OFFSET
0,8
COMMENTS
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.
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.
Columns are linear recurrence sequences with signature (3*k, -3*k^2, k^3).
REFERENCES
Louis H. Kauffman, Formal Knot Theory, Princeton University Press, 1983.
LINKS
Louis H. Kauffman, State models and the Jones polynomial, Topology, Vol. 26 (1987), 395-407.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Alexander Stoimenow, Everywhere Equivalent 2-Component Links, Symmetry Vol. 7 (2015), 365-375.
Wikipedia, Pretzel link
FORMULA
T(n,k) = k^n + k^(n - 2)*binomial(n, 2)*(2*binomial(k, 2) + 1), k > 0.
T(n,k) = (3*k)*T(n-1,k) - (3*k^2)*T(n-2,k) + (k^3)*T(n-3,k), n > 3.
T(n,1) = A152947(n+1).
T(n,2) = A300451(n).
T(2,n) = A130883(n).
G.f. for columns: (1 - 2*k*x + (1 - k + 2*k^2)*x^2 )/(1 - k*x)^3.
E.g.f. for columns: ((1 - k + k^2)*x^2 + 2)*exp(k*x)/2.
EXAMPLE
Square array begins:
1, 1, 1, 1, 1, 1, 1, ...
0, 1, 2, 3, 4, 5, 6, ...
1, 2, 7, 16, 29, 46, 67, ...
0, 4, 26, 90, 220, 440, 774, ...
0, 7, 88, 459, 1504, 3775, 7992, ...
0, 11, 272, 2133, 9344, 29375, 74736, ...
0, 16, 784, 9234, 54016, 212500, 649296, ...
0, 22, 2144, 37908, 295936, 1456250, 5342112, ...
...
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
abc: 001101, 001110;
acb: 000111, 001011;
cab: 010011, 100011;
bac: 110001, 110010;
bca: 110100, 111000;
cba: 011100, 101100;
bbc: 111101, 111110;
bcb: 110111, 111011;
cbb: 011111, 101111;
ccc: 010101, 101010, 010110, 011001, 100101, 101001, 100110, 011010.
MATHEMATICA
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]];
Table[Table[T[n - k, k], {k, 0, n}], {n, 0, 10}]//Flatten
PROG
(Maxima) t(n, k) := k^n + k^(n - 2)*binomial(n, 2)*(2*binomial(k, 2) + 1)$
u(n) := if n = 0 or n = 1 then 1 else 0$
T(n, k) := if k = 0 then u(n) else t(n, k)$
tabl(nn) := for n:0 thru 10 do print(makelist(T(n, k), k, 0, nn))$
CROSSREFS
KEYWORD
AUTHOR
Franck Maminirina Ramaharo, Oct 14 2018
STATUS
approved