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%I #28 Nov 12 2019 10:09:05
%S 1,1,0,2,1,0,4,4,1,0,8,12,6,1,0,16,32,24,8,1,0,32,80,80,40,10,1,0,64,
%T 192,240,160,60,12,1,0,128,448,672,560,280,84,14,1,0,256,1024,1792,
%U 1792,1120,448,112,16,1,0,512,2304,4608,5376,4032,2016,672,144,18,1,0
%N Triangle T(n,k), read by rows, given by (1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
%C A071919*A007318 as infinite lower triangular matrices.
%C A129186*A038207 as infinite lower triangular matrices.
%C From _Paul Curtz_, Nov 12 2019: (Start)
%C If a new main diagonal of 0's is added to the triangle, then for this variant the following propositions hold:
%C The first column is A166444.
%C The second column is A139756.
%C The antidiagonal sums are A000129 (Pell numbers).
%C The row sums are (-1)^n*A141413.
%C The signed row sums are 0 followed by 1's, autosequence companion to A054977.
%C (End)
%F T(n,k) = 2*T(n-1,k) + T(n-1,k-1) with T(0,0)=T(1,0)=1 and T(1,1)=0 .
%F G.f.: (1-(1+y)*x)/(1-(2+y)*x).
%F Sum_{k, 0<=k<=n} T(n,k)*x^k = A019590(n+1), A000012(n), A011782(n), A133494(n) for x = -2, -1, 0, 1 respectively.
%F Sum_{k, 0<=k<=n} T(n,k)*x^(n-k) = A000007(n), A133494(n), A020699(n) for x = 0, 1, 2 respectively.
%F T(2n,n) = A069720(n).
%e Triangle begins:
%e 1;
%e 1, 0;
%e 2, 1, 0;
%e 4, 4, 1, 0;
%e 8, 12, 6, 1, 0;
%e 16, 32, 24, 8, 1, 0;
%e 32, 80, 80, 40, 10, 1, 0;
%e ...
%p T:= proc(n, k) option remember; `if`(k<0 or k>n, 0,
%p `if`(n<2, 1-k, 2*T(n-1, k) +T(n-1, k-1)))
%p end:
%p seq(seq(T(n,k), k=0..n), n=0..12); # _Alois P. Heinz_, Nov 08 2019
%t T[n_, k_] /; 0 <= k <= n := T[n, k] = 2 T[n-1, k] + T[n-1, k-1];
%t T[0, 0] = T[1, 0] = 1; T[1, 1] = 0; T[_, _] = 0;
%t Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 08 2019 *)
%Y Essentially the same as A038207, A062715, A065109.
%Y Cf. A001787, A001788, A139756, A000129 (antidiagonals sums).
%Y Cf. A141413, A166444, A054977.
%K nonn,tabl
%O 0,4
%A _Philippe Deléham_, Dec 14 2011
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