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A117584
Generalized Pellian triangle.
3
1, 1, 2, 1, 3, 5, 1, 4, 7, 12, 1, 5, 9, 17, 29, 1, 6, 11, 22, 41, 70, 1, 7, 13, 27, 53, 99, 169, 1, 8, 15, 32, 65, 128, 239, 408, 1, 9, 17, 37, 77, 157, 309, 577, 985, 1, 10, 19, 42, 89, 186, 379, 746, 1393, 2378
OFFSET
1,3
COMMENTS
Diagonals of the triangle are composed of the infinite set of Pellian sequences. Right border = A000129. Next diagonal going to the left = A001333 starting (1, 3, 7, 17, ...). A048654 = (1, 4, 9, ...). A048655 = (1, 5, 11, ...). A048693 = (1, 6, 13, ...); and so on.
FORMULA
Antidiagonals of the generalized Pellian array. First row of the array = A000129: (1, 2, 5, 12, ...). n-th row of the array starts (1, n+1, ...); as a Pellian sequence.
From G. C. Greubel, Jul 05 2021: (Start)
T(n, k) = P(k) + (n-1)*P(k-1), where P(n) = A000129(n) (square array).
Sum_{k=1..n} T(n-k+1, k) = A117185(n). (End)
EXAMPLE
First few rows of the triangle are:
1;
1, 2;
1, 3, 5;
1, 4, 7, 12;
1, 5, 9, 17, 29;
1, 6, 11, 22, 41, 70;
1, 7, 13, 27, 53, 99, 169;
...
The triangle rows are antidiagonals of the generalized Pellian array:
1, 2, 5, 12, 29, ...
1, 3, 7, 17, 41, ...
1, 4, 9, 22, 53, ...
1, 5, 11, 27, 65, ...
...
For example, in the row (1, 5, 11, 27, 65, ...), 65 = 2*27 + 11.
MATHEMATICA
T[n_, k_]:= Fibonacci[k, 2] + (n-1)*Fibonacci[k-1, 2];
Table[T[n-k+1, k], {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jul 05 2021 *)
PROG
(Magma)
P:= func< n | Round( ((1+Sqrt(2))^n - (1-Sqrt(2))^n)/(2*Sqrt(2)) ) >;
T:= func< n, k | P(k) + (n-1)*P(k-1) >;
[T(n-k+1, k): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 05 2021
(Sage)
def T(n, k): return lucas_number1(k, 2, -1) + (n-1)*lucas_number1(k-1, 2, -1)
flatten([[T(n-k+1, k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Jul 05 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Gary W. Adamson, Mar 29 2006
STATUS
approved