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

Author

Omar E. Pol, Jan 23 2014

Keywords

• nonn
• tabf
• look

Comments

These are the squares of the entries of the triangle in A235791: T(n,k) = (A235791(n,k))^2.

Row n has length A003056(n) hence the first element of column k is in row A000217(k).

Columns 1-3 (including the initial zeros) are A000290, A008794, A211547.

Also column k lists the partial sums of the k-th column of triangle A196020 which gives an identity for sigma.

Since all the elements of this sequence are squares, we can draw an illustration of the alternating sum of row n step by step, and a symmetric diagram for A000203, A024916, A004125; see example.

For more information about the diagram see A237593.

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:
--------------------------------------------------
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 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

Cf. A000203, A000217, A000290, A001227, A003056, A008794, A024916, A004125, A196020, A211343, A228813, A231345, A231347, A235791, A235794, A235799, A236106, A236112, A236540, A237270, A237591, A237593, A239660, A244050, A245092, A262626, A286000.

Programs

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