A236631
Triangle read by rows: T(j,k), j>=1, k>=1, in which column k lists the positive squares repeated k-1 times, except the column 1 which is A123327. The elements of the even-indexed columns are multiplied by -1. The first element of column k is in row k(k+1)/2.
Original entry on oeis.org
1, 3, 5, -1, 8, -1, 10, -4, 15, -4, 1, 16, -9, 1, 23, -9, 1, 25, -16, 4, 31, -16, 4, -1, 34, -25, 4, -1, 45, -25, 9, -1, 42, -36, 9, -1, 55, -36, 9, -4, 60, -49, 16, -4, 1, 67, -49, 16, -4, 1, 69, -64, 16, -4, 1, 86, -64, 25, -9, 1, 84, -81, 25, -9, 1, 103
Offset: 1
Written as an irregular triangle the sequence begins:
1;
3;
5, -1;
8, -1;
10, -4;
15, -4, 1;
16, -9, 1;
23, -9, 1;
25, -16, 4;
31, -16, 4, -1;
34, -25, 4, -1;
45, -25, 9, -1;
42, -36, 9, -1;
55, -36, 9, -4;
60, -49, 16, -4, 1;
67, -49, 16, -4, 1;
69, -64, 16, -4, 1;
86, -64, 25, -9, 1;
84, -81, 25, -9, 1;
103, -81, 25, -9, 4;
102, -100, 36, -9, 4, -1;
113, -100, 36, -16, 4, -1;
122, -121, 36, -16, 4, -1;
145, -121, 49, -16, 4, -1;
...
For j = 15 the divisors of 15 are 1, 3, 5, 15, therefore the sum of divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand the 15th row of triangle is 60, -49, 16, -4, 1, therefore the row sum is 60 - 49 + 16 - 4 + 1 = 24, equalling the sum of divisors of 15.
Cf.
A000203,
A000217,
A000290,
A003056,
A004125,
A024916,
A008794,
A123327,
A196020,
A211547,
A236104,
A236630,
A237593.
A236109
Triangle read by rows: another version of A048158, only here the representation of A004125 is symmetric, as in the representation of A024916 and A000203.
Original entry on oeis.org
0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 2, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 3, 3, 0, 0, 0, 0, 0, 2, 3, 3, 0, 0, 0, 0, 0, 2, 2, 4, 4, 0, 0, 0, 0, 0, 0, 2, 3, 4, 4, 0, 0, 0, 0, 0, 0, 3, 4, 5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 2, 2, 3, 5, 5, 0, 0, 0, 0, 0, 0, 0, 3
Offset: 1
Triangle begins:
0;
0, 0;
0, 0, 1;
0, 0, 0, 1;
0, 0, 0, 2, 2;
0, 0, 0, 0, 1, 2;
0, 0, 0, 0, 2, 3, 3;
0, 0, 0, 0, 0, 2, 3, 3;
0, 0, 0, 0, 0, 2, 2, 4, 4;
0, 0, 0, 0, 0, 0, 2, 3, 4, 4;
0, 0, 0, 0, 0, 0, 3, 4, 5, 5, 5;
0, 0, 0, 0, 0, 0, 0, 2, 2, 3, 5, 5;
...
For the symmetric representation of A000203, A024916, A004125 in the fourth quadrant using a diagram which arises from the sequence A236104 see below:
--------------------------------------------------
n A000203 A024916 Diagram
--------------------------------------------------
. _ _ _ _ _ _ _ _ _ _ _ _
1 1 1 |_| | | | | | | | | | | |
2 3 4 |_ _|_| | | | | | | | | |
3 4 8 |_ _| _|_| | | | | | | |
4 7 15 |_ _ _| _|_| | | | | |
5 6 21 |_ _ _| _| _ _|_| | | |
6 12 33 |_ _ _ _| _| | _ _|_| |
7 8 41 |_ _ _ _| |_ _|_| _ _|
8 15 56 |_ _ _ _ _| _| |* *
9 13 69 |_ _ _ _ _| | _|* *
10 18 87 |_ _ _ _ _ _| _ _|* * *
11 12 99 |_ _ _ _ _ _| |* * * * *
12 28 127 |_ _ _ _ _ _ _|* * * * *
.
The 12th row is ........ 0,0,0,0,0,0,0,2,2,3,5,5
.
The total number of cells in the first n set of symmetric regions of the diagram equals A024916(n). It appears that the total number of cells in the n-th set of symmetric regions of the diagram equals sigma(n) = A000203(n). Example: for n = 12 the 12th row of triangle is 144, 25, 9, 1, hence the alternating sums is 144 - 25 + 9 - 1 = 127. On the other hand we have that A000290(12) - A004125(12) = 144 - 17 = A024916(12) = 127, equaling the total number of cells in the diagram after 12 stages. The number of cells in the 12th set of symmetric regions of the diagram is sigma(12) = A000203(12) = 28. Note that in this case there is only one region. The number of "*"'s is A004125(12) = 17.
Cf.
A000203,
A004125,
A024916,
A048158,
A196020,
A235799,
A236104,
A236630,
A236631,
A237591,
A237593,
A237270.
Showing 1-2 of 2 results.
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