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A318274
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Triangle read by rows: T(n,k) = n for 0 < k < n and T(n,0) = T(n,n) = 1.
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3
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1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 4, 1, 1, 5, 5, 5, 5, 1, 1, 6, 6, 6, 6, 6, 1, 1, 7, 7, 7, 7, 7, 7, 1, 1, 8, 8, 8, 8, 8, 8, 8, 1, 1, 9, 9, 9, 9, 9, 9, 9, 9, 1, 1, 10, 10, 10, 10, 10, 10, 10, 10, 10, 1, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1, 1, 12
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OFFSET
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0,5
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COMMENTS
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T(n,k) is the number of binary bitonic words of length n having k letters 1.
Draw a circular rosette such that all the circles contain the rosette's center. Then T(n,k) is also the number of regions in the plane located inside k circles. In fact, a region can be encoded by a binary bitonic word as follows: label each circle from 1 to n in clockwise or counterclockwise order, then write a length n binary word such that the i-th letter indicates whether the concerned region does (write 1) or does not (write 0) lie inside the i-th circle.
Row n is a partition of A014206(n-1) for n > 0.
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LINKS
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FORMULA
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The n-th row are the coefficients in the expansion of 1 + x^n + n*x*(1 - x^(n - 1))/(1 - x), n > 0.
G.f. for column k > 0: (((1 - k)*x^2 - (1 - k)*x + 1)*x^k)/(x - 1)^2.
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EXAMPLE
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Triangle begins:
n\k| 0 1 2 3 4 5 6 7 8
---+--------------------------
0 | 1
1 | 1 1
2 | 1 2 1
3 | 1 3 3 1
4 | 1 4 4 4 1
5 | 1 5 5 5 5 1
6 | 1 6 6 6 6 6 1
7 | 1 7 7 7 7 7 7 1
8 | 1 8 8 8 8 8 8 8 1
...
For n = 5, the binary bitonic words are
(k = 0) 00000;
(k = 1) 10000, 01000, 00100, 00010, 00001;
(k = 2) 11000, 01100, 00110, 00011, 10001;
(k = 3) 11100, 01110, 00111, 10011, 11001;
(k = 4) 11110, 01111, 10111, 11011, 11101;
(k = 5) 11111.
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MATHEMATICA
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Table[If[k == n || k == 0, 1, n], {n, 0, 20}, {k, 0, n}] // Flatten
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PROG
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(Maxima)
T(n, k) := if k = 0 or k = n then 1 else if k < n then n else 0$
for n:0 thru 10 do print(makelist(T(n, k), k, 0, n));
(PARI) T(n, k) = if ((k==0) || (k==n), 1, n);
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Aug 25 2018
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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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