

A297824


The number of iterations to remove all runs from the binary string 11011100...n (formed by concatenating the first n binary numbers, see A058935(n)).


2



0, 1, 1, 1, 1, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 9, 10, 10, 10, 10, 10, 10, 9, 8, 8, 10, 7, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 9, 10, 9, 10, 9, 10, 10, 8, 9, 9, 10, 10, 9, 7, 9, 10, 10, 10, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
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OFFSET

1,6


COMMENTS

Each iteration removes all runs of two or more identical bits that appear at the beginning of that iteration. By definition, the bits of the final string will be an initial segment, possibly empty, of 0, 1, 0, 1, ... (A000035) or 1, 0, 1, 0, ... (A059841). A297825 gives the lengths of the final strings; the sign of each nonzero term indicates which case occurs.


LINKS

Rick L. Shepherd, Table of n, a(n) for n = 1..10000


EXAMPLE

a(21) = 6 because 1101110010111011110001001101010111100110111101111100001000110010100111010010101 > 01001010100011010110101 > 01101010100101 > 0010101101 > 101001 > 1011 > 10, where each arrow points to the result of one iteration.


PROG

(PARI)
{remove_runs(v) = my(w, run_found = 0);
if(#v == 1, w = v, w = []);
for(k = 2, #v,
if(v[k1] == v[k],
run_found = 1,
if(run_found == 0, w = concat(w, v[k1]), run_found = 0);
if(k == #v, w = concat(w, v[k]))
)
); w}
{a(n) = my(v = [], L, c = 0); \\ remove "write(...); " if don't need other bfile
for(k = 1, n, v = concat(v, binary(k)));
L = #v;
while(1,
v = remove_runs(v);
if(#v == L, write("b297825.txt", n, " ", L*(if(L == 0, 0, 2*v[1]  1))); break, L = #v);
c++
); c}
for(n = 1, 10000, write("b297824.txt", n, " ", a(n))) \\ created two bfiles


CROSSREFS

Cf. A000035, A059841, A058935, A297825.
Sequence in context: A157901 A327441 A335855 * A281796 A072376 A131883
Adjacent sequences: A297821 A297822 A297823 * A297825 A297826 A297827


KEYWORD

nonn,base


AUTHOR

Rick L. Shepherd, Jan 06 2018


STATUS

approved



