A236104 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, and the first element of column k is in row k(k+1)/2.
1, 4, 9, 1, 16, 1, 25, 4, 36, 4, 1, 49, 9, 1, 64, 9, 1, 81, 16, 4, 100, 16, 4, 1, 121, 25, 4, 1, 144, 25, 9, 1, 169, 36, 9, 1, 196, 36, 9, 4, 225, 49, 16, 4, 1, 256, 49, 16, 4, 1, 289, 64, 16, 4, 1, 324, 64, 25, 9, 1, 361, 81, 25, 9, 1, 400, 81, 25, 9, 4
Offset: 1
Examples
Triangle begins:
1;
4;
9, 1;
16, 1;
25, 4;
36, 4, 1;
49, 9, 1;
64, 9, 1;
81, 16, 4;
100, 16, 4, 1;
121, 25, 4, 1;
144, 25, 9, 1;
169, 36, 9, 1;
196, 36, 9, 4;
225, 49, 16, 4, 1;
256, 49, 16, 4, 1;
289, 64, 16, 4, 1;
324, 64, 25, 9, 1;
361, 81, 25, 9, 1;
400, 81, 25, 9, 4;
441, 100, 36, 9, 4, 1;
484, 100, 36, 16, 4, 1;
529, 121, 36, 16, 4, 1;
576, 121, 49, 16, 4, 1;
...
For n = 6 the sum of all divisors of all positive integers <= 6 is [1] + [1+2] + [1+3] + [1+2+4] + [1+5] + [1+2+3+6] = 1 + 3 + 4 + 7 + 6 + 12 = 33. On the other hand the 6th row of triangle is 36, 4, 1, therefore the alternating row sum is 36 - 4 + 1 = 33, equaling the sum of all divisors of all positive integers <= 6.
Illustration of the alternating sum of the 6th row as the area of a polygon (or the number of cells), step by step, in the fourth quadrant:
. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
. | | | | | |
. | | | | | |
. | | | | | |
. | | | _ _| | _|
. | | | | | _|
. |_ _ _ _ _ _| |_ _ _ _| |_ _ _ _|
.
. 36 36 - 4 = 32 36 - 4 + 1 = 33
.
Then using this method we can draw a symmetric diagram for A000203, A024916, A004125, as shown below:
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n A000203 A024916 Diagram
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. _ _ _ _ _ _ _ _ _ _ _ _
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 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. Finally, the number of *'s is A004125(12) = 17.
Note that the diagram is also the top view of the stepped pyramid described in A245092. - _Omar E. Pol_, Feb 12 2018
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10075 (rows 1 <= n <= 500).
Crossrefs
Programs
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Mathematica
Table[Ceiling[(n + 1)/k - (k + 1)/2]^2, {n, 20}, {k, Floor[(Sqrt[8 n + 1] - 1)/2]}] // Flatten (* Michael De Vlieger, Feb 10 2018, after Hartmut F. W. Hoft at A235791 *) -
Python
from sympy import sqrt import math def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) for n in range(1, 21): print([T(n, k)**2 for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 25 2017
Formula
Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A024916(n). [Although this was stated as a fact, as far as I can tell, no proof was known. However, Don Reble has recently found a proof, which will be added here soon. - N. J. A. Sloane, Nov 23 2020]
A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k) - T(n-1,k)), assuming that T(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 10 2018
Comments