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%I #7 May 01 2014 02:44:55
%S 0,1,2,5,8,3,64,37,18,9,1024,7,32768,65,10,549,2097152,19,268435456,
%T 13,66,1025,68719476736,39,136,32769,274,69,35184372088832,11,
%U 36028797018963968,16933,1026,2097153,72,23,73786976294838206464
%N Prime factorization of n encoded by interleaving successive prime exponents in unary to bit-positions given by columns of A001477.
%C Here we store the exponent e_i of p_i (p1=2, p2=3, p3=5, ...) in the factorization of n to the bit positions given by the column i-1 of A001477 viewed as a table (the exponent of 2 is thus stored to bit positions 0, 2, 5, 9, 14, 20, ..., exponent of 3 to 1, 4, 8, 13, 19, ..., exponent of 5 to 3, 7, 12, 18, 25, ...) using unary system, i.e. we actually store 2^(e_i) - 1 in binary.
%C This injective mapping from N to N offers an example of the proof given in Cameron's book that any distributive lattice can be represented as a sublattice of the power-set lattice P(X) of some set X. With this we can implement GCD (A003989) with bitwise AND (A004198) and LCM (A003990) with bitwise OR (A003986). Also, to test whether x divides y, it is enough to check that ((a(x) OR a(y)) XOR a(y)) = A003987(A003986(a(x),a(y)),a(y)) is zero.
%D P. J. Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press, 1998, page 191. (12.3. Distributive lattices)
%e a(24) = 39 because 24 = 2^3 * 3^1 so we add the binary words 100101 and 10 to get 100111 in binary = 39 in decimal and a(25) = 136 because 25 = 5^2 so we form a binary word 10001000 = 136 in decimal.
%Y Variant: A075173. Inverse: A075176.
%Y A003989(x, y) = A075176(A004198(a(x), a(y))), A003990(x, y) = A075176(A003986(a(x), a(y))).
%K nonn
%O 1,3
%A _Antti Karttunen_, Sep 13 2002
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