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%I #28 Aug 11 2021 05:49:52
%S 0,1100,11100100,11110000,111100100100,110011100100,111110000100,
%T 111110010000,111001110000,1100111100100100,1111100100100100,
%U 1111000011100100,1110011100100100,1100111110000100,11100100111100100100,111111000000,1111110000100100,1100111001110000,1111110010000100
%N 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'.
%C Consider the prime factorization of n, replace each factor prime(i)^e_i with the parenthesized expression [i [e_i]], iterate this process on the indices i and exponents e_i, and finally replace '[' and ']' with digits '1' and '0'.
%C See A300561 for the decimal representation of these binary numbers.
%C There is redundancy: trailing '0's can be removed without loss of information; then each term ends in a digit 1 which can also be removed. This more condensed version is given in A300562, the decimal representation in A300563(n) = (m/2^valuation(m,2) - 1)/2 with m = a(n) [read in binary] = A300561(n).
%C The initial a(1) = 0 represents the empty string of binary digits.
%H J. Awbrey, <a href="https://oeis.org/wiki/Riffs_and_Rotes#Selected_Sequences">Riffs and Rotes, Selected Sequences</a>, OEIS wiki, Feb. 2010.
%e The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.
%e 2 = prime(1)^1 => (1(1)) => (()) => 1100 = a(2). (The 1's disappear, having empty factorization.)
%e 3 = prime(2)^1 => (2(1)) => ((())()) [using 2 => (())] => 11100100 = a(3).
%e 4 = prime(1)^2 => (1(2)) => (((()))) => 11110000 = a(4).
%e 5 = prime(3)^1 => (3(1)) => (((())())()) => 111100100100 = a(5).
%e 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.)
%e 7 = prime(4)^1 => (4(1)) => ((((())))()) => 111110000100 = a(7).
%e 8 = prime(1)^3 => (1(3)) => ((((())()))) => 111110010000 = a(8), and so on.
%e 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.
%e Examples: 1100 = (()) = (x(y)) with x = y = 1, so (()) = prime_1^1 = 2.
%e 110011100100 = _(())_(_(())_()) = 2 (2()) = 2 prime_2^1 = 6.
%e 111110010000 = (((_(())_()))) = ((_(2())_)) = ((3)) = prime_1^3 = 8.
%o (PARI) 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))~))
%Y Cf. A300561, A300562, A300563.
%Y Cf. A061396, A062504, A062860.
%K nonn
%O 1,2
%A _M. F. Hasler_, Mar 08 2018
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