[UFO] "commutative number theory " f(g) = g(f) tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 03/24/2023, 11:36 PM A UFO topic ; unidentified flying observations. ( Got the tag idea from Gottfried ) Hard to put in a box what kind of math this is. Let n  > 1 be an integer. Let f(n) and g(n) be integer function that are strictly increasing. We want  a) f(n+1) > f(n) , g(n+1) > g(n) b) f(g(n)) = g(f(n)) c) f and g grow faster than linear. d) f(n) + 1 < g(n) < f(f(n)) - 1 e) f and g are not both chebychev functions nor both of the form A X^B. f) fixpoints are not allowed : f(n) =/= n , g(n) =/= n. Notice these conditions cannot be satisfied by both f and g being polynomials. Hence not algebra , but rather number theory. We have no fixpoint and the integer condition so traditional dynamics is propably not easy to apply either. Therefore the name : commutative number theory. The construction might not be so easy and questions about existance and uniqueness or extra conditions are just around the corner. But lets try  f(1) = 3 f := {3,5,8,12,17,23,30,38,47,...} f(17) =  155 This f is a polynomial of degree 2 : 3+2=5 5+3=8 8+4=12 12+5=17 etc g :={5,9,17,29,47,73,109,155,...} and you can check that f(g(n)) = g(f(n)) and the other conditions hold. the sequence g might not be unique and i assume no mistake has been made. But it is not so easy to see. The sequence g resembles or equals this perhaps : https://oeis.org/A034329 But it might not relate or be coincidence. I invite you to think about this. One of the ideas is that g(n) is just floor ( f^[r](n) ) ; in other words g(n) is just the rounded down number of some real iterate of f(n). But that is just a conjecture. f(n) = 1/2 (n^2 + n + 4) f(z) = z  has the solutions z = 1/2 + sqrt(-15)/2 << For those who care , it might relate ?? I know Z(sqrt(-15)) is not a UFD by heart, but this one has no integer minimum polynomial ( with integer coef ). >> regards tommy1729 « Next Oldest | Next Newest »

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