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A192928 The Gi1 and Gi2 sums of Losanitsch's triangle A034851 6

%I #20 Jun 09 2020 17:54:27

%S 1,1,1,1,2,2,3,3,5,6,9,11,16,20,29,37,53,69,98,130,183,245,343,463,

%T 646,877,1220,1664,2310,3161,4381,6009,8319,11430,15811,21751,30070,

%U 41405,57216,78836,108906,150130,207346

%N The Gi1 and Gi2 sums of Losanitsch's triangle A034851

%C The Gi1 and Gi2 sums, see A180662 for their definitions, of Losanitsch's triangle A034851 equal this sequence.

%C From _Johannes W. Meijer_, Aug 26 2013: (Start)

%C The a(n) are also the Ca1 and Ca2 sums of McGarvey’s triangle A102541.

%C Furthermore they are the Kn11 and Kn12 sums of triangle A228570.

%C And finally the terms of this sequence are the row sums of triangle A228572. (End)

%H R. J. Mathar, <a href="http://arxiv.org/abs/1311.6135">Paving rectangular regions with rectangular tiles,....</a>, arXiv:1311.6135 [math.CO], Table 25.

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1,1,0,-1,0,1,-1,0,0,-1).

%F G.f.: (-1/2)*(1/(x^4+x-1) + (1+x+x^4)/(x^8+x^2-1))= -(1+x)*(x^7-x^6+x^5+x-1) / ( (x^4+x-1)*(x^8+x^2-1) ).

%F a(n) = (A003269(n+1)+x(n)+x(n-1)+x(n-4))/2 with x(2*n) = A003269(n+1) and x(2*n+1) = 0.

%F From _Johannes W. Meijer_, Aug 26 2013: (Start)

%F a(n) = sum(A228572(n, k), k=0..n)

%F a(n) = sum(A228570(n-k, k), k=0..floor(n/2))

%F a(n) = sum(A102541(n-2*k, k), k=0..floor(n/3))

%F a(n) = sum(A034851(n-3*k, k), k=0..floor(n/4)) (End)

%p A192928 := proc(n): (A003269(n+1)+x(n)+x(n-1)+x(n-4))/2 end: A003269 := proc(n): sum(binomial(n-1-3*j, j), j=0..(n-1)/3) end: x:=proc(n): if type(n,even) then A003269(n/2+1) else 0 fi: end: seq(A192928(n),n=0..42);

%t LinearRecurrence[{1, 1, -1, 1, 0, -1, 0, 1, -1, 0, 0, -1}, {1, 1, 1, 1, 2, 2, 3, 3, 5, 6, 9, 11}, 43] (* _Jean-François Alcover_, Nov 16 2017 *)

%Y Cf. A003269, A034851, A102541, A180662, A228570, A228572.

%K nonn,easy

%O 0,5

%A _Johannes W. Meijer_, Jul 14 2011

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