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A205573
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Array M read by antidiagonals in which successive rows evidently converge to A001405 (central binomial coefficients).
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4
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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, 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, 61, 55, 1, 1, 1, 2, 3, 6, 10, 20, 35, 69, 108, 89, 1
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OFFSET
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0,9
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COMMENTS
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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.
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
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LINKS
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FORMULA
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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.
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].
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).
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EXAMPLE
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Array begins
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,...
1, 1, 2, 3, 6, 10, 19, 33, 61, 108, 197,...
1, 1, 2, 3, 6, 10, 20, 35, 69, 124, 241,...
1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 251,...
1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252,...
...
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).
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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