with 3^(2*71 * k) = 3^(142k), with k being an integer, we will find the factor p*m in the decomposition of the term at index 1 and we will find the factor m in the decomposition of the term at index 2. I'm not quite sure how to try to demonstrate this conjecture yet, I haven't spent any time on it. Either she's already known. Otherwise, it shouldn't be very difficult to prove it for someone who's used to this kind of problem...  