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A245099 Triangle read by rows: T(n,k) = A024916(k)*A002865(n-k).

%I A245099 #17 Jul 17 2014 19:55:44
%S A245099 1,0,4,1,0,8,1,4,0,15,2,4,8,0,21,2,8,8,15,0,33,4,8,16,15,21,0,41,4,16,
%T A245099 16,30,21,33,0,56,7,16,32,30,42,33,41,0,69,8,28,32,60,42,66,41,56,0,
%U A245099 87,12,32,56,60,84,66,82,56,69,0,99,14,48,64
%N A245099 Triangle read by rows: T(n,k) = A024916(k)*A002865(n-k).
%C A245099 Row sums give A066186.
%C A245099 Column 1 is A002865.
%C A245099 Leading diagonal is A024916.
%C A245099 Since A024916(k) has a symmetric representation then both T(n,k) and the partial sums of row n can be represented by symmetric polycubes - for more information see A237593 and A237270. For another version see A221529.
%e A245099 Triangle begins:
%e A245099 1;
%e A245099 0,   4;
%e A245099 1,   0,  8;
%e A245099 1,   4,  0, 15;
%e A245099 2,   4,  8,  0, 21;
%e A245099 2,   8,  8, 15,  0, 33;
%e A245099 4,   8, 16, 15, 21,  0, 41;
%e A245099 4,  16, 16, 30, 21, 33,  0, 56;
%e A245099 7,  16, 32, 30, 42, 33, 41,  0, 69;
%e A245099 8,  28, 32, 60, 42, 66, 41, 56,  0, 87;
%e A245099 12, 32, 56, 60, 84, 66, 82, 56, 69,  0, 99;
%e A245099 ...
%e A245099 For n = 6:
%e A245099 -------------------------
%e A245099 k   A024916        T(6,k)
%e A245099 -------------------------
%e A245099 1      1  *  2   =    2
%e A245099 2      4  *  2   =    8
%e A245099 3      8  *  1   =    8
%e A245099 4     15  *  1   =   15
%e A245099 5     21  *  0   =    0
%e A245099 6     33  *  1   =   33
%e A245099 .         A002865
%e A245099 -------------------------
%e A245099 So row 6 is [2, 8, 8, 15, 0, 33] and the sum of row 6 is 2+8+8+15+0+33 = 66 equaling A066186(6) = 6*A000041(6) = 6*11 = 66.
%Y A245099 Cf. A000041, A000203, A002865, A024916, A066186, A221529, A221530, A245095.
%K A245099 nonn,tabl
%O A245099 1,3
%A A245099 _Omar E. Pol_, Jul 13 2014