%I M0074 #124 Jul 17 2024 08:47:24
%S 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,
%T 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,
%U 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
%N Linus sequence: a(n) "breaks the pattern" by avoiding the longest doubled suffix.
%C 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
%C 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
%D N. S. Hellerstein, Letter to _N. J. A. Sloane_ (1978).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, Robert Israel, and Hugo van der Sanden, <a href="/A006345/b006345.txt">Table of n, a(n) for n = 1..50000</a> (1..1000 from T. D. Noe, 1001..20000 from Robert Israel)
%H P. Balister, S. Kalikow, and A. Sarkar, <a href="http://www2.ims.nus.edu.sg/preprints/2007-19.pdf">The Linus sequence</a>, Preprint May 2007; Combinatorics, Probability and Computing, Volume 19, Issue 1 January 2010 , pp. 21-46..
%H N. Hellerstein, <a href="/A006345/a006345.pdf">Letter to N. J. A. Sloane, 1978</a>
%H N. Hellerstein, M. Gardner, and S. Kim, <a href="/A006345/a006345_1.pdf">Correspondence related to the Linus and Sally sequences, 1977</a>
%H N. J. A. Sloane, <a href="/A006345/a006345.html">Illustration of initial terms</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LinusSequence.html">Linus Sequence.</a>
%e 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.
%p LDS:= proc(L)
%p local Cands, r,m;
%p Cands:= {$1..floor(nops(L)/2)};
%p r:= 0;
%p for m from 1 while nops(Cands) > 0 do
%p Cands:= select(c -> L[-m] = L[-c-m], Cands);
%p if min(Cands) = m then
%p r:= m;
%p Cands:= subs(m=NULL,Cands);
%p fi
%p od;
%p r
%p end proc:
%p A:= 1:
%p for n from 2 to 10^3 do
%p if LDS([A,1]) < LDS([A,2]) then A:= A,1 else A:= A,2 fi;
%p od:
%p seq(A[i],i=1..10^3); # _Robert Israel_, Jun 22 2015
%t LDS[L_] := Module[{Cands, r, m}, Cands = Range[Floor[Length[L]/2]]; r = 0; For[m = 1, Length[Cands] > 0, m++, Cands = Select[Cands, L[[-m]] == L[[-# - m]]&]; If[Min[Cands] == m, r = m; Cands = ReplaceAll[Cands, m -> Nothing]]]; r]; A = {1}; For[n = 2, n <= 10^3, n++, If[LDS[Append[A, 1]] < LDS[Append[A, 2]], A = Append[A, 1], A = Append[A, 2]]];
%t Table[A[[i]], {i, 1, 105}] (* _Jean-François Alcover_, Jul 17 2024, after _Robert Israel_ *)
%o (Perl) -le 'print$_.=3**/(.*)(.)\1$/-$2for($_)x99' # (Ton Hospel/Phil Carmody) [An example of Perl golfing: use as few (key)strokes as possible]
%o (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 */
%o (Perl)
%o =Comment on calculating this sequence and A006346 with Perl, from _Hugo van der Sanden_, Jun 23 2015:
%o 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:
%o =cut
%o my $string = "";
%o digit('1', 0);
%o for (2 .. $limit) {
%o my($repeat, $digit) = ($string =~ m{ ^ (.*) ([12]) \1 }x) or die;
%o digit($digit eq '1' ? '2' : '1', length($repeat) + 1);
%o }
%o sub digit {
%o my($digit, $repeat) = @_;
%o $string = $digit . $string;
%o # n A6345(n) A6346(n)
%o printf "%s %s %s\n", length($string), $digit, $repeat;
%o }
%o # This takes about 45s to calculate 50000 terms of both sequences.
%Y Cf. A006346, A094840, A157238 (A006345(n) - 1).
%K nonn,easy,nice
%O 1,2
%A _N. J. A. Sloane_
%E More terms from _Naohiro Nomoto_, May 21 2001
%E Additional comments from _Mitch Harris_, Dec 31 2003