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Array M read by antidiagonals in which successive rows evidently converge to A001405 (central binomial coefficients).
4

%I #29 Apr 20 2013 04:47:36

%S 1,1,1,1,1,1,1,1,2,1,1,1,2,3,1,1,1,2,3,5,1,1,1,2,3,6,8,1,1,1,2,3,6,10,

%T 13,1,1,1,2,3,6,10,19,21,1,1,1,2,3,6,10,20,33,34,1,1,1,2,3,6,10,20,35,

%U 61,55,1,1,1,2,3,6,10,20,35,69,108,89,1

%N Array M read by antidiagonals in which successive rows evidently converge to A001405 (central binomial coefficients).

%C CONJECTURE 1. Let M(n,k) (n,k >= 0) denote the entry in row n and column k of the array. For all n, M(n,j) = A001405(j), j=0,...,2*n+1; hence row n of M -> A001405 as n -> infinity.

%C Taking finite differences of even numbered columns from the top -> down yields triangle A205946 with row sums A000984, central binomial coefficients; while odd numbered columns yield triangle A205945 with row sums A001700. A205946 and A205945 represent the bisection of A191314. - Gary W. Adamson, Feb 01 2012

%H L. E. Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>

%F Let N=2*n+3. For each n>0, define the (n+1) X (n+1) tridiagonal unit-primitive matrix (see [Jeffery]) B_n = A_{N,1} = [0,1,0,...,0; 1,0,1,0,...,0; 0,1,0,1,0,...,0; ...; 0,...,0,1,0,1; 0,...,0,1,1], and put B_0 = [1]. Then, for all n, M(n,k)=[(B_n)^k]_{n+1,n+1}, k=0,1,..., where X_{n+1,n+1} denotes the lower right corner entry of X.

%F CONJECTURE 2 (Rows of M). Let S(n,i) denote term i in row n of A115139, i=0,...,floor(n/2), and let T(n,j) denote term j in row n of A108299, j=0,...,n. The generating function for row n of M is of the form F_n(x) =sum[i=0,...,floor(n/2) S(n,i)*x^(2*i)]/sum[j=0,...,n T(n,j)*x^j].

%F CONJECTURE 3 (Columns of M). Let D(m,k) denote term m in column k of A191314, m=0,...,floor(k/2). The generating function for column k of M is of the form G_k(x)=sum[m=0,...,floor(k/2) D(m,k)*x^m]/(1-x).

%e Array begins

%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...

%e 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,...

%e 1, 1, 2, 3, 6, 10, 19, 33, 61, 108, 197,...

%e 1, 1, 2, 3, 6, 10, 20, 35, 69, 124, 241,...

%e 1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 251,...

%e 1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252,...

%e ...

%e According to Conjecture 2, row n=3 has g.f. F_3(x)=(1-2*x^2)/(1-x-3*x^2+2*x^3+x^4).

%Y Cf. A001405, A108299, A115139, A191314.

%Y Cf. also A205945, A205946, A001700, A000984.

%K nonn,tabl

%O 0,9

%A _L. Edson Jeffery_, Jan 29 2012