A300563
Condensed deep factorization of n, A300562(n) written in decimal: floor of odd part of A300561(n) divided by 2.
Original entry on oeis.org
0, 1, 28, 7, 484, 412, 496, 124, 115, 6628, 7972, 7708, 7396, 6640, 117220, 31, 8068, 1651, 8080, 123364, 117232, 106276, 7792, 127516, 1939, 105700, 1852, 123376, 118564, 1690084, 129316, 2020, 1875748, 106372, 1985008, 30835, 127204, 106384, 1875172, 2040292, 124708
Offset: 1
The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.
The binary-coded deep factorization is restored as follows (and a(n) calculated from this going the opposite direction):
a(2) = 1, append a bit 1 or do 1 X 2 + 1 = 3 = 11[2]. This has 2 bits 1, no bit 0 so append 2 bits 0 => A300560(2) = 1100 in binary, or 12 = A300561(2) in decimal.
a(3) = 28 = 11100[2], append a bit 1 or do 28 X 2 + 1 = 57 = 111001[2]. This has 4 bits 1 and 2 bits 0, so append two more of the latter => A300560(3) = 11100100 in binary or A300561(3) = 228 in decimal.
a(4) = 7 = 111[2], append a bit 1 or do 7 X 2 + 1 = 15 = 1111[2]. This has 4 bits 1 and no bit 0 so append 4 0's => 11110000 = A300560(4) or A300561(4) = 240 in decimal.
See A300560 for conversion of this binary coding of the deep factorization into the ordinary factorization.
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.
-
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))~))
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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