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A119915 Number of ternary words of length n and having exactly one run of 0's of odd length.

%I A119915 #12 Jan 15 2020 00:15:35
%S A119915 0,1,4,13,40,117,332,921,2512,6761,18004,47525,124536,324317,840092,
%T A119915 2166065,5562272,14232273,36300196,92321085,234192584,592695109,
%U A119915 1496810732,3772761289,9492450672,23844342073,59804611060,149787196117
%N A119915 Number of ternary words of length n and having exactly one run of 0's of odd length.
%C A119915 Column 1 of A119914.
%H A119915 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-4,-1).
%F A119915 a(n) = [z^n] z*(1 - z^2)/(1 - 2*z - z^2)^2.
%F A119915 a(n) = A006645(n+1) - A006645(n-1). - _R. J. Mathar_, Aug 07 2015
%F A119915 From _Peter Luschny_, Jan 14 2020: (Start)
%F A119915 a(n) = Sum_{k=0..n} A193737(n, k)*k.
%F A119915 Let h(k) = (1 + k)*exp((1 + k)*x)*(1 + x - 1/k)/4 then
%F A119915 a(n) = n!*[x^n](h(sqrt(2)) + h(-sqrt(2))).  (End)
%e A119915 a(3) = 13 because we have 000, 011, 012, 021, 022, 101, 102, 110, 120, 201, 202, 210 and 220 (for example, 001, 020 do not qualify).
%p A119915 g := z*(1-z^2)/(1-2*z-z^2)^2:
%p A119915 gser := series(g,z=0,34):
%p A119915 seq(coeff(gser,z,n), n=0..30);
%t A119915 LinearRecurrence[ {4, -2, -4, -1}, {0, 1, 4, 13}, 28] (* _Peter Luschny_, Jan 14 2020 *)
%Y A119915 Cf. A119914, A193737, A006645.
%K A119915 nonn,easy
%O A119915 0,3
%A A119915 _Emeric Deutsch_, May 29 2006