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 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. 203
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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. LINKS Michael De Vlieger, Table of n, a(n) for n = 1..10075 (rows 1 <= n <= 500). 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 EXAMPLE 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 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 *) PROG (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 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. Sequence in context: A200393 A070438 A070638 * A152205 A129861 A055491 Adjacent sequences:  A236101 A236102 A236103 * A236105 A236106 A236107 KEYWORD nonn,tabf,look AUTHOR Omar E. Pol, Jan 23 2014 STATUS approved

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Last modified May 5 19:09 EDT 2021. Contains 343573 sequences. (Running on oeis4.)