

A188106


Triangle T(n,k) with the coefficient [x^k] of 1/(12*xx^2+x^3)^(nk+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, 9414, 5521, 1429, 1, 18, 152, 777, 2625, 6032, 9366, 9414, 5521, 1429
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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/(1x), 2*x/(1x)^2, x^2*(5x)/(1x)^3, x^3*(112*xx^2)/(1x)^4, x^4*(256*x3*x^2)/(1x)^5, ..., for which the sum of the signed coefficients in the nth 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/(12*xx^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.


LINKS

Table of n, a(n) for n=0..55.


FORMULA

sum_{k=0..n} T(n,k)=A033505(n).
T(n,0) = 1. T(n,2) = A014106(n1). T(n,3) = (n2)*(4*n^2+2*n9)/3. T(n,4) = (n2)*(n3)*(2*n+7)*(2*n3)/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/(12*xx^2+x^3)^(nk+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

Cf. A006054, A033505, A189426, A189427.
Sequence in context: A161135 A237274 A038730 * A050166 A124959 A081281
Adjacent sequences: A188103 A188104 A188105 * A188107 A188108 A188109


KEYWORD

nonn,tabl


AUTHOR

L. Edson Jeffery, Mar 20 2011


STATUS

approved



