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

1, 1, 1, 1, 2, 2, 3, 3, 5, 6, 9, 11, 16, 20, 29, 37, 53, 69, 98, 130, 183, 245, 343, 463, 646, 877, 1220, 1664, 2310, 3161, 4381, 6009, 8319, 11430, 15811, 21751, 30070, 41405, 57216, 78836, 108906, 150130, 207346

Offset: 0

Author

Johannes W. Meijer, Jul 14 2011

Keywords

• nonn
• easy

Comments

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

From Johannes W. Meijer, Aug 26 2013: (Start)

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

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

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

Links

• R. J. Mathar, Paving rectangular regions with rectangular tiles,...., arXiv:1311.6135 [math.CO], Table 25.
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,0,-1,0,1,-1,0,0,-1).

Crossrefs

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

Programs

Maple

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);

Mathematica

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 *)

Formula

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) ).

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.

From Johannes W. Meijer, Aug 26 2013: (Start)

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

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

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

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