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A193922
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Mirror of the triangle A193921.
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2
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1, 1, 1, 2, 2, 1, 4, 4, 3, 2, 7, 7, 6, 5, 3, 12, 12, 11, 10, 8, 5, 20, 20, 19, 18, 16, 13, 8, 33, 33, 32, 31, 29, 26, 21, 13, 54, 54, 53, 52, 50, 47, 42, 34, 21, 88, 88, 87, 86, 84, 81, 76, 68, 55, 34, 143, 143, 142, 141, 139, 136, 131, 123, 110, 89, 55, 232, 232, 231
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table;
graph;
refs;
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text;
internal format)
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OFFSET
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0,4
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COMMENTS
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Also, triangle read by rows: T(n,k) = Fibonacci(n+2) - Fibonacci(k+1) with T(0,0) = 1, 0 <= k <= n. - Arkadiusz Wesolowski, Aug 05 2012
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LINKS
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FORMULA
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Write w(n,k) for the triangle at A193921. The triangle at A193922 is then given by w(n,n-k).
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EXAMPLE
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First six rows:
1
1 1
2 2 1
4 4 3 2
7 7 6 5 3
12 12 11 10 8 5
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MATHEMATICA
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z = 12;
p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n - k), {k, 0, n}];
q[n_, x_] := x*q[n - 1, x] + 1; q[0, n_] := 1;
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193921 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* this sequence *)
Factor[w[7, x]]
Factor[w[8, x]]
Table[Expand[p[n, x]], {n, 0, 4}]
Table[Expand[q[n, x]], {n, 0, 4}]
Prepend[Flatten@Table[Fibonacci[n + 2] - Fibonacci[k + 1], {n, 10}, {k, 0, n}], 1] (* Arkadiusz Wesolowski, Aug 05 2012 *)
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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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