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A231345
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers interleaved with k-1 zeros but T(n,1) = -1 and the first element of column k is in row k(k+1)/2.
13
-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
OFFSET
1,8
COMMENTS
Gives an identity for the abundance of n. Alternating sum of row n equals the abundance of n, i.e., sum_{k=1..A003056(n))} (-1)^(k-1)*T(n,k) = A033880(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
LINKS
Eric Weisstein's World of Mathematics, Abundance
Eric Weisstein's World of Mathematics, Quasiperfect Number
FORMULA
T(n,1) = -1; T(n,k) = A196020(n,k), for k >= 2.
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;
...
For n = 15 the divisors of 15 are 1, 3, 5, 15 hence the abundance of 15 is 1 + 3 + 5 + 15 - 2*15 = 1 + 3 + 5 - 15 = -6. On the other hand the 15th row of triangle is -1, 13, 7, 0, 1, hence the alternating row sum is -1 - 13 + 7 - 0 + 1 = -6, equalling the abundance of 15.
If n is even then the alternating sum of the n-th row of triangle is simpler than the sum of divisors of n minus 2*n. Example: the sum of divisors of 24 minus 2*24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 - 2*24 = 60 - 48 = 12, and the alternating sum of the 24th row of triangle is -1 - 0 + 13 - 0 + 0 - 0 = 12.
KEYWORD
sign,tabf,nice
AUTHOR
Omar E. Pol, Dec 26 2013
STATUS
approved