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 A065937 a(n) is the integer (reduced squarefree) under the square root obtained when the inverse of Minkowski's question mark function is applied to the n-th ratio A007305(n+1)/A047679(n-1) in the full Stern-Brocot tree and zero when it results a rational value. 4
 0, 0, 0, 5, 5, 0, 0, 0, 2, 2, 0, 5, 5, 0, 0, 2, 3, 0, 3, 3, 0, 3, 2, 0, 2, 2, 0, 5, 5, 0, 0, 5, 13, 17, 2, 17, 37, 5, 13, 13, 5, 37, 17, 2, 17, 13, 5, 2, 3, 0, 3, 3, 0, 3, 2, 0, 2, 2, 0, 5, 5, 0, 0, 3, 17, 3, 37, 21, 13, 10, 37, 3, 401, 6, 13, 10, 401, 0, 17, 17, 0, 401, 10, 13, 6, 401, 3, 37 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Note: the underlying function N2Q (see the Maple code) maps natural numbers 1, 2, 3, 4, 5, ..., through all the positive rationals 1/1, 1/2, 2/1, 1/3, 2/3, 3/2, 3/1, 1/4, ... bijectively to the union of positive rationals and quadratic surds. In his "On Numbers and Games", Conway denotes Minkowski's question mark function with x enclosed in a box. REFERENCES J. H. Conway, On Numbers and Games, 2nd ed. Natick, MA: A. K. Peters, pp. 82-86 (First ed.), 2000. LINKS Robert Hill, An article in sci.math newsgroup Eric Weisstein's World of Mathematics, Minkowski's Question Mark Function. Wikipedia, Minkowski's question mark function EXAMPLE The first few values for this mapping are N2Q(1) = Inverse_of_MinkowskisQMark(1) = 1, N2Q(2) = Inverse_of_MinkowskisQMark(1/2) = 1/2, N2Q(3) = Inverse_of_MinkowskisQMark(2) = 2, N2Q(4) = Inverse_of_MinkowskisQMark(1/3) = (3-sqrt(5))/2, N2Q(5) = Inverse_of_MinkowskisQMark(2/3) = (sqrt(5)-1)/2, N2Q(6) = Inverse_of_MinkowskisQMark(3/2) = 3/2, N2Q(7) = Inverse_of_MinkowskisQMark(3) = 3, N2Q(8) = Inverse_of_MinkowskisQMark(1/4) = 1/3, N2Q(9) = Inverse_of_MinkowskisQMark(2/5) = sqrt(2)-1, N2Q(10) = Inverse_of_MinkowskisQMark(3/5) = 2-sqrt(2). MAPLE [seq(find_sqrt(N2Q(j)), j=1..512)]; N2Q := n -> Inverse_of_MinkowskisQMark(A007305(m+1)/A047679(m-1)); Inverse_of_MinkowskisQMark := proc(r) local x, y, b, d, k, s, i, q; x := numer(r); y := denom(r); if(1 = y) then RETURN(x/y); fi; if(2 = y) then RETURN(x/y); fi; b := []; d := []; k := 0; s := 0; i := 0; while(x <> 0) do q := floor(x/y); if(i > 0) then b := [op(b), q]; d := [op(d), x]; fi; x := 2*(x-(q*y)); if(member(x, d, 'k') and (k > 1) and (b[k] <> b[k-1]) and (q <> floor(x/y))) then s := eval_periodic_confrac_tail(list2runcounts(b[k..nops(b)])); b := b[1..(k-1)]; break; fi; i := i+1; od; if(0 = k) then b := b[1..(nops(b)-1)]; b := [op(b), b[nops(b)]]; fi; RETURN(factor(eval_confrac([floor(r), op(list2runcounts([0, op(b)]))], s))); end; eval_confrac := proc(c, z) local x, i; x := z; for i in reverse(c) do x := (`if`((0=x), x, (1/x)))+i; od; RETURN(x); end; eval_periodic_confrac_tail := proc(c) local x, i, u, r; x := (eval_confrac(c, u) - u) = 0; r := [solve(x, u)]; RETURN(max(r, r)); end; # Note: I am not sure if the larger root is always the correct one for the inverse of Minkowski's question mark function. However, whichever root we take, it does not change this sequence, as the integer under the square root is same in both cases. - Antti Karttunen, Aug 26 2006 list2runcounts := proc(b) local a, p, y, c; if(0 = nops(b)) then RETURN([]); fi; a := []; c := 0; p := b; for y in b do if(y <> p) then a := [op(a), c]; c := 0; p := y; fi; c := c+1; od; RETURN([op(a), c]); end; find_sqrt := proc(x) local n, i, y; n := nops(x); if(n < 2) then RETURN(0); fi; if((2 = n) and (`^` = op(0, x)) and (1/2 = op(2, x))) then RETURN(op(1, x)); else for i from 0 to n do y := find_sqrt(op(i, x)); if(y <> 0) then RETURN(y); fi; od; RETURN(0); fi; end; # This returns an integer under the square-root expression in Maple. CROSSREFS a(n) = A065936(A065935(n)). Positions of sqrt(n) in this mapping: A065939. Sequence in context: A266668 A043299 A144776 * A197738 A189232 A247667 Adjacent sequences: A065934 A065935 A065936 * A065938 A065939 A065940 KEYWORD nonn AUTHOR Antti Karttunen, Dec 07 2001 EXTENSIONS Description clarified by Antti Karttunen, Aug 26 2006 STATUS approved

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Last modified March 21 19:39 EDT 2023. Contains 361410 sequences. (Running on oeis4.)