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A236112
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k+1 copies of the squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.
16
0, 0, 1, 0, 1, 0, 4, 0, 4, 1, 0, 9, 1, 0, 9, 1, 0, 16, 4, 0, 16, 4, 1, 0, 25, 4, 1, 0, 25, 9, 1, 0, 36, 9, 1, 0, 36, 9, 4, 0, 49, 16, 4, 1, 0, 49, 16, 4, 1, 0, 64, 16, 4, 1, 0, 64, 25, 9, 1, 0, 81, 25, 9, 1, 0, 81, 25, 9, 4, 0, 100, 36, 9, 4, 1, 0, 100, 36, 16, 4, 1, 0, 121, 36, 16, 4, 1, 0, 121, 49, 16, 4, 1, 0
OFFSET
1,7
COMMENTS
Gives an identity for the sum of remainders of n mod k, for k = 1,2,3,...,n. Alternating sum of row n equals A004125(n), i.e., sum_{k=1..A003056(n))} (-1)^(k-1)*T(n,k) = A004125(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
EXAMPLE
Triangle begins:
0;
0;
1, 0;
1, 0;
4, 0;
4, 1, 0;
9, 1, 0;
9, 1, 0;
16, 4, 0;
16, 4, 1, 0;
25, 4, 1, 0;
25, 9, 1, 0;
36, 9, 1, 0;
36, 9, 4, 0;
49, 16, 4, 1, 0;
49, 16, 4, 1, 0;
64, 16, 4, 1, 0;
64, 25, 9, 1, 0;
81, 25, 9, 1, 0;
81, 25, 9, 4, 0;
100, 36, 9, 4, 1, 0;
100, 36, 16, 4, 1, 0;
121, 36, 16, 4, 1, 0;
121, 49, 16, 4, 1, 0;
...
For n = 24 the 24th row of triangle is 121, 49, 16, 4, 1, 0 therefore the alternating row sum is 121 - 49 + 16 - 4 + 1 - 0 = 85 equaling A004125(24).
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Jan 23 2014
STATUS
approved