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%I #21 Apr 29 2018 18:17:22
%S 1,1,1,-1,1,0,1,-3,1,0,-1,1,-5,0,1,0,0,1,-7,-3,1,0,0,-1,1,-9,0,0,1,0,
%T -5,0,1,-11,0,0,1,0,0,-3,1,-13,-7,0,-1,1,0,0,0,0,1,-15,0,0,0,1,0,-9,
%U -5,0,1,-17,0,0,0,1,0,0,0,-3,1,-19,-11,0,0,-1,1,0,0,-7,0,0,1,-21,0,0,0,0,1,0,-13,0,0,0
%N Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers multiplied by -1, interleaved with k-1 zeros, but T(n,1) = 1 and the first element of column k is in row k(k+1)/2.
%C Gives an identity for the deficiency of n. Alternating sum of row n equals the deficiency of n, i.e., sum_{k=1..A003056(n))} (-1)^(k-1)*T(n,k) = A033879(n).
%C Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%C The number of nonzero elements of row n is A001227(n).
%C If T(n,k) is the second nonzero term in column k then T(n+1,k+1) = -1 is the first element of column k+1.
%F T(n,k) = -1*A231345(n,k).
%F T(n,k) = -1*A196020(n,k), if k >= 2.
%e Triangle begins:
%e 1;
%e 1;
%e 1, -1;
%e 1, 0;
%e 1, -3;
%e 1, 0, -1;
%e 1, -5, 0;
%e 1, 0, 0;
%e 1, -7, -3;
%e 1, 0, 0, -1;
%e 1, -9, 0, 0;
%e 1, 0, -5, 0;
%e 1, -11, 0, 0;
%e 1, 0, 0, -3;
%e 1, -13, -7, 0, -1;
%e 1, 0, 0, 0, 0;
%e 1, -15, 0, 0, 0;
%e 1, 0, -9, -5, 0;
%e 1, -17, 0, 0, 0;
%e 1, 0, 0, 0, -3;
%e 1, -19, -11, 0, 0, -1;
%e 1, 0, 0, -7, 0, 0;
%e 1, -21, 0, 0, 0, 0;
%e 1, 0, -13, 0, 0, 0;
%e 1, -23, 0, 0, -5, 0;
%e 1, 0 0, -9, 0, 0;
%e 1, -25, -15, 0, 0, -3;
%e 1, 0, 0, 0, 0, 0, -1;
%e ...
%e For n = 24 the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24 so the deficiency of 24 is 24 - 12 - 8 - 6 - 4 - 3 - 2 - 1 = -12. On the other hand the 24th row of triangle is 1, 0, -13, 0, 0, 0, and the alternating row sum is 1 - 0 +(-13) - 0 + 0 - 0 = -12, equaling the deficiency of 24; A033879(24) = -12, so 24 is an abundant number (A005101).
%e For n = 27 the divisors of 27 are 1, 3, 9, 27 so the deficiency of 27 is 27 - 9 - 3 - 1 = 14. On the other hand the 27th row of triangle is 1, -25, -15, 0, 0, -3, and the alternating row sum is 1 -(-25) +(-15) - 0 + 0 -(-3) = 14, equalling the deficiency of 27; A033879(27) = 14, so 27 is a deficient number (A005100).
%e For n = 28 the divisors of 28 are 1, 2, 4, 7, 14, 28 so the deficiency of 28 is 28 - 14 - 7 - 4 - 2 - 1 = 0. On the other hand the 28th row of triangle is 1, 0, 0, 0, 0, 0, -1, and the alternating row sum is 1 - 0 + 0 - 0 + 0 - 0 +(-1) = 0, equaling the deficiency of 28; A033879(28) = 0, so 28 is a perfect number (A000396).
%Y Column 1 is A000012. Column 2 is -1*A193356.
%Y Cf. A000203, A000217, A000396, A001227, A003056, A033879, A033880, A005100, A005101, A196020, A231345, A237593.
%K sign,tabf
%O 1,8
%A _Omar E. Pol_, Apr 19 2016
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