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A320530
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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.
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2
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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, 272, 459, 220, 46, 6, 1, 0, 22, 784, 2133, 1504, 440, 67, 7, 1, 0, 29, 2144, 9234, 9344, 3775, 774, 92, 8, 1, 0, 37, 5632, 37908, 54016, 29375, 7992, 1246, 121, 9
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OFFSET
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0,8
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COMMENTS
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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).
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REFERENCES
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Louis H. Kauffman, Formal Knot Theory, Princeton University Press, 1983.
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LINKS
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FORMULA
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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.
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.
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EXAMPLE
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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.
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MATHEMATICA
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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
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PROG
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(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))$
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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