%I #5 May 01 2014 02:49:14
%S 4,16,1,22,136,1,64,3,2,1,8,64,25,2,1,160,19,4,3,76,1,1,6,5,4,3,2,1,
%T 256,7,97,5,4,3328,2,1,32,256,13,6,167772160,4,3,2,1,67,1054,8,7,6,5,
%U 4,3,2,1,34,4,9,130,7,97,5,4,3,1249,1,1279,40,10,9,8,7,6,5,4,3,2,1,10,12
%N The table of permutations of N, each row induced by the rotation (to the left) of the n-th node in the infinite binary "decimal" fraction tree.
%C See the comment at A065658.
%H <a href="/index/St#Stern">Index entries for sequences related to Stern's sequences</a>
%p [seq(RotateBinFracLeftTable(j),j=0..119)]; RotateBinFracLeftTable := n -> RotateBinFracNodeLeft(1+(n-((trinv(n)*(trinv(n)-1))/2)),(((trinv(n)-1)*(((1/2)*trinv(n))+1))-n)+1);
%p RotateBinFracNodeLeft := (t,n) -> frac2position_in_0_1_SB_tree(RotateBinFracNodeLeft_x(t,SternBrocot0_1frac(n)));
%p RotateBinFracNodeLeft_x := proc(t,x) local num,den; den := 2^(1+floor_log_2(t)); num := (2*(t-(den/2)))+1; if((x <= (num-1)/den) or (x >= (num+1)/den)) then RETURN(x); fi; if(x >= ((2*num)+1)/(2*den)) then RETURN(((num-1)/den) + (2*(x - (num/den)))); fi; if(x > (num/den)) then RETURN(x - (1/(2*den))); fi; RETURN(((num-1)/den) + ((x-((num-1)/den))/2)); end;
%p SternBrocot0_1frac := proc(n) local m; m := n + 2^floor_log_2(n); SternBrocotTreeNum(m)/SternBrocotTreeDen(m); end;
%p frac2position_in_0_1_SB_tree := r -> RETURN(ReflectBinTreePermutation(cfrac2binexp(convert(1/r,confrac))));
%Y The first row (rotate the top node left): A065661, 2nd row (rotate the top node's left child): A065663, 3rd row (rotate the top node's right child): A065665, 4th row: A065667, 5th row: A065669, 6th row: A065671, 7th row: A065673. Cf. also A065674-A065676. For the other needed Maple procedures follow A065658 which gives the inverse permutations.
%K nonn,tabl
%O 0,1
%A _Antti Karttunen_, Nov 22 2001