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A117918
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Difference row triangle of the Pell sequence.
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4
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1, 1, 2, 2, 3, 5, 2, 4, 7, 12, 4, 6, 10, 17, 29, 4, 8, 14, 24, 41, 70, 8, 12, 20, 34, 58, 99, 169, 8, 16, 28, 48, 82, 140, 239, 408, 16, 24, 40, 68, 116, 198, 338, 577, 985, 16, 32, 56, 96, 164, 280, 478, 816, 1393, 2378, 32, 48, 80, 136, 232, 396, 676, 1154, 1970, 3363, 5741
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OFFSET
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1,3
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COMMENTS
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Leftmost column (1, 1, 2, 2, 4, 4, ...), (A016116); is the inverse binomial transform of the Pell sequence.
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REFERENCES
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Raymond Lebois, "Le théorème de Pythagore et ses implications", p. 123, Editions PIM, (1979).
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LINKS
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FORMULA
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Difference rows of the Pell sequence A000129 starting (1, 2, 5, 12, ...) become the diagonals of the triangle.
T(n, k) = T(n, k-1) + T(n-1, k-1) with T(n, 1) = 2^floor((n-1)/2).
T(n, k) = Sum_{j=0..n-k} (-1)^j*binomial(n-k, j)*Pell(n-j), where Pell(n) = A000129(n).
Sum_{k=1..n} T(n, k) = Pell(n+1) -2^floor(n/2)*((1 + (-1)^n)/2) - 2^floor((n - 1)/2)*((1 - (-1)^n)/2). (End)
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EXAMPLE
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First difference row (1, 3, 7, 17, 41, ...) is the next diagonal.
First few rows of the triangle are:
1;
1, 2;
2, 3, 5;
2, 4, 7, 12;
4, 6, 10, 17, 29;
4, 8, 14, 24, 41, 70;
8, 12, 20, 34, 58, 99, 169;
...
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MATHEMATICA
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T[n_, k_]:= T[n, k]= If[k==1, 2^Floor[(n-1)/2], T[n, k-1] + T[n-1, k-1]];
Table[T[n, k], {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Oct 22 2021 *)
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PROG
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(Magma)
Pell:= func< n | Round(((1+Sqrt(2))^n -(1-Sqrt(2))^n)/(2*Sqrt(2))) >;
T:= func< n, k | (&+[ (-1)^j*Binomial(n-k, j)*Pell(n-j): j in [0..n-k]]) >;
[T(n, k): k in [1..n], n in [1..12]]; // G. C. Greubel, Oct 23 2021
(Sage)
def A117918(n, k): return sum( (-1)^j*binomial(n-k, j)*lucas_number1(n-j, 2, -1) for j in (0..n) )
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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