%I #13 Feb 06 2021 20:04:01

%S 1,1,1,0,1,1,0,2,2,1,0,4,7,3,1,0,7,26,16,4,1,0,11,88,90,29,5,1,0,16,

%T 272,459,220,46,6,1,0,22,784,2133,1504,440,67,7,1,0,29,2144,9234,9344,

%U 3775,774,92,8,1,0,37,5632,37908,54016,29375,7992,1246,121,9

%N T(n,k) = k^n + k^(n - 2)*n*(n - 1)*(k*(k - 1) + 1)/2 for 0 < k <= n and T(n,0) = A154272(n+1), square array read by antidiagonals upwards.

%C Construct a length n ternary word over the alphabet {a, b, c} as follows: letters from the set {a, b} are only used in pairs of at most one, and consist of either (a,b), (b,a) or (b,b). Next, replace each occurrence of a, b and c with a length k binary word such that 'a' has exactly two letters 1, 'b' contains no 0's and 'c' has exactly one letter 0 (empty words otherwise, respectively). Then T(n,k) gives the number of length n*k binary words resulting from this substitution. First column follows from the next definition.

%C In Kauffman's language, T(n,k) is the number of ways of splitting the crossings of the Pretzel knot shadow P(k, k, ..., k) having n tangles, of k half-twists respectively, such that the final diagram consists of two Jordan curves. This result can be achieved by assigning each tangle of the Pretzel knot a length k binary words in a way that letters 1 and 0 indicate the adequate choice for splitting the crossings.

%C Columns are linear recurrence sequences with signature (3*k, -3*k^2, k^3).

%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) = k^n + k^(n - 2)*binomial(n, 2)*(2*binomial(k, 2) + 1), k > 0.

%F T(n,k) = (3*k)*T(n-1,k) - (3*k^2)*T(n-2,k) + (k^3)*T(n-3,k), n > 3.

%F T(n,1) = A152947(n+1).

%F T(n,2) = A300451(n).

%F T(2,n) = A130883(n).

%F G.f. for columns: (1 - 2*k*x + (1 - k + 2*k^2)*x^2 )/(1 - k*x)^3.

%F E.g.f. for columns: ((1 - k + k^2)*x^2 + 2)*exp(k*x)/2.

%e Square array begins:

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

%e 0, 1, 2, 3, 4, 5, 6, ...

%e 1, 2, 7, 16, 29, 46, 67, ...

%e 0, 4, 26, 90, 220, 440, 774, ...

%e 0, 7, 88, 459, 1504, 3775, 7992, ...

%e 0, 11, 272, 2133, 9344, 29375, 74736, ...

%e 0, 16, 784, 9234, 54016, 212500, 649296, ...

%e 0, 22, 2144, 37908, 295936, 1456250, 5342112, ...

%e ...

%e T(3,2) = 2^3 + 2^(3 - 2)*3*(3 - 1)*(2*(2 - 1) + 1)/2 = 26. The corresponding ternary words are abc, acb, cab, bac, bca, cba, bbc, bcb, cbb, ccc. Next, let a = {00}, b = {11} and c = {01, 10}. The resulting binary words are

%e abc: 001101, 001110;

%e acb: 000111, 001011;

%e cab: 010011, 100011;

%e bac: 110001, 110010;

%e bca: 110100, 111000;

%e cba: 011100, 101100;

%e bbc: 111101, 111110;

%e bcb: 110111, 111011;

%e cbb: 011111, 101111;

%e ccc: 010101, 101010, 010110, 011001, 100101, 101001, 100110, 011010.

%t T[n_, k_] = If[k > 0, k^n + k^(n - 2)*n*(n - 1)*(k*(k - 1) + 1)/2, If[k == 0 && (n == 0 || n == 1), 1, 0]];

%t Table[Table[T[n - k, k], {k, 0, n}], {n, 0, 10}]//Flatten

%o (Maxima) t(n, k) := k^n + k^(n - 2)*binomial(n, 2)*(2*binomial(k, 2) + 1)$

%o u(n) := if n = 0 or n = 1 then 1 else 0$

%o T(n, k) := if k = 0 then u(n) else t(n,k)$

%o tabl(nn) := for n:0 thru 10 do print(makelist(T(n, k), k, 0, nn))$

%Y Column 1 is column 2 of A300453.

%Y Column 2 is column 2 of A300184.

%Y Cf. A300401, A303273, A320530.

%K nonn,tabl,easy

%O 0,8

%A _Franck Maminirina Ramaharo_, Oct 14 2018