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Largest part minus the number of parts > 1 in the n-th region of the set of partitions of j, if 1 <= n <= A000041(j).
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%I #29 Mar 11 2014 01:34:19

%S 1,1,2,1,2,2,3,1,2,2,2,2,3,3,3,1,2,2,2,4,3,1,2,3,3,3,2,4,4,1,1,2,2,2,

%T 4,3,1,3,5,5,4,-2,2,3,3,3,2,4,4,1,4,3,5,6,5,-3,1,2,2,2,4,3,1,3,5,5,4,

%U -2,2,4,4,5,3,6,6,5,-9

%N Largest part minus the number of parts > 1 in the n-th region of the set of partitions of j, if 1 <= n <= A000041(j).

%C Also triangle read by rows: T(j,k) = largest part minus the numbers of parts > 1 in the k-th region of the last section of the set of partitions of j. It appears that the sum of row j is equal to A000041(j-1). For the definition of "region" of the set of partitions of j see A206437. See also A135010.

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

%F a(n) = A141285(n) - A194448(n).

%e The 7th region of the shell model of partitions is [5, 2, 1, 1, 1, 1, 1]. The largest part is 5 and the number of parts > 1 is 2, so a(7) = 5 - 2 = 3 (see an illustration in the link section).

%e Written as an irregular triangle T(j,k) begins:

%e 1;

%e 1;

%e 2;

%e 1,2;

%e 2,3;

%e 1,2,2,2;

%e 2,3,3,3;

%e 1,2,2,2,4,3,1;

%e 2,3,3,3,2,4,4,1;

%e 1,2,2,2,4,3,1,3,5,5,4,-2;

%e 2,3,3,3,2,4,4,1,4,3,5,6,5,-3;

%e 1,2,2,2,4,3,1,3,5,5,4,-2,2,4,4,5,3,6,6,5,-9;

%Y Row j has length A187219(j).

%Y Cf. A000041, A135010, A138121, A138137, A138879, A186114, A186412, A193870, A194436, A194437, A194438, A194439, A194446, A194447, A206437.

%K sign,tabf

%O 1,3

%A _Omar E. Pol_, Dec 10 2011