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%I #4 Oct 18 2018 03:05:21
%S 0,1,0,0,1,0,0,2,1,0,0,3,4,1,0,0,4,12,6,1,0,0,5,32,27,8,1,0,0,6,80,
%T 108,48,10,1,0,0,7,192,405,256,75,12,1,0,0,8,448,1458,1280,500,108,14,
%U 1,0,0,9,1024,5103,6144,3125,864,147,16,1,0,0,10,2304
%N T(n,k) = n*k^(n - 1), k > 0, with T(n,0) = A063524(n), square array read by antidiagonals upwards.
%C T(n,k) is the number of length n*k binary words of n consecutive blocks of length k, respectively, one of the blocks having exactly k letters 1, and the other having exactly one letter 0. First column follows from the next definition.
%C In Kauffman's language, T(n,k) is the total number of Jordan trails that are obtained by placing state markers at the crossings of the Pretzel universe P(k, k, ..., k) having n tangles, of k half-twists respectively. In other words, T(n,k) is the number of ways of splitting the crossings of the Pretzel knot shadow P(k, k, ..., k) such that the final diagram is a single Jordan curve. The aforementionned binary words encode these operations by assigning each tangle a length k binary words with the adequate choice for splitting the crossings.
%C Columns are linear recurrence sequences with signature (2*k, -k^2).
%D Louis H. Kauffman, Formal Knot Theory, Princeton University Press, 1983.
%H Louis H. Kauffman, <a href="https://doi.org/10.1016/0040-9383(87)90009-7">State models and the Jones polynomial</a>, Topology, Vol. 26 (1987), 395-407.
%H Franck Ramaharo, <a href="https://arxiv.org/abs/1805.10680">A generating polynomial for the pretzel knot</a>, arXiv:1805.10680 [math.CO], 2018.
%H Alexander Stoimenow, <a href="https://doi.org/10.3390/sym7020365">Everywhere Equivalent 2-Component Links</a>, Symmetry Vol. 7 (2015), 365-375.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Pretzel_link">Pretzel link</a>
%F T(n,k) = (2*k)*T(n-1,k) - (k^2)*T(n-2,k).
%F G.f. for columns: x/(1 - k*x)^2.
%F E.g.f. for columns: x*exp(k*x).
%F T(n,1) = A001477(n).
%F T(n,2) = A001787(n).
%F T(n,3) = A027471(n+1).
%F T(n,4) = A002697(n).
%F T(n,5) = A053464(n).
%F T(n,6) = A053469(n), n > 0.
%F T(n,7) = A027473(n), n > 0.
%F T(n,8) = A053539(n).
%F T(n,9) = A053540(n), n > 0.
%F T(n,10) = A053541(n), n > 0.
%F T(n,11) = A081127(n).
%F T(n,12) = A081128(n).
%e Square array begins:
%e 0, 0, 0, 0, 0, 0, 0, 0, ...
%e 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 0, 2, 4, 6, 8, 10, 12, 14, ... A005843
%e 0, 3, 12, 27, 48, 75, 108, 147, ... A033428
%e 0, 4, 32, 108, 256, 500, 864, 1372, ... A033430
%e 0, 5, 80, 405, 1280, 3125, 6480, 12005, ... A269792
%e 0, 6, 192, 1458, 6144, 18750, 46656, 100842, ...
%e 0, 7, 448, 5103, 28672, 109375, 326592, 823543, ...
%e ...
%e T(3,2) = 3*2^(3 - 1) = 12. The corresponding binary words are 110101, 110110, 111001, 111010, 011101, 011110, 101101, 101110, 010111, 011011, 100111, 101011.
%t T[n_, k_] = If [k > 0, n*k^(n - 1), If[k == 0 && n == 1, 1, 0]];
%t Table[Table[T[n - k, k], {k, 0, n}], {n, 0, 12}]//Flatten
%o (Maxima)
%o T(n, k) := if k > 0 then n*k^(n - 1) else if k = 0 and n = 1 then 1 else 0$
%o tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, nn))$
%Y Antidiagonal sums: A101495.
%Y Column 1 is column 2 of A300453.
%Y Column 2 is column 1 of A300184.
%Y Cf. A104002, A320530.
%K nonn,easy,tabl
%O 0,8
%A _Franck Maminirina Ramaharo_, Oct 14 2018
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