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%I #8 Dec 17 2019 05:34:31
%S 1,1,1,2,1,1,2,1,1,1,3,2,0,1,1,4,2,2,0,1,1,4,2,2,1,0,1,1,4,2,0,2,1,0,
%T 1,1,4,2,2,1,1,1,0,1,1,5,3,2,0,1,1,1,0,1,1,6,3,1,1,1,0,1,1,0,1,1,6,3,
%U 3,3,0,1,0,1,1,0,1,1,6,3,2,2,2,1,0,0,1,1,0,1,1,6,3,1,0,2,1,1,0,0,1,1,0,1,1
%N Triangle T(n,k) read by rows, defined by: T(1,1)=1; n > 1 and k=1: T(n,1) = T(n-1,2) + T(n,2); k=2: T(n,2) = A000196(n-1); k > 2: T(n,k) = (Sum_{i=1..k-1} T(n-i,k-1)) - (Sum_{i=1..k-1} T(n-i,k)).
%C The second column, sequence A000196, is the initial condition for the recurrence in this triangle. See A051731, formula entered on Feb 16 2010 for the more pure form of this recurrence.
%F T(1,1)=1; n > 1 and k=1: T(n,1) = T(n-1,2) + T(n,2); k=2: T(n,2) = A000196(n-1); k > 2: T(n,k) = (Sum_{i=1..k-1} T(n-i,k-1)) - (Sum_{i=1..k-1} T(n-i,k)).
%e Triangle begins:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 2, 1, 1, 1;
%e 3, 2, 0, 1, 1;
%e 4, 2, 2, 0, 1, 1;
%e 4, 2, 2, 1, 0, 1, 1;
%e 4, 2, 0, 2, 1, 0, 1, 1;
%e 4, 2, 2, 1, 1, 1, 0, 1, 1;
%e 5, 3, 2, 0, 1, 1, 1, 0, 1, 1;
%e 6, 3, 1, 1, 1, 0, 1, 1, 0, 1, 1;
%o (Excel) Using European dot comma style:
%o =if(and(row()=1;column()=1);1;if(row()>=column();if(column()=1;indirect(address(row()-1;column()+1))+indirect(address(row();column()+1));if(column()=2;floor(((row()-1)^0,5);1);if(row()>=column();sum(indirect(address(row()-column()+1;column()-1;4)&":"&address(row()-1;column()-1;4);4))-sum(indirect(address(row()-column()+1;column();4)&":"&address(row()-1;column();4);4));0)));0))
%Y Cf. A179286, A179287, A059571, A051731.
%K nonn,tabl
%O 1,4
%A _Mats Granvik_, Jul 09 2010
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