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%I #13 Sep 08 2022 08:45:24
%S 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,
%T 1,8,15,32,65,128,239,408,1,9,17,37,77,157,309,577,985,1,10,19,42,89,
%U 186,379,746,1393,2378
%N Generalized Pellian triangle.
%C 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.
%H G. C. Greubel, <a href="/A117584/b117584.txt">Antidiagonal rows n = 0..100, flattened</a>
%F 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.
%F From _G. C. Greubel_, Jul 05 2021: (Start)
%F T(n, k) = P(k) + (n-1)*P(k-1), where P(n) = A000129(n) (square array).
%F Sum_{k=1..n} T(n-k+1, k) = A117185(n). (End)
%e First few rows of the triangle are:
%e 1;
%e 1, 2;
%e 1, 3, 5;
%e 1, 4, 7, 12;
%e 1, 5, 9, 17, 29;
%e 1, 6, 11, 22, 41, 70;
%e 1, 7, 13, 27, 53, 99, 169;
%e ...
%e The triangle rows are antidiagonals of the generalized Pellian array:
%e 1, 2, 5, 12, 29, ...
%e 1, 3, 7, 17, 41, ...
%e 1, 4, 9, 22, 53, ...
%e 1, 5, 11, 27, 65, ...
%e ...
%e For example, in the row (1, 5, 11, 27, 65, ...), 65 = 2*27 + 11.
%t T[n_, k_]:= Fibonacci[k, 2] + (n-1)*Fibonacci[k-1, 2];
%t Table[T[n-k+1, k], {n,12}, {k,n}]//Flatten (* _G. C. Greubel_, Jul 05 2021 *)
%o (Magma)
%o P:= func< n | Round( ((1+Sqrt(2))^n - (1-Sqrt(2))^n)/(2*Sqrt(2)) ) >;
%o T:= func< n,k | P(k) + (n-1)*P(k-1) >;
%o [T(n-k+1, k): k in [1..n], n in [1..12]]; // _G. C. Greubel_, Jul 05 2021
%o (Sage)
%o def T(n,k): return lucas_number1(k,2,-1) + (n-1)*lucas_number1(k-1,2,-1)
%o flatten([[T(n-k+1, k) for k in (1..n)] for n in (1..12)]) # _G. C. Greubel_, Jul 05 2021
%Y Cf. A000129, A001333, A048654, A048655, A048693, A117185.
%K nonn
%O 1,3
%A _Gary W. Adamson_, Mar 29 2006
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