A300562 - OEIS

%I #8 Mar 17 2018 17:42:01

%S 0,1,11100,111,111100100,110011100,111110000,1111100,1110011,

%T 1100111100100,1111100100100,1111000011100,1110011100100,

%U 1100111110000,11100100111100100,11111,1111110000100,11001110011,1111110010000,11110000111100100

%N Condensed deep factorization of n, in binary. (Remove all trailing 0's and one trailing 1 from A300560.)

%C The binary representation of the deep factorization of n, A300560, is obtained by recursively replacing any factor prime(i)^e_i by the expression [i [e_i]], and finally taking '[' and ']' as binary digits 1 and 0.

%C This always ends in trailing 0's which can be safely removed without loss of information; then there is a final binary digit 1 that can also be dropped. The result is a(n), in binary, equal to A300563(n) when converted to decimal.

%C The initial a(1) = 0 results from the empty factorization of 1.

%C To reconstruct the deep factorization of n > 1, append a digit 1 and then as many 0's (namely: A145037(a(n))+1) as to have the same number of as 1's.

%F a(n) = A007088(A300563(n)), see there for an expression in terms of A300560. - _M. F. Hasler_, Mar 16 2018

%e The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.

%e 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:

%e a(2) = 1, append a 1 => 11; append two 0's => 1100 = A300560(2).

%e a(3) = 11100, append a 1 => 111001; append two 0's => 11100100 = A300560(3).

%e a(4) = 111, append a 1 => 1111; append four 0's => 11110000 = A300560(4).

%o (PARI) A300562(n)=(n=eval(A300560(n)))\10^valuation(10*n+!n,10)

%Y Cf. A300560, A300561, A300563.

%Y Cf. A061396, A062504, A062860.

%K nonn

%O 1,3

%A _M. F. Hasler_, Mar 08 2018