OFFSET
1,8
COMMENTS
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).
The number of nonzero elements of row n is A001227(n).
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.
EXAMPLE
Triangle begins:
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, -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, -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;
1, -23, 0, 0, -5, 0;
1, 0 0, -9, 0, 0;
1, -25, -15, 0, 0, -3;
1, 0, 0, 0, 0, 0, -1;
...
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).
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).
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).
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Omar E. Pol, Apr 19 2016
STATUS
approved