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A323942
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Irregular triangle read by rows giving the total number of isomers (nonisomorphic systems) of unbranched k-4-catafusenes.
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2
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1, 1, 1, 2, 3, 3, 1, 4, 7, 9, 3, 1, 10, 23, 29, 16, 5, 1, 25, 69, 99, 62, 27, 5, 1, 70, 229, 351, 275, 132, 39, 7, 1, 196, 731, 1249, 1121, 643, 221, 55, 7, 1, 574, 2385, 4437, 4584, 2997, 1278, 367, 72, 9, 1, 1681, 7657, 15597, 18012, 13458, 6678, 2322, 540, 93, 9, 1
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OFFSET
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2,4
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COMMENTS
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Let I(r, k) be the total number of isomers (nonisomorphic systems) of unbranched k-4-catafusenes, which are generated from catafusenes by converting k of its r hexagons to tetragons. According to Cyvin et al. (1996), for r >= k, we have I(r, k) = 1/4 *(binomial(r, k) + (r - 2)! * (r^2 + (4 * k - 1) * r + 4 * k * (k - 2)) * 3^(r - k - 2)/(k! * (r - k)!) + (2 + (-1)^k - (-1)^r) * (binomial(floor(r/2), floor(k/2)) + 2 * binomial(floor(r/2) - 1, floor(k/2) - 1)) * 3^(floor(r/2) - floor(k/2) - 1)). See Eq. (48) on p. 503 in the paper.
Letting k = 0 - 10, we get the eleven columns of Table 2 on p. 501 of Cyvin et al. (1996). (We need r >= max(k, 2) because the number of hexagons r should be greater than or equal to the number of converted polygons k.)
(End)
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LINKS
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FORMULA
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For the element T(n, k) in row n >= 2 and column k >= 0 (such that max(k, 2) <= n), we have T(n, k) = I(r = n, k), where I(r, k) is given above in the comments. - Petros Hadjicostas, May 26 2019
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EXAMPLE
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Triangle begins (rows start at n = 2 and columns at k = 0):
1, 1, 1;
2, 3, 3, 1;
4, 7, 9, 3, 1;
10, 23, 29, 16, 5, 1;
25, 69, 99, 62, 27, 5, 1;
70, 229, 351, 275, 132, 39, 7, 1;
196, 731, 1249, 1121, 643, 221, 55, 7, 1;
574, 2385, 4437, 4584, 2997, 1278, 367, 72, 9, 1;
1681, 7657, 15597, 18012, 13458, 6678, 2322, 540, 93, 9, 1;
...
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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