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Triangle read by rows which arises from A195831, in the same way as A175003 arises from A195310. Column k starts at row A195160(k).
9

%I #11 Jun 10 2012 06:31:36

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

%T 5,3,-1,7,4,-1,10,4,-2,12,4,-3,13,4,-4,13,4,-4,13,4,-4,13,5,-4,14,7,

%U -4,-1,16,10,-4,-1,21,12,-5,-1,27,13,-7,-1

%N Triangle read by rows which arises from A195831, in the same way as A175003 arises from A195310. Column k starts at row A195160(k).

%C The sum of terms of row n is equal to the leftmost term of row n+1. This sequence is related to the generalized hendecagonal numbers A195160, A195831 and A195851 in the same way as A175003 is related to the generalized pentagonal numbers A001318, A195310 and A000041. See comments in A195825.

%e Written as a triangle:

%e 1;

%e 1;

%e 1;

%e 1;

%e 1;

%e 1;

%e 1;

%e 1, 1;

%e 2, 1;

%e 3, 1;

%e 4, 1, -1;

%e 4, 1, -1;

%e 4, 1, -1;

%e 4, 1, -1;

%e 4, 1, -1;

%e 4, 2, -1;

%e 5, 3, -1;

%e 7, 4, -1;

%Y Row sums give A195851.

%Y Cf. A001082, A175003, A195828, A195836, A195837, A195838, A195839, A195840.

%K sign,tabf

%O 1,10

%A _Omar E. Pol_, Sep 24 2011