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A300561
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Deep factorization of n, A300560, converted from binary to decimal. (Binary digits obtained by recursively replacing each factor p^e with [primepi(p) [e]], then '[' = 1, ']' = 0.)
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4
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0, 12, 228, 240, 3876, 3300, 3972, 3984, 3696, 53028, 63780, 61668, 59172, 53124, 937764, 4032, 64548, 52848, 64644, 986916, 937860, 850212, 62340, 1020132, 62064, 845604, 59280, 987012, 948516, 13520676, 1034532, 64656, 15005988, 850980, 15880068, 986736, 1017636
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OFFSET
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1,2
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COMMENTS
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Convert to decimal the binary numbers A300560, which represent the deep factorization of n: each factor prime(i)^e_i is replaced by the expression [i [e_i]], recursively for indices i and exponents e_i, and finally '[' and ']' are considered as binary digits 1 and 0.
The initial a(1) = 0 represents the empty string of binary digits.
All terms are multiples of 4, and some of a higher power of 2, which represent the trailing closing parentheses of the deep factorization. These factors of 2 can be removed without loss of information; then all terms (except for n = 1) are odd, and we can consider (x-1)/2. This more condensed version is A300563(n) = (a(n)/2^valuation(a(n),2) - 1)/2, with binary representation given in A300562(n).
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LINKS
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EXAMPLE
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The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.
2 = prime(1)^1 => (1(1)) => (()) => 1100_2 = 12 = a(2).
3 = prime(2)^1 => (2(1)) => ((())()) => 11100100_2 = 228 = a(3).
4 = prime(1)^2 => (1(2)) => (((()))) => 11110000_2 = 240 = a(4).
5 = prime(3)^1 => (3(1)) => (((())())()) => 111100100100_2 = 3876 = a(5).
6 = prime(1)^1*prime(2)^1 => (1(1))(2(1)) => (())((())()) => 110011100100_2 = 3300 = a(6).
7 = prime(4)^1 => (4(1)) => ((((())))()) => 111110000100_2 = 3972 = a(7).
8 = prime(1)^3 => (1(3)) => ((((())()))) => 111110010000_2 = 3984 = a(8), and so on.
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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