A271343 Triangle read by rows: T(n,k) = A196020(n,k) - A266537(n,k), n>=1, k>=1.

1, 1, 5, 1, 1, 0, 9, 3, 1, -2, 1, 13, 5, 0, 1, 0, 0, 17, 7, 3, 1, -6, 0, 1, 21, 9, 0, 0, 1, 0, 3, 0, 25, 11, 0, 0, 1, -10, 0, 3, 29, 13, 7, 0, 1, 1, 0, 0, 0, 0, 33, 15, 0, 0, 0, 1, -14, 3, 5, 0, 37, 17, 0, 0, 0, 1, 0, 0, -2, 3, 41, 19, 11, 0, 0, 1, 1, -18, 0, 7, 0, 0, 45, 21, 0, 0, 0, 0, 1, 0, 3, 0, 0, 0

Offset: 1

Author

Omar E. Pol, Apr 06 2016

Keywords

• sign
• tabf

Comments

Gives an identity for A000593. Alternating sum of row n equals the sum of odd divisors of n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A000593(n).

Row n has length A003056(n) hence the column k starts in row A000217(k).

Since the odd-indexed rows of the triangle A266537 contain all zeros then odd-indexed rows of this triangle are the same as the odd-indexed rows of the triangle A196020.

If T(n,k) is the second odd number in the column k then T(n+1,k+1) = 1 is the first element in the column k+1.

Alternating row sums of A196020 give A000203.

Alternating row sums of A266537 give A146076.

Examples

			Triangle begins:
1;
1;
5,   1;
1,   0;
9,   3;
1,  -2,  1;
13,  5,  0;
1,   0,  0;
17,  7,  3;
1,  -6,  0,  1;
21,  9,  0,  0;
1,   0,  3,  0;
25, 11,  0,  0;
1, -10,  0,  3;
29, 13,  7,  0,  1;
1,   0,  0,  0,  0;
33, 15,  0,  0,  0;
1, -14,  3,  5,  0;
37, 17,  0,  0,  0;
1,   0,  0, -2,  3;
41, 19, 11,  0,  0,  1;
1, -18,  0,  7,  0,  0;
45, 21,  0,  0,  0,  0;
1,   0,  3,  0,  0,  0;
49, 23,  0,  0,  5,  0;
1, -22,  0,  9,  0,  0;
53, 25, 15,  0,  0,  3;
1,   0,  0, -6,  0,  0,  1;
...
For n = 18 the divisors of 18 are 1, 2, 3, 6, 9, 18 and the sum of odd divisors of 18 is 1 + 3 + 9 = 13. On the other hand, the 18th row of the triangle is 1, -14, 3, 5, 0, so the alternating row sum is 1 -(-14) + 3 - 5 + 0 = 13, equaling the sum of odd divisors of 18.
		

Crossrefs

Column 1 is A266072.

Cf. A000217, A000593, A001227, A003056, A146076, A196020, A236104, A237593, A261699, A266537.