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A117918
Difference row triangle of the Pell sequence.
4
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
OFFSET
1,3
COMMENTS
Leftmost column (1, 1, 2, 2, 4, 4, ...), (A016116); is the inverse binomial transform of the Pell sequence.
REFERENCES
Raymond Lebois, "Le théorème de Pythagore et ses implications", p. 123, Editions PIM, (1979).
FORMULA
Difference rows of the Pell sequence A000129 starting (1, 2, 5, 12, ...) become the diagonals of the triangle.
T(n, n) = A000129(n).
T(n, n-1) = A000129(n) - A000129(n-1).
From G. C. Greubel, Oct 23 2021: (Start)
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)
EXAMPLE
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;
...
MATHEMATICA
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 *)
PROG
(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) )
flatten([[A117918(n, k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Oct 23 2021
CROSSREFS
Sequence in context: A175908 A152430 A297495 * A302495 A368255 A368256
KEYWORD
nonn,easy
AUTHOR
Gary W. Adamson, Apr 02 2006
STATUS
approved