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A338198
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Triangle read by rows, T(n,k) = ((k+1)*2^(n-k)-(k-2)*(-1)^(n-k))/3 for 0 <= k <= n.
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0
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1, 0, 1, 2, 1, 1, 2, 3, 2, 1, 6, 5, 4, 3, 1, 10, 11, 8, 5, 4, 1, 22, 21, 16, 11, 6, 5, 1, 42, 43, 32, 21, 14, 7, 6, 1, 86, 85, 64, 43, 26, 17, 8, 7, 1, 170, 171, 128, 85, 54, 31, 20, 9, 8, 1, 342, 341, 256, 171, 106, 65, 36, 23, 10, 9, 1, 682, 683, 512, 341, 214, 127, 76, 41, 26, 11, 10, 1
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OFFSET
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0,4
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COMMENTS
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This triangle is related to the Jacobsthal numbers (A001045).
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LINKS
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FORMULA
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T(n,n) = 1 for n >= 0; T(n,n-1) = n-1 for n > 0.
T(n,k) = T(n-1,k) + 2 * T(n-2,k) for 0 <= k <= n-2.
T(n,k) = 2 * T(n-1,k) - (k-2) * (-1)^(n-k) for 0 <= k < n.
T(n,k) = T(n+1-k,1) + (k-1) * T(n-k,1) for 0 <= k < n.
T(n+1,k) * T(n-1,k) - T(n,k+1) * T(n,k-1) = T(n-k,1)^2 for 0 < k < n.
Row sums are A083579(n+1) for n >= 0.
G.f. of column k >= 0: (1+(k-1)*t) * t^k / (1-t-2*t^2).
G.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = (1 - (1+x)*t + 2*x*t^2) / ((1 - x*t)^2 * (1 - t - 2*t^2)).
Conjecture: Let M(n,k) be the matrix inverse of T(n,k), seen as a matrix. Then M(i,j) = 0 if j < 0 or j > i, M(n,n) = 1 for n >= 0, M(n,n-1) = 1-n for n > 0, and M(n,k) = (-1)^(n-k) * (k^2-2) * (n-2)! / k! for 0 <= k <= n-2.
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EXAMPLE
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The triangle T(n,k) for 0 <= k <= n starts:
n\k : 0 1 2 3 4 5 6 7 8 9
======================================================
0 : 1
1 : 0 1
2 : 2 1 1
3 : 2 3 2 1
4 : 6 5 4 3 1
5 : 10 11 8 5 4 1
6 : 22 21 16 11 6 5 1
7 : 42 43 32 21 14 7 6 1
8 : 86 85 64 43 26 17 8 7 1
9 : 170 171 128 85 54 31 20 9 8 1
etc.
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MATHEMATICA
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Table[((k + 1)*2^(n - k) - (k - 2)*(-1)^(n - k))/3, {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 15 2020 *)
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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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