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%I #10 Nov 11 2017 19:59:59
%S 1,1,1,1,1,3,1,5,2,1,7,7,1,9,16,3,1,11,29,15,1,13,46,43,5,1,15,67,95,
%T 30,1,17,92,179,104,8,1,19,121,303,271,58,1,21,154,475,591,235,13,1,
%U 23,191,703,1140,705,109,1,25,232,995,2010,1746,506,21,1,27,277,1359,3309,3780
%N Triangle read by rows: T(n,k) = number of k-matchings in the graph obtained by a zig-zag triangulation of a convex n-gon, T(0,0)=T(1,0)=T(2,0)=T(2,1)=1 (n > 2, 0 <= k <= floor(n/2)).
%H Andrew Howroyd, <a href="/A101350/b101350.txt">Table of n, a(n) for n = 0..649</a>
%F G.f.: 1/(1 - z - tz^2 - tz^3 - t^2z^4).
%e T(5,2)=7 because in the triangulation of the convex pentagon ABCDEA with diagonals AD and AC we have seven 2-matchings: {AB,CD},{AB,DE},{BC,AD},{BC,DE},{BC,EA},{CD,EA} and {DE,AC}.
%e Triangle begins:
%e 1;
%e 1;
%e 1, 1;
%e 1, 3;
%e 1, 5, 2;
%e 1, 7, 7;
%e 1, 9, 16, 3;
%e 1, 11, 29, 15;
%e 1, 13, 46, 43, 5;
%e ...
%p G:=1/(1-z-t*z^2-t*z^3-t^2*z^4):Gserz:=simplify(series(G,z=0,18)):P[0]:=1: for n from 1 to 16 do P[n]:=sort(coeff(Gserz,z^n)) od:for n from 0 to 16 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od;# yields the sequence in triangular form
%o (PARI)
%o s(n) = 1/(1-x-y*x^2-y*x^3-y^2*x^4) + O(x^n);
%o my(gf=Pol(s(20))); for(n=0, poldegree(gf), my(p=polcoeff(gf,n)); for(k=0, poldegree(p), print1(polcoeff(p,k), ", ")); print) \\ _Andrew Howroyd_, Nov 04 2017
%Y Row sums yield A000078 (the tetranacci numbers). T(2n+1, n) = A023610(n) (n > 0). T(2n, n) = A000045(n+1) (the Fibonacci numbers).
%Y Cf. A000078, A023610, A000045.
%K nonn,tabf
%O 0,6
%A _Emeric Deutsch_, Dec 25 2004
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