This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A236104 #137 Nov 05 2024 05:39:51
%S A236104 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,
%T A236104 144,25,9,1,169,36,9,1,196,36,9,4,225,49,16,4,1,256,49,16,4,1,289,64,
%U A236104 16,4,1,324,64,25,9,1,361,81,25,9,1,400,81,25,9,4
%N 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.
%C A236104 These are the squares of the entries of the triangle in A235791: T(n,k) = (A235791(n,k))^2.
%C A236104 Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%C A236104 Columns 1-3 (including the initial zeros) are A000290, A008794, A211547.
%C A236104 Also column k lists the partial sums of the k-th column of triangle A196020 which gives an identity for sigma.
%C A236104 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.
%C A236104 For more information about the diagram see A237593.
%H A236104 Michael De Vlieger, <a href="/A236104/b236104.txt">Table of n, a(n) for n = 1..10075</a> (rows 1 <= n <= 500).
%F A236104 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]
%F A236104 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
%e A236104 Triangle begins:
%e A236104 1;
%e A236104 4;
%e A236104 9, 1;
%e A236104 16, 1;
%e A236104 25, 4;
%e A236104 36, 4, 1;
%e A236104 49, 9, 1;
%e A236104 64, 9, 1;
%e A236104 81, 16, 4;
%e A236104 100, 16, 4, 1;
%e A236104 121, 25, 4, 1;
%e A236104 144, 25, 9, 1;
%e A236104 169, 36, 9, 1;
%e A236104 196, 36, 9, 4;
%e A236104 225, 49, 16, 4, 1;
%e A236104 256, 49, 16, 4, 1;
%e A236104 289, 64, 16, 4, 1;
%e A236104 324, 64, 25, 9, 1;
%e A236104 361, 81, 25, 9, 1;
%e A236104 400, 81, 25, 9, 4;
%e A236104 441, 100, 36, 9, 4, 1;
%e A236104 484, 100, 36, 16, 4, 1;
%e A236104 529, 121, 36, 16, 4, 1;
%e A236104 576, 121, 49, 16, 4, 1;
%e A236104 ...
%e A236104 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.
%e A236104 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:
%e A236104 . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
%e A236104 . | | | | | |
%e A236104 . | | | | | |
%e A236104 . | | | | | |
%e A236104 . | | | _ _| | _|
%e A236104 . | | | | | _|
%e A236104 . |_ _ _ _ _ _| |_ _ _ _| |_ _ _ _|
%e A236104 .
%e A236104 . 36 36 - 4 = 32 36 - 4 + 1 = 33
%e A236104 .
%e A236104 Then using this method we can draw a symmetric diagram for A000203, A024916, A004125, as shown below:
%e A236104 --------------------------------------------------
%e A236104 n A000203 A024916 Diagram
%e A236104 --------------------------------------------------
%e A236104 . _ _ _ _ _ _ _ _ _ _ _ _
%e A236104 1 1 1 |_| | | | | | | | | | | |
%e A236104 2 3 4 |_ _|_| | | | | | | | | |
%e A236104 3 4 8 |_ _| _|_| | | | | | | |
%e A236104 4 7 15 |_ _ _| _|_| | | | | |
%e A236104 5 6 21 |_ _ _| _| _ _|_| | | |
%e A236104 6 12 33 |_ _ _ _| _| | _ _|_| |
%e A236104 7 8 41 |_ _ _ _| |_ _|_| _ _|
%e A236104 8 15 56 |_ _ _ _ _| _| |* *
%e A236104 9 13 69 |_ _ _ _ _| | _|* *
%e A236104 10 18 87 |_ _ _ _ _ _| _ _|* * *
%e A236104 11 12 99 |_ _ _ _ _ _| |* * * * *
%e A236104 12 28 127 |_ _ _ _ _ _ _|* * * * *
%e A236104 .
%e A236104 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.
%e A236104 Note that the diagram is also the top view of the stepped pyramid described in A245092. - _Omar E. Pol_, Feb 12 2018
%t A236104 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 *)
%o A236104 (Python)
%o A236104 from sympy import sqrt
%o A236104 import math
%o A236104 def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))
%o A236104 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
%Y A236104 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.
%K A236104 nonn,tabf,look
%O A236104 1,2
%A A236104 _Omar E. Pol_, Jan 23 2014