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%I #8 Jan 07 2018 23:30:42
%S 1,-1,-1,-1,-4,-2,-2,-2,-2,2,4,4,2,3,3,3,3,4,5,7,2,4,6,6,7,6,5,5,5,5,
%T 5,5,5,6,7,6,4,6,7,9,7,1,-3,-2,-1,-1,0,0,1,2,2,0,3,1,1,1,0,0,0,0,0,0,
%U 0,0,0,0,1,0,1,0,1,0,0,2,1,1,-2,0,1,3,1,0,-2,-6,1,-2,-5,-4,-4,-3,-4,-4,-3,-3,-3,-3,-4,-5,-5,-6
%N a(n) is the (negative) length of the final string after iteratively removing all runs from the binary string 11011100...n (formed by concatenating the first n binary numbers, see A058935(n)).
%C There are exactly three cases: The final string may have length zero, in which case a(n) = 0. If the final string begins 0101... (bits are an initial segment of A000035), a(n) is negative. Otherwise, the final string begins 1010... (bits are an initial segment of A059841) and a(n) is positive. Thus |a(n)| always gives the actual string length (number of terms in the segment).
%C Of interest is the frequency of sign changes and 0 terms as n becomes large. The largest values of n in the current b-file such that a(n) = 0 are 8612 and 9899.
%H Rick L. Shepherd, <a href="/A297825/b297825.txt">Table of n, a(n) for n = 1..10000</a>
%e a(21) = 2 because the final string after iteratively removing all runs of two or more identical bits from 11011100...10101 is 10 of length two (as shown in the A297824 example). This term is positive because the first bit of 10 is 1.
%o (PARI) \\ See the program given in A297824.
%Y Cf. A000035, A059841, A058935, A297824.
%K sign,base
%O 1,5
%A _Rick L. Shepherd_, Jan 07 2018
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