OFFSET
0,8
COMMENTS
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.
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.
Columns are linear recurrence sequences with signature (2*k, -k^2).
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) = (2*k)*T(n-1,k) - (k^2)*T(n-2,k).
G.f. for columns: x/(1 - k*x)^2.
E.g.f. for columns: x*exp(k*x).
T(n,1) = A001477(n).
T(n,2) = A001787(n).
T(n,3) = A027471(n+1).
T(n,4) = A002697(n).
T(n,5) = A053464(n).
T(n,6) = A053469(n), n > 0.
T(n,7) = A027473(n), n > 0.
T(n,8) = A053539(n).
T(n,9) = A053540(n), n > 0.
T(n,10) = A053541(n), n > 0.
T(n,11) = A081127(n).
T(n,12) = A081128(n).
EXAMPLE
Square array begins:
0, 0, 0, 0, 0, 0, 0, 0, ...
1, 1, 1, 1, 1, 1, 1, 1, ...
0, 2, 4, 6, 8, 10, 12, 14, ... A005843
0, 3, 12, 27, 48, 75, 108, 147, ... A033428
0, 4, 32, 108, 256, 500, 864, 1372, ... A033430
0, 5, 80, 405, 1280, 3125, 6480, 12005, ... A269792
0, 6, 192, 1458, 6144, 18750, 46656, 100842, ...
0, 7, 448, 5103, 28672, 109375, 326592, 823543, ...
...
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.
MATHEMATICA
T[n_, k_] = If [k > 0, n*k^(n - 1), If[k == 0 && n == 1, 1, 0]];
Table[Table[T[n - k, k], {k, 0, n}], {n, 0, 12}]//Flatten
PROG
(Maxima)
T(n, k) := if k > 0 then n*k^(n - 1) else if k = 0 and n = 1 then 1 else 0$
tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, nn))$
CROSSREFS
KEYWORD
AUTHOR
Franck Maminirina Ramaharo, Oct 14 2018
STATUS
approved