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A239313
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers interleaved with k-1 zeros, except the first column which lists 0 together with the nonnegative integers, and the first element of column k is in row k*(k+1)/2.
1
0, 0, 1, 1, 2, 0, 3, 3, 4, 0, 1, 5, 5, 0, 6, 0, 0, 7, 7, 3, 8, 0, 0, 1, 9, 9, 0, 0, 10, 0, 5, 0, 11, 11, 0, 0, 12, 0, 0, 3, 13, 13, 7, 0, 1, 14, 0, 0, 0, 0, 15, 15, 0, 0, 0, 16, 0, 9, 5, 0, 17, 17, 0, 0, 0, 18, 0, 0, 0, 3, 19, 19, 11, 0, 0, 1, 20, 0, 0, 7, 0, 0
OFFSET
1,5
COMMENTS
Alternating row sums give the Chowla's function, i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A048050(n).
Row n has length A003056(n) hence column k starts in row A000217(k).
Column 1 gives 0 together with A001477.
Column 2 is A193356.
The number of positive terms in row n is A001227(n), if n >= 3. - Omar E. Pol, Apr 18 2016
FORMULA
T(n,k) = A196020(n,k), if k >= 2. - Omar E. Pol, Apr 18 2016
EXAMPLE
Triangle begins (row n = 1..24):
0;
0;
1, 1;
2, 0;
3, 3;
4, 0, 1;
5, 5, 0;
6, 0, 0;
7, 7, 3;
8, 0, 0, 1;
9, 9, 0, 0;
10, 0, 5, 0;
11, 11, 0, 0;
12, 0, 0, 3;
13, 13, 7, 0, 1;
14, 0, 0, 0, 0;
15, 15, 0, 0, 0;
16, 0, 9, 5, 0;
17, 17, 0, 0, 0;
18, 0, 0, 0, 3;
19, 19, 11, 0, 0, 1;
20, 0, 0, 7, 0, 0;
21, 21, 0, 0, 0, 0;
22, 0, 13, 0, 0, 0;
...
For n = 15 the divisors of 15 are 1, 3, 5, 15 therefore the sum of divisors of 15 except 1 and 15 is 3 + 5 = 8. On the other hand the 15th row of triangle is 13, 13, 7, 0, 1, hence the alternating row sum is 13 - 13 + 7 - 0 + 1 = 8, equalling the sum of divisors of 15 except 1 and 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, except 1 and n. Example: the sum of divisors of 24 except 1 and 24 is 2 + 3 + 4 + 6 + 8 + 12 = 35, and the alternating sum of the 24th row of triangle is 22 - 0 + 13 - 0 + 0 - 0 = 35.
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Mar 15 2014
STATUS
approved