A000253 - OEIS

A000253

a(n) = 2*a(n-1) - a(n-2) + a(n-3) + 2^(n-1).

0, 1, 4, 11, 27, 63, 142, 312, 673, 1432, 3015, 6295, 13055, 26926, 55284, 113081, 230572, 468883, 951347, 1926527, 3894878, 7863152, 15855105, 31936240, 64269135, 129234351, 259690239, 521524126, 1046810092, 2100221753, 4212028452, 8444387067

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OFFSET

0,3

COMMENTS

From Holger Petersen (petersen(AT)informatik.uni-stuttgart.de), May 29 2006: (Start)

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).

(End)

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Mathematics Stack Exchange, Recurrence relations - binary substrings

Index entries for linear recurrences with constant coefficients, signature (4,-5,3,-2).

FORMULA

From Ralf Stephan, Aug 19 2004: (Start)

a(n) = (1/3)*(4*2^n + A077941(n-1) - 2*A077941(n+1)).

G.f.: x/((1-2*x)*(1 - 2*x + x^2 - x^3)). (End)

a(n) = A000079(n+2) - A005251(n+5). - Alois P. Heinz, Apr 03 2012

MAPLE

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;