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%I #17 Feb 13 2022 23:10:33
%S 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,
%T 169,8,16,28,48,82,140,239,408,16,24,40,68,116,198,338,577,985,16,32,
%U 56,96,164,280,478,816,1393,2378,32,48,80,136,232,396,676,1154,1970,3363,5741
%N Difference row triangle of the Pell sequence.
%C Leftmost column (1, 1, 2, 2, 4, 4, ...), (A016116); is the inverse binomial transform of the Pell sequence.
%D Raymond Lebois, "Le théorème de Pythagore et ses implications", p. 123, Editions PIM, (1979).
%H G. C. Greubel, <a href="/A117918/b117918.txt">Rows n = 1..50 of the triangle, flattened</a>
%F Difference rows of the Pell sequence A000129 starting (1, 2, 5, 12, ...) become the diagonals of the triangle.
%F T(n, n) = A000129(n).
%F T(n, n-1) = A000129(n) - A000129(n-1).
%F From _G. C. Greubel_, Oct 23 2021: (Start)
%F T(n, k) = T(n, k-1) + T(n-1, k-1) with T(n, 1) = 2^floor((n-1)/2).
%F T(n, k) = Sum_{j=0..n-k} (-1)^j*binomial(n-k, j)*Pell(n-j), where Pell(n) = A000129(n).
%F 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)
%e First difference row (1, 3, 7, 17, 41, ...) is the next diagonal.
%e First few rows of the triangle are:
%e 1;
%e 1, 2;
%e 2, 3, 5;
%e 2, 4, 7, 12;
%e 4, 6, 10, 17, 29;
%e 4, 8, 14, 24, 41, 70;
%e 8, 12, 20, 34, 58, 99, 169;
%e ...
%t T[n_, k_]:= T[n, k]= If[k==1, 2^Floor[(n-1)/2], T[n, k-1] + T[n-1, k-1]];
%t Table[T[n, k], {n,12}, {k,n}]//Flatten (* _G. C. Greubel_, Oct 22 2021 *)
%o (Magma)
%o Pell:= func< n | Round(((1+Sqrt(2))^n -(1-Sqrt(2))^n)/(2*Sqrt(2))) >;
%o T:= func< n,k | (&+[ (-1)^j*Binomial(n-k,j)*Pell(n-j): j in [0..n-k]]) >;
%o [T(n,k): k in [1..n], n in [1..12]]; // _G. C. Greubel_, Oct 23 2021
%o (Sage)
%o def A117918(n,k): return sum( (-1)^j*binomial(n-k, j)*lucas_number1(n-j, 2,-1) for j in (0..n) )
%o flatten([[A117918(n,k) for k in (1..n)] for n in (1..12)]) # _G. C. Greubel_, Oct 23 2021
%Y Cf. A000129, A016116.
%K nonn,easy
%O 1,3
%A _Gary W. Adamson_, Apr 02 2006
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