OFFSET
0,4
COMMENTS
Consider the following rooted trees useful in the combinatorial game theory: each tree has zero or more subtrees at its left side and zero or more subtrees at its right side. The orientation of the subtrees among the other branches of the same side is not distinguished and all the subtrees of the same side are distinct from each other. These kinds of trees map bijectively to nonnegative integers by the following map f: f(empty tree) = 0 and f(tree with left subtrees Tl1, ..., Tlj and right subtrees Tr1, ..., Trk) = 2^(2*f(Tl1)) + ... + 2^(2*f(Tlj)) + 2^((2*f(Tr1))+1) + ... + 2^((2*f(Trk))+1). The ten game trees given on page 40 of "Winning Ways" thus translate to values Game 0 -> 0, Game 1 -> 1, Game -1 -> 2, Game 2 -> 4, Game 1/2 -> 9, Game 1/4 -> 524289, Game * -> 3, Game 1* -> 12, Game *2 -> 195, Game ^ (up) -> 129. However, this correspondence is not bijective with the computed equivalence classes of games, as many integers map to game trees with dominated o r reversible options. Here a(n) gives the total number of edges in the tree f(n).
REFERENCES
E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Second Edition, Vol. 1, A K Peters, 2001, p. 40.
EXAMPLE
3 = 2^0 + 2^1 = 2^(2*0) + 2^((2*0)+1) encodes the CGT tree \/ which has two edges, thus a(3)=2.
64 = 2^6 = 2^(2*3), i.e., it encodes the CGT tree
\/
.\
which has three edges, so a(64)=3.
PROG
(MIT Scheme:) (define (A106486 n) (cond ((zero? n) 0) (else (fold-left (lambda (x y) (+ x y 1)) 0 (map A106486 (map shr (on-bit-indices n))))))) (define (shr n) (if (odd? n) (/ (- n 1) 2) (/ n 2))) (define (on-bit-indices n) (let loop ((n n) (i 0) (c (list))) (cond ((zero? n) (reverse! c)) ((odd? n) (loop (/ (- n 1) 2) (1+ i) (cons i c))) (else (loop (/ n 2) (1+ i) c)))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 21 2005
STATUS
approved