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A300562 Condensed deep factorization of n, in binary. (Remove all trailing 0's and one trailing 1 from A300560.) 4
0, 1, 11100, 111, 111100100, 110011100, 111110000, 1111100, 1110011, 1100111100100, 1111100100100, 1111000011100, 1110011100100, 1100111110000, 11100100111100100, 11111, 1111110000100, 11001110011, 1111110010000, 11110000111100100 (list; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A204758 A211686 A224463 * A253378 A253371 A204226
KEYWORD
nonn
AUTHOR
M. F. Hasler, Mar 08 2018
STATUS
approved

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Last modified February 28 16:50 EST 2024. Contains 370400 sequences. (Running on oeis4.)