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%I #4 Mar 30 2012 17:25:36
%S 1,1,1,0,1,2,0,0,2,3,-1,0,0,3,5,0,-1,0,0,5,7,-1,0,-2,0,0,7,11,0,-1,0,
%T -3,0,0,11,15,0,0,-2,0,-5,0,15,22,0,0,0,-3,0,-7,0,0,22,30,0,0,0,0,-5,
%U 0,-11,0,0,30,42,1,0,0,0,0,-7,0,-15,0,0,42,56,0,1,0,0,0,0,-11,0,-22,0,0,0,0
%N Triangle read by rows, a partition number generator; A145006 * the diagonalized variant of A000041, (A174712).
%C Row sums = the partition numbers, A000041 starting with offset 1.
%C The triangle demonstrates an equivalency to Euler's pentagonal recurrence
%C relation, such that sum of n-th row terms = rightmost term of next row, a
%C partition number.
%C Contribution from _Gary W. Adamson_, Mar 28 2010: (Start)
%C A174739 is equivalent to Euler's pentagonal theorem in triangular form.
%C For example, row 9 = (0, 0, -2, 0, -5, 0, 0, 15, 22) or: p(9) = 30 = p(8)
%C + p(7) - p(4) - p(2). (End)
%F Given triangle A145006, delete the first "1", = triangle Q. With M = A174712,
%F the diagonalize variant of the partition numbers, perform Q*M as infinite lower
%F triangular matrices.
%e First few rows of the triangle =
%e 1;
%e 1, 1;
%e 0, 1, 2;
%e 0, 0, 2, 3;
%e -1, 0, 0, 3, 5;
%e 0, -1, 0, 0, 5, 7;
%e -1, 0, -2, 0, 0, 7, 11;
%e 0, -1, 0, -3, 0, 0, 11, 15;
%e 0, 0, -2, 0, -5, 0, 0, 15, 22;
%e 0, 0, 0, -3, 0, -7, 0, 0, 22, 30;
%e 0, 0, 0, 0, -5, 0, -11, 0, 0, 30, 42;
%e 1, 0, 0, 0, 0, -7, 0, -15, 0, 0, 42, 56;
%e 0, 1, 0, 0, 0, 0, -11, 0, 22, 0, 0, 56, 77;
%e 0, 0, 2, 0, 0, 0, 0, -15, 0, -30, 0, 0, 77, 101;
%e 1, 0, 0, 3, 0, 0, 0, 0, -22, 0, -42, 0, 0, 101, 135;
%e 0, 1, 0, 0, 5, 0, 0, 0, 0, -30, 0, -56, 0, 0, 135, 176;
%e 0, 0, 2, 0, 0, 7, 0, 0, 0, 0, -42, 0, -77, 0, 0, 176, 231;
%e 0, 0, 0, 3, 0, 0, 11, 0, 0, 0, 0, -56, 0, -101, 0, 0, 231, 297;
%e ...
%Y Cf. A000041, A145006, A174712
%K tabl,sign
%O 1,6
%A _Gary W. Adamson_, Mar 28 2010
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