%I #29 Mar 11 2014 01:34:20

%S 1,1,2,1,2,3,2,3,4,5,2,4,5,6,7,4,6,8,9,10,11,4,8,10,12,13,14,15,7,11,

%T 15,17,19,20,21,22,8,15,19,23,25,27,28,29,30,12,20,27,31,35,37,39,40,

%U 41,42,14,26,34,41,45,49,51,53,54,55,56,21,35,47

%N Triangle read by rows: T(n,k) = total number of regions in the last k shells of n.

%C For the definition of "region of n" see A206437. For the definition of "last section of n" see A135010.

%C Apparently differs from A027300 at the right border.

%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polpar02.jpg">Illustration of the seven regions of 5</a>

%F T(n,k) = A000041(n) - A000041(n-k), if 1<k<n.

%F T(n,k) = A000041(n), if k = n.

%F T(n,k) = Sum_{j=1..k} A187219(n-j+1,1).

%e For n = 5 and k = 2 we have that the 4th shell of 5 contains two regions: [2] and [4,2,1,1,1]. Then we can see that the 5th shell of 5 contains two regions: [3] and [5,2,1,1,1,1,1]. Therefore the total number of regions in the last two shells of 5 is T(5,2) = 2+2 = 4 (see illustration in the link section).

%e Triangle begins:

%e 1;

%e 1, 2;

%e 1, 2, 3;

%e 2, 3, 4, 5;

%e 2, 4, 5, 6, 7;

%e 4, 6, 8, 9, 10, 11;

%e 4, 8, 10, 12, 13, 14, 15;

%e 7, 11, 15, 17, 19, 20, 21, 22;

%e 8, 15, 19, 23, 25, 27, 28, 29, 30;

%e 12, 20, 27, 31, 35, 37, 39, 40, 41, 42;

%e 14, 26, 34, 41, 45, 49, 51, 53, 54, 55, 56;

%e 21, 35, 47, 55, 62, 66, 70, 72, 74, 75, 76, 77;

%Y Mirror of triangle A211980. Column 1 is A187219. Right border gives A000041, n >= 1.

%Y Cf. A027300, A135010, A138121, A182703, A186114, A206437, A212010, A212011.

%K nonn,tabl

%O 1,3

%A _Omar E. Pol_, Apr 27 2012