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A188106
Triangle T(n,k) with the coefficient [x^k] of 1/(1-2*x-x^2+x^3)^(n-k+1) in row n, column k.
3
1, 1, 2, 1, 4, 5, 1, 6, 14, 11, 1, 8, 27, 42, 25, 1, 10, 44, 101, 119, 56, 1, 12, 65, 196, 342, 322, 126, 1, 14, 90, 335, 770, 1080, 847, 283, 1, 16, 119, 526, 1495, 2772, 3248, 2180, 636, 1, 18, 152, 777, 2625, 6032, 9366, 9414, 5521, 1429, 1, 20, 189, 1096, 4284, 11718, 22590, 30148, 26517, 13804, 3211
OFFSET
0,3
COMMENTS
Modified versions of the generating function for D(0)={1,2,5,11,...}=A006054(m+2), m=0,1,2,..., are related to rhombus substitution tilings (see A187068, A187069 and A187070). The columns of the triangle have generating functions 1/(1-x), 2*x/(1-x)^2, x^2*(5-x)/(1-x)^3, x^3*(11-2*x-x^2)/(1-x)^4, x^4*(25-6*x-3*x^2)/(1-x)^5, ..., for which the sum of the signed coefficients in the n-th numerator equals 2^n. The diagonals {1,2,5,...}, {1,4,14,...}, ..., are generated by successive series expansion of F(n+1,x), n=0,1,..., where F(n,x)=1/(1-2*x-x^2+x^3)^n. For example, the second diagonal is {T{1,0},T{2,1},...}={1,4,14,...}=A189426, for which successive partial sums give A189427 (excluding the zero terms). Moreover, the diagonals correspond to successive convolutions of A006054 (= the first diagonal) with itself.
FORMULA
Sum_{k=0..n} T(n,k) = A033505(n).
T(n,0) = 1.
T(n,2) = A014106(n-1).
T(n,3) = (n-2)*(4*n^2+2*n-9)/3.
T(n,4) = (n-2)*(n-3)*(2*n+7)*(2*n-3)/6.
EXAMPLE
1;
1, 2;
1, 4, 5;
1, 6, 14, 11;
1, 8, 27, 42, 25;
1, 10, 44, 101, 119, 56;
1, 12, 65, 196, 342, 322, 126;
1, 14, 90, 335, 770, 1080, 847, 283;
1, 16, 119, 526, 1495 ...
MAPLE
A188106 := proc(n, k) 1/(1-2*x-x^2+x^3)^(n-k+1) ; coeftayl(%, x=0, k) ; end proc:
seq(seq(A188106(n, k), k=0..n), n=0..10) ; # R. J. Mathar, Mar 22 2011
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
L. Edson Jeffery, Mar 20 2011
EXTENSIONS
a(43) and following corrected by Georg Fischer, Oct 14 2023
STATUS
approved