0,3

Comment from Holger Petersen (petersen(AT)informatik.uni-stuttgart.de), May 29 2006: "Also number of binary strings of length n+2 containing the pattern 010. Proof: Clear for n = 0, 1, 2. For n > 2 each string with pattern 010 of length n-1 gives 2 strings of length n with the property by appending a symbol. In addition each string of length n-1 without 010 and ending in 01 contributes one new string. Denote by c_w(m) the number of strings of length m without 010 and ending in w.

"Since there is a total of 2^m strings of length m, we have c_01(m) = c_0(m-1) = (2^{m-1} - a(m-3)) - c_1(m-1) = (2^{m-1} - a(m-3)) - (2^{m-2} - a(m-4)) = 2^{m-2} - a(m-3) + a(m-4) (the first and third equalities follow from the fact that appending a 1 will not generate the pattern). The recurrence is a(n) = 2a(n-1) + c_01(n+1) = 2a(n-1) + 2^{n-1} - a(n-2) + a(n-3)"

a(n) = (1/3) [4*2^n + A077941(n-1) - 2*A077941(n+1)]. G.f.: x/[(1-2x)(1-2x+x^2+x^3)]. - R. Stephan, Aug 19 2004

f := proc(n) option remember; if n<=1 then n else if n<=3 then 7*n-10; else 2*f(n-1)-f(n-2)+f(n-3)+2^(n-1); fi; fi; end;

Sequence in context: A119706 A034345 A036890 this_sequence A047859 A100335 A080869

Adjacent sequences: A000250 A000251 A000252 this_sequence A000254 A000255 A000256

nonn

Jason Howald (jahowald(AT)umich.edu)