|
|
A236540
|
|
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k copies of the positive squares in nondecreasing order, except the first column which lists the triangular numbers, and the first element of column k is in row k(k+1)/2.
|
|
3
|
|
|
0, 1, 3, 1, 6, 1, 10, 4, 15, 4, 1, 21, 9, 1, 28, 9, 1, 36, 16, 4, 45, 16, 4, 1, 55, 25, 4, 1, 66, 25, 9, 1, 78, 36, 9, 1, 91, 36, 9, 4, 105, 49, 16, 4, 1, 120, 49, 16, 4, 1, 136, 64, 16, 4, 1, 153, 64, 25, 9, 1, 171, 81, 25, 9, 1, 190, 81, 25, 9, 4, 210, 100, 36, 9, 4, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
Gives an identity for the sum of all aliquot divisors of all positive integers <= n.
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
Column k lists the partial sums of the k-th column of triangle A231347 which gives an identity for the sum of aliquot divisors of n. - Omar E. Pol, Nov 11 2014
|
|
LINKS
|
|
|
EXAMPLE
|
Triangle begins:
0;
1;
3, 1;
6, 1;
10, 4;
15, 4, 1;
21, 9, 1;
28, 9, 1;
36, 16, 4;
45, 16, 4, 1;
55, 25, 4, 1;
66, 25, 9, 1;
78, 36, 9, 1;
91, 36, 9, 4;
105, 49, 16, 4, 1;
120, 49, 16, 4, 1;
136, 64, 16, 4, 1;
153, 64, 25, 9, 1;
171, 81, 25, 9, 1;
190, 81, 25, 9, 4;
210, 100, 36, 9, 4, 1;
231, 100, 36, 16, 4, 1;
253, 121, 36, 16, 4, 1;
276, 121, 49, 16, 4, 1;
...
For n = 6 the divisors of all positive integers <= 6 are [1], [1, 2], [1, 3], [1, 2, 4], [1, 5], [1, 2, 3, 6] hence the sum of all aliquot divisors is [0] + [1] + [1] + [1+2] + [1] + [1+2+3] = 0 + 1 + 1 + 3 + 1 + 6 = 12. On the other hand the 6th row of triangle is 15, 4, 1, therefore the alternating row sum is 15 - 4 + 1 = 12, equaling the sum of all aliquot divisors of all positive integers <= 6.
|
|
CROSSREFS
|
Cf. A000203, A000217, A001065, A008794, A003056, A153485, A196020, A211547, A211343, A228813, A231345, A231347, A235791, A235794, A235799, A236104, A236106, A236112, A237591, A237593, A286001.
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|