%I #27 Jun 26 2023 16:09:49
%S 2,4,7,12,16,24,29,40,46,60,67,84,92,112,121,144,154,180,191,220,232,
%T 264,277,312,326,364,379,420,436,480,497,544,562,612,631,684,704,760,
%U 781,840,862,924,947,1012,1036,1104,1129,1200,1226,1300,1327
%N Expansion of g.f. x*(-2 - 2*x + x^2 - x^3)/((1 + x)^2 *(-1 + x)^3).
%C a(n) gives the number of vertices encountered along the shortest walk that encounters every edge at least once on the graph with n vertices where the graph is both complete and every node also has an edge to itself.
%C a(n) can be thought of as the length of a list made up using n distinct elements where every element is next to every other element (including a copy of itself) at least once. Such a list could be used forwards and backward when kerning a font as a way to minimize the number of characters typed in total.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F a(n) = n + (n mod 2) + (n * (n - (n mod 2)))/2.
%F a(2*n) = 2*n + 2*n^2;
%F a(2*n - 1) = 1 - n + 2*n^2.
%F E.g.f.: (2 + x)*(exp(x)*x + sinh(x))/2. - _Stefano Spezia_, May 07 2023
%e G.f.: 2*x + 4*x^2 + 7*x^3 + 12*x^4 + 16*x^5 + 24*x^6 + 29*x^7 + 40*x^8 + 46*x^9 + ...
%t CoefficientList[Series[x(-2-2x+x^2-x^3)/((1+x)^2(-1+x)^3), {x, 0, 50}], x]
%t (* or *)
%t LinearRecurrence[{1, 2, -2, -1, 1}, {2, 4, 7, 12, 16}, 50]
%o (Python) def a(n: int): return n + (n & 1) + n * ( n >> 1 )
%Y Cf. A053439.
%K nonn,easy
%O 1,1
%A _Jonathon Priestley_, Apr 28 2023