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.
Original entry on oeis.org
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
Offset: 1
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.
Cf.
A000203,
A000217,
A001065,
A008794,
A003056,
A153485,
A196020,
A211547,
A211343,
A228813,
A231345,
A231347,
A235791,
A235794,
A235799,
A236104,
A236106,
A236112,
A237591,
A237593,
A286001.
A239446
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the elements of A004273 interleaved with k zeros, and the first element of column k is in row k*(k+1)/2.
Original entry on oeis.org
0, 0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 5, 0, 0, 0, 0, 0, 7, 3, 0, 0, 0, 1, 0, 9, 0, 0, 0, 0, 5, 0, 0, 11, 0, 0, 0, 0, 0, 3, 0, 13, 7, 0, 1, 0, 0, 0, 0, 0, 0, 15, 0, 0, 0, 0, 0, 9, 5, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0, 3, 0, 19, 11, 0, 0, 1, 0, 0, 0, 7, 0, 0, 0, 21, 0
Offset: 1
Triangle begins:
0;
0;
1, 0;
0, 0;
3, 0;
0, 1, 0;
5, 0, 0;
0, 0, 0;
7, 3, 0;
0, 0, 1, 0;
9, 0, 0, 0;
0, 5, 0, 0;
11, 0, 0, 0;
0, 0, 3, 0;
13, 7, 0, 1, 0;
0, 0, 0, 0, 0;
15, 0, 0, 0, 0;
0, 9, 5, 0, 0;
17, 0, 0, 0, 0;
0, 0, 0, 3, 0;
19, 11, 0, 0, 1, 0;
0, 0, 7, 0, 0, 0;
21, 0, 0, 0, 0, 0;
0, 13, 0, 0, 0, 0;
23, 0, 0, 5, 0, 0;
...
For n = 15 the 15th row of triangle is 13, 7, 0, 1, and the alternating sum is 13 - 7 + 0 - 1 = A235796(15) = 5.
Cf.
A000203,
A000217,
A003056,
A004125,
A004273,
A196020,
A231345,
A231347,
A235791,
A235794,
A235796,
A236106,
A236104,
A236112,
A237048,
A237588,
A237591,
A239313.
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.
Original entry on oeis.org
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
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.
Cf.
A000203,
A000217,
A001065,
A001227,
A001477,
A193356,
A196020,
A211343,
A212119,
A228813,
A231345,
A231347,
A235791,
A235794,
A236104,
A236106,
A236112.
A380231
Alternating row sums of triangle A237591.
Original entry on oeis.org
1, 2, 1, 2, 1, 4, 3, 4, 5, 4, 3, 6, 5, 4, 7, 8, 7, 8, 7, 10, 9, 8, 7, 10, 11, 10, 9, 12, 11, 14, 13, 14, 13, 12, 15, 16, 15, 14, 13, 16, 15, 18, 17, 16, 19, 18, 17, 20, 21, 22, 21, 20, 19, 22, 21, 24, 23, 22, 21, 24, 23, 22, 25, 26, 25, 28, 27, 26, 25, 28, 27, 32, 31, 30, 29, 28, 31, 30, 29
Offset: 1
For n = 14 the 14th row of A237591 is [8, 3, 1, 2] hence the alternating row sum is 8 - 3 + 1 - 2 = 4, so a(14) = 4.
On the other hand the 14th row of A237593 is the 14th row of A237591 together with the 14 th row of A237591 in reverse order as follows: [8, 3, 1, 2, 2, 1, 3, 8].
Then with the terms of the 14th row of A237593 we can draw a Dyck path in the first quadrant of the square grid as shown below:
.
(y axis)
.
.
. (4,14) (14,14)
._ _ _ . _ _ _ _ .
. |
. |
. |_
. |
. |_ _
. C |_ _ _
. |
. |
. |
. |
. . (14,4)
. |
. |
. . . . . . . . . . . . . . | . . . (x axis)
(0,0)
.
In the example the point C is the point (9,9).
The three line segments [(4,14),(9,9)], [(14,4),(9,9)] and [(14,14),(9,9)] have the same length.
The points (14,14), (9,9) and (4,14) are the vertices of a virtual isosceles right triangle.
The points (14,14), (9,9) and (14,4) are the vertices of a virtual isosceles right triangle.
The points (4,14), (14,14) and (14,4) are the vertices of a virtual isosceles right triangle.
Other alternating row sums (ARS) related to the Dyck paths of
A237593 and the stepped pyramid described in
A245092 are as follows:
-
row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
a(n) = my(orow = concat(row235791(n), 0)); vecsum(vector(#orow-1, i, (-1)^(i+1)*(orow[i] - orow[i+1]))); \\ Michel Marcus, Apr 13 2025
Comments