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A075173 Prime factorization of n encoded by interleaving successive prime exponents in unary to bit-positions given by columns of A075300. 10
0, 1, 2, 5, 8, 3, 128, 21, 34, 9, 32768, 7, 2147483648, 129, 10, 85, 9223372036854775808, 35, 170141183460469231731687303715884105728, 13, 130, 32769 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
As in A059884, here also 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 A075300 (the exponent of 2 is thus stored to bit positions 0, 2, 4, ..., exponent of 3 to 1, 5, 9, 13, ..., exponent of 5 to 3, 11, 19, 27, 35, ...), but using unary instead of binary system, i.e. we actually store 2^(e_i) - 1 in binary.
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. This allows us to 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.
REFERENCES
P. J. Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge University Press, 1998, page 191. (12.3. Distributive lattices)
LINKS
EXAMPLE
a(24) = 23 because 24 = 2^3 * 3^1 so we add the binary words 10101 and 10 to get 10111 in binary = 23 in decimal and a(25) = 2056 because 25 = 5^2 so we form a binary word 100000001000 = 2056 in decimal.
CROSSREFS
Variant: A075175. Inverse: A075174. Cf. A059884.
A003989(x, y) = A075174(A004198(a(x), a(y))), A003990(x, y) = A075174(A003986(a(x), a(y))).
Sequence in context: A152248 A248906 A075175 * A163337 A341488 A114550
KEYWORD
nonn
AUTHOR
Antti Karttunen, Sep 13 2002
STATUS
approved

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Last modified March 18 22:34 EDT 2024. Contains 370951 sequences. (Running on oeis4.)