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 A235791 Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2. 150

%I

%S 1,2,3,1,4,1,5,2,6,2,1,7,3,1,8,3,1,9,4,2,10,4,2,1,11,5,2,1,12,5,3,1,

%T 13,6,3,1,14,6,3,2,15,7,4,2,1,16,7,4,2,1,17,8,4,2,1,18,8,5,3,1,19,9,5,

%U 3,1,20,9,5,3,2,21,10,6,3,2,1,22,10,6,4,2,1,23,11,6,4,2,1,24,11,7,4,2,1

%N Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

%C The alternating sum of the squares of the elements of the n-th row equals the sum of all divisors of all positive integers <= n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*(T(n,k))^2 = A024916(n).

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

%C The sum of row n gives A060831(n), the sum of the number of odd divisors of all positive integers <= n. - _Omar E. Pol_, Mar 01 2014

%C Comments from _Franklin T. Adams-Watters_ on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)

%C The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.

%C You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.

%C Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).

%C Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)

%C From _Hartmut F. W. Hoft_, Apr 07 2014: (Start)

%C Mathematica function has been written to check the first property up to n = 20000.

%C T(n,(sqrt(8n+1)-1)/2+1) = 0 for all n >= 1, which is useful for formulas for A237591 and A237593. (End)

%C Alternating row sums give A240542. - _Omar E. Pol_, Apr 16 2014

%C Conjecture: T(n,k) is also the total number of partitions of all positive integers <= n into exactly k consecutive parts, i.e., the partial column sum of A285898. - _Omar E. Pol_, Apr 28 2017

%H G. C. Greubel, <a href="/A235791/b235791.txt">Table of n, a(n) for the first 150 rows, flattened</a>

%F T(n,k) = ceiling((n+1)/k - (k+1)/2) for 1 <= n, 1 <= k <= floor((sqrt(8n+1)-1)/2) = A003056(n). - _Hartmut F. W. Hoft_, Apr 07 2014

%F T(n,k) = Sum_{j=1..n} A237048(j,k). - _Omar E. Pol_, May 18 2017

%F T(n,k) = sqrt(A236104(n,k)). - _Omar E. Pol_, Feb 14 2018

%F Sigma(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k)^2 - T(n-1,k)^2), assuming that T(k*(k+1)/2-1,k) = 0. - _Omar E. Pol_, Oct 10 2018

%e Triangle begins:

%e 1;

%e 2;

%e 3, 1;

%e 4, 1;

%e 5, 2;

%e 6, 2, 1;

%e 7, 3, 1;

%e 8, 3, 1;

%e 9, 4, 2;

%e 10, 4, 2, 1;

%e 11, 5, 2, 1;

%e 12, 5, 3, 1;

%e 13, 6, 3, 1;

%e 14, 6, 3, 2;

%e 15, 7, 4, 2, 1;

%e 16, 7, 4, 2, 1;

%e 17, 8, 4, 2, 1;

%e 18, 8, 5, 3, 1;

%e 19, 9, 5, 3, 1;

%e 20, 9, 5, 3, 2;

%e 21, 10, 6, 3, 2, 1;

%e 22, 10, 6, 4, 2, 1;

%e 23, 11, 6, 4, 2, 1;

%e 24, 11, 7, 4, 2, 1;

%e 25, 12, 7, 4, 3, 1;

%e 26, 12, 7, 5, 3, 1;

%e 27, 13, 8, 5, 3, 2;

%e 28, 13, 8, 5, 3, 2, 1;

%e ...

%e For n = 10 the 10th row of triangle is 10, 4, 2, 1, so we have that 10^2 - 4^2 + 2^2 - 1^2 = 100 - 16 + 4 - 1 = 87, the same as A024916(10) = 87, the sum of all divisors of all positive integers <= 10.

%e From _Omar E. Pol_, Nov 19 2015: (Start)

%e Illustration of initial terms in the third quadrant:

%e . y

%e Row _|

%e 1 _|1|

%e 2 _|2 _|

%e 3 _|3 |1|

%e 4 _|4 _|1|

%e 5 _|5 |2 _|

%e 6 _|6 _|2|1|

%e 7 _|7 |3 |1|

%e 8 _|8 _|3 _|1|

%e 9 _|9 |4 |2 _|

%e 10 _|10 _|4 |2|1|

%e 11 _|11 |5 _|2|1|

%e 12 _|12 _|5 |3 |1|

%e 13 _|13 |6 |3 _|1|

%e 14 _|14 _|6 _|3|2 _|

%e 15 _|15 |7 |4 |2|1|

%e 16 _|16 _|7 |4 |2|1|

%e 17 _|17 |8 _|4 _|2|1|

%e 18 _|18 _|8 |5 |3 |1|

%e 19 _|19 |9 |5 |3 _|1|

%e 20 _|20 _|9 _|5 |3|2 _|

%e 21 _|21 |10 |6 _|3|2|1|

%e 22 _|22 _|10 |6 |4 |2|1|

%e 23 _|23 |11 _|6 |4 |2|1|

%e 24 _|24 _|11 |7 |4 _|2|1|

%e 25 _|25 |12 |7 _|4|3 |1|

%e 26 _|26 _|12 _|7 |5 |3 _|1|

%e 27 _|27 |13 |8 |5 |3|2 _|

%e 28 |28 |13 |8 |5 |3|2|1|

%e ...

%e T(n,k) is also the number of cells between the k-th vertical line segment (from left to right) and the y-axis in the n-th row of the structure.

%e Note that the number of horizontal line segments in the n-th row of the structure equals A001227(n), the number of odd divisors of n.

%e Also the diagram represents the left part of the front view of the pyramid described in A245092. (End)

%t row[n_] := Floor[(Sqrt[8*n + 1] - 1)/2]; f[n_, k_] := Ceiling[(n + 1)/k - (k + 1)/2]; Table[f[n, k], {n, 1, 150}, {k, 1, row[n]}] // Flatten (* _Hartmut F. W. Hoft_, Apr 07 2014 *)

%o (PARI) row(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i); \\ _Michel Marcus_, Mar 27 2014

%o (Python)

%o from sympy import sqrt

%o import math

%o def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))

%o for n in xrange(1, 21): print [T(n, k) for k in xrange(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)] # _Indranil Ghosh_, Apr 25 2017

%Y Columns 1..3: A000027, A008619, A008620.

%Y Cf. A000203, A000217, A001227, A003056, A024916, A196020, A211343, A228813, A231345, A231347, A235794, A236104, A236106, A236112, A237048, A237270, A237271, A237591, A237593, A239660, A245092, A261699, A262626, A286000.

%K nonn,tabf

%O 1,2

%A _Omar E. Pol_, Jan 23 2014

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Last modified October 15 20:04 EDT 2019. Contains 328037 sequences. (Running on oeis4.)