%I #40 Apr 02 2017 14:50:01
%S 1,1,1,2,1,3,2,5,3,-1,7,5,-1,11,7,-2,-1,15,11,-3,-1,22,15,-5,-2,30,22,
%T -7,-3,42,30,-11,-5,56,42,-15,-7,1,77,56,-22,-11,1,101,77,-30,-15,2,
%U 135,101,-42,-22,3,1,176,135,-56,-30,5,1,231,176,-77,-42,7,2
%N Triangle read by rows demonstrating Euler's pentagonal theorem for partition numbers.
%C Row sums = A000041 starting with offset 1.
%C Sum of n-th row terms = leftmost term of next row, such that terms in each row demonstrate Euler's pentagonal theorem.
%C Let Q = triangle A027293 with partition numbers in each column.
%C Let M = a diagonalized variant of A080995 as the characteristic function of the generalized pentagonal numbers starting with offset 1: (1, 1, 0, 0, 1,...)
%C Sign the 1's: (++--++...) getting (1, 1, 0, 0, -1, 0, -1,...) which is the diagonal of matrix M, (as an infinite lower triangular matrix with the rest zeros).
%C Triangle A175003 = Q*M, with deleted zeros.
%C Column k starts at row A001318(k). - _Omar E. Pol_, Sep 21 2011
%C From _Omar E. Pol_, Apr 22 2014: (Start)
%C Row n has length A235963(n).
%C For Euler's pentagonal theorem for the sum of divisors see A238442.
%C Note that both of Euler's pentagonal theorems refer to generalized pentagonal numbers (A001318), not to pentagonal numbers (A000326). (End)
%F T(n,k) = A057077(k-1)*A000041(A195310(n,k)), n >= 1, k >= 1. - _Omar E. Pol_, Sep 21 2011
%e Triangle begins:
%e 1;
%e 1, 1;
%e 2, 1;
%e 3, 2;
%e 5, 3, -1;
%e 7, 5, -1;
%e 11, 7, -2, -1;
%e 15, 11, -3, -1;
%e 22, 15, -5, -2;
%e 30, 22, -7, -3;
%e 42, 30, -11, -5;
%e 56, 42, -15, -7, 1;
%e 77, 56, -22, -11, 1;
%e 101, 77, -30, -15, 2;
%e ...
%Y Cf. A000041, A080995, A027293, A238442.
%K tabf,sign
%O 1,4
%A _Gary W. Adamson_, Apr 03 2010
%E Corrected and extended by _Omar E. Pol_, Feb 14 2013