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%I #25 Apr 10 2016 10:01:32
%S 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,
%T 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,
%U 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
%N Triangle read by rows: T(n,k) = A196020(n,k) - A266537(n,k), n>=1, k>=1.
%C 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).
%C Row n has length A003056(n) hence the column k starts in row A000217(k).
%C 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.
%C 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.
%C Alternating row sums of A196020 give A000203.
%C Alternating row sums of A266537 give A146076.
%e Triangle begins:
%e 1;
%e 1;
%e 5, 1;
%e 1, 0;
%e 9, 3;
%e 1, -2, 1;
%e 13, 5, 0;
%e 1, 0, 0;
%e 17, 7, 3;
%e 1, -6, 0, 1;
%e 21, 9, 0, 0;
%e 1, 0, 3, 0;
%e 25, 11, 0, 0;
%e 1, -10, 0, 3;
%e 29, 13, 7, 0, 1;
%e 1, 0, 0, 0, 0;
%e 33, 15, 0, 0, 0;
%e 1, -14, 3, 5, 0;
%e 37, 17, 0, 0, 0;
%e 1, 0, 0, -2, 3;
%e 41, 19, 11, 0, 0, 1;
%e 1, -18, 0, 7, 0, 0;
%e 45, 21, 0, 0, 0, 0;
%e 1, 0, 3, 0, 0, 0;
%e 49, 23, 0, 0, 5, 0;
%e 1, -22, 0, 9, 0, 0;
%e 53, 25, 15, 0, 0, 3;
%e 1, 0, 0, -6, 0, 0, 1;
%e ...
%e 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.
%Y Column 1 is A266072.
%Y Cf. A000217, A000593, A001227, A003056, A146076, A196020, A236104, A237593, A261699, A266537.
%K sign,tabf
%O 1,3
%A _Omar E. Pol_, Apr 06 2016
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