A300562 Condensed deep factorization of n, in binary. (Remove all trailing 0's and one trailing 1 from A300560.)

0, 1, 11100, 111, 111100100, 110011100, 111110000, 1111100, 1110011, 1100111100100, 1111100100100, 1111000011100, 1110011100100, 1100111110000, 11100100111100100, 11111, 1111110000100, 11001110011, 1111110010000, 11110000111100100

Offset

1,3

Comments

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.

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.

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

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.

Links

Table of n, a(n) for n=1..20.

Formula

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

Example

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).

Prog

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

Crossrefs

Cf. A300560, A300561, A300563.

Cf. A061396, A062504, A062860.

Sequence in context: A204758 A211686 A224463 * A253378 A253371 A204226

Adjacent sequences: A300559 A300560 A300561 * A300563 A300564 A300565

Keyword

nonn

Author

M. F. Hasler, Mar 08 2018

Status

approved