A300560
Deep factorization of n, written in binary: replace each factor p^e with the expression [primepi(p) [ e ]], iterate on these numbers, finally replace '[' and ']' with '1' and '0'.
Original entry on oeis.org
0, 1100, 11100100, 11110000, 111100100100, 110011100100, 111110000100, 111110010000, 111001110000, 1100111100100100, 1111100100100100, 1111000011100100, 1110011100100100, 1100111110000100, 11100100111100100100, 111111000000, 1111110000100100, 1100111001110000, 1111110010000100
Offset: 1
The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.
2 = prime(1)^1 => (1(1)) => (()) => 1100 = a(2). (The 1's disappear, having empty factorization.)
3 = prime(2)^1 => (2(1)) => ((())()) [using 2 => (())] => 11100100 = a(3).
4 = prime(1)^2 => (1(2)) => (((()))) => 11110000 = a(4).
5 = prime(3)^1 => (3(1)) => (((())())()) => 111100100100 = a(5).
6 = prime(1)^1*prime(2)^1 => (1(1))(2(1)) => (())((())()) => 110011100100 = a(6) (= concatenation of a(2) and a(3), since 6 = 2*3.)
7 = prime(4)^1 => (4(1)) => ((((())))()) => 111110000100 = a(7).
8 = prime(1)^3 => (1(3)) => ((((())()))) => 111110010000 = a(8), and so on.
To convert back to the usual factorization, replace 0 and 1 by ')' and '(', then iteratively replace any (x(y)) by prime_x^y, where an empty x or y means 1.
Examples: 1100 = (()) = (x(y)) with x = y = 1, so (()) = prime_1^1 = 2.
110011100100 = _(())_(_(())_()) = 2 (2()) = 2 prime_2^1 = 6.
111110010000 = (((_(())_()))) = ((_(2())_)) = ((3)) = prime_1^3 = 8.
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A300560(n)=(n=factor(n))||return("");n[,1]=apply(primepi,n[,1]);concat(apply(t->Str("1"t[1]"1"t[2]"00"),Col(apply(A300560,n))~))
A300561
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.)
Original entry on oeis.org
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
Offset: 1
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.
A300562
Condensed deep factorization of n, in binary. (Remove all trailing 0's and one trailing 1 from A300560.)
Original entry on oeis.org
0, 1, 11100, 111, 111100100, 110011100, 111110000, 1111100, 1110011, 1100111100100, 1111100100100, 1111000011100, 1110011100100, 1100111110000, 11100100111100100, 11111, 1111110000100, 11001110011, 1111110010000, 11110000111100100
Offset: 1
The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.
To reconstruct the full deep factorization A300560(n), append a digit 1 and then as many 0's as to balance the number of 1's:
a(2) = 1, append a 1 => 11; append two 0's => 1100 = A300560(2).
a(3) = 11100, append a 1 => 111001; append two 0's => 11100100 = A300560(3).
a(4) = 111, append a 1 => 1111; append four 0's => 11110000 = A300560(4).
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