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A231345
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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.
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13
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-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
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OFFSET
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1,8
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COMMENTS
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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).
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LINKS
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Eric Weisstein's World of Mathematics, Abundance
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FORMULA
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T(n,1) = -1; T(n,k) = A196020(n,k), for k >= 2.
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EXAMPLE
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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.
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CROSSREFS
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Cf. A000203, A000217, A000396, A003056, A005100, A005843, A033879, A033880, A069283, A196020, A212119, A228813, A231347, A235791, A235794, A236104, A236106, A236112, A237593.
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KEYWORD
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sign,tabf,nice
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AUTHOR
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STATUS
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approved
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