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A065937
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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.
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4
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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; internal format)
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OFFSET
| 1,4
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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 the Minkowski's question mark function with x enclosed in a box.
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REFERENCES
| J. H. Conway, On Numbers and Games, 2nd ed. Natick, MA: A. K. Peters, pp. 82-86 (First ed.), 2000.
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LINKS
| Robert Hill, An article in sci.math newsgroup
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
Wikipedia, Minkowski's question mark function
Index entries for sequences related to Minkowski's question mark function
Index entries for sequences related to Stern's sequences
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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)
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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[1], r[2])); end; # NB: 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. - AK, Aug 26 2006.
list2runcounts := proc(b) local a, p, y, c; if(0 = nops(b)) then RETURN([]); fi; a := []; c := 0; p := b[1]; 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.
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CROSSREFS
| a(n) = A065936(A065935(n)). Positions of sqrt(n) in this mapping: A065939.
Sequence in context: A186639 A043299 A144776 * A197738 A189232 A115144
Adjacent sequences: A065934 A065935 A065936 * A065938 A065939 A065940
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KEYWORD
| nonn
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AUTHOR
| Antti Karttunen Dec 07 2001. Description clarified by Antti Karttunen (his-firstname.his-surname(AT)gmail.com), Aug 26 2006
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