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A006345
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Linus sequence: a(n) "breaks the pattern" by avoiding the longest doubled suffix.
(Formerly M0074)
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9
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1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2
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OFFSET
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1,2
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COMMENTS
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To find a(n), consider either a 1 or a 2. For each, find the longest repeated suffix, that is, for each of a(n)=1,2, find the longest sequence s with the property that the sequence a(1),...,a(n) ends with ss. Use the digit that results in the shorter such suffix. a(1) = 1. The empty sequence of length 0 is the shortest possible suffix and is trivially doubled. Note that this doesn't result in exactly Linus's choices. - K. Ramsey, kramsey(AT)aol.com
On average, it seems that (# of 1s up to n) - (# of 2s up to n) -> infinity as n -> infinity (as O(log n)?), while the asymptotic density of either 1s or 2s appears to be 1/2. - Daniel Forgues, Mar 01 2017
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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P. Balister, S. Kalikow, A. Sarkar, The Linus sequence, Preprint May 2007; Combinatorics, Probability and Computing, Volume 19, Issue 1 January 2010 , pp. 21-46..
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EXAMPLE
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After 1,2,1,1,2,2,1,2, if we put a 1, the suffix {2,1} repeats, but if we put a 2 the longer suffix {1,2,2} repeats, so the next term is 1.
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MAPLE
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LDS:= proc(L)
local Cands, r, m;
Cands:= {$1..floor(nops(L)/2)};
r:= 0;
for m from 1 while nops(Cands) > 0 do
Cands:= select(c -> L[-m] = L[-c-m], Cands);
if min(Cands) = m then
r:= m;
Cands:= subs(m=NULL, Cands);
fi
od;
r
end proc:
A:= 1:
for n from 2 to 10^3 do
if LDS([A, 1]) < LDS([A, 2]) then A:= A, 1 else A:= A, 2 fi;
od:
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MATHEMATICA
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a[1]=1; a[2]=2; suffix[lst_] := If[MatchQ[lst, {___, b__, b__}], lst /. {___, b__, b__} :> {b}, {}]; a[n_] := a[n] = Module[{aa, lg1, lg2}, aa = Array[a, n-1]; lg1 = suffix[Append[aa, 1]] // Length; lg2 = suffix[Append[aa, 2]] // Length; If[lg1 <= lg2, 1, 2]]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Dec 11 2014 *)
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PROG
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(Perl) -le 'print$_.=3**/(.*)(.)\1$/-$2for($_)x99' (Ton Hospel/Phil Carmody) [An example of Perl golfing: use as few (key)strokes as possible]
(PARI) {a(n)=local(A, t); if(n<2, n>0, A=[1]; for(i=2, n, forstep(j=i\2-1, 0, -1, for(k=1, j, if(A[i-j-k-1]!=A[i-k], next(2))); t=j; break); A=concat(A, [3-A[i-t-1]])); A[n])} /* Michael Somos, May 04 2006 */
The approach I used was to take advantage of Perl's regular expression capabilities, coupled with the realization that Perl can optimize patterns anchored to the start far better than those anchored to the end - reversing the string to allow that gave a speedup of several orders of magnitude:
my $string = "";
digit('1', 0);
for (2 .. $limit) {
my($repeat, $digit) = ($string =~ m{ ^ (.*) ([12]) \1 }x) or die;
digit($digit eq '1' ? '2' : '1', length($repeat) + 1);
}
sub digit {
my($digit, $repeat) = @_;
$string = $digit . $string;
# n A6345(n) A6346(n)
printf "%s %s %s\n", length($string), $digit, $repeat;
}
This takes about 45s to calculate 50000 terms of both sequences. (End)
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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