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%I #12 Jun 14 2022 14:11:33
%S 102,51,51,23,28,23,12,14,14,12,3,4,8,4,3,1,2,4,4,2,1
%N Finite table read by antidiagonals: T(n, m) is the number of conics passing through n points, tangent to m lines, and tangent to k=5-n-m conics in general position, divided by 2^k, with 0 <= n+m <= 5.
%C Row (i. e. antidiagonal) sums are: {102, 102, 74, 52, 22, 14}.
%C T(0, 0) * 2^5 = A328148(2) = 3264 answers the Steiner's problem: How many conics are simultaneously tangent to five fixed conics?
%H Andrew Bashelor, Amy Ksir, and Will Traves, <a href="http://www.maa.org/programs/maa-awards/writing-awards/hasse-prizes/enumerative-algebraic-geometry-of-conics-0">Enumerative Algebraic Geometry of Conics</a>, The American Mathematical Monthly, vol. 115, no. 8, October 2008, pp. 701-728.
%e The complete triangle:
%e {{102},
%e {51, 51},
%e {23, 28, 23},
%e {12, 14, 14, 12},
%e {3, 4, 8, 4, 3},
%e {1, 2, 4, 4, 2, 1}}
%e Without dividing by 2^k, the triangle becomes:
%e {3264}
%e {816, 816}
%e {184, 224, 184}
%e {48, 56, 56, 48}
%e {6, 8, 16, 8, 6}
%e {1, 2, 4, 4, 2, 1}
%Y Cf. A328148.
%K nonn,tabl,fini,full
%O 1,1
%A _Roger L. Bagula_ and _Gary W. Adamson_, Oct 09 2008
%E Edited by _Andrey Zabolotskiy_, Jun 14 2022
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