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The sequence is closely related to the alternating harmonic series.

Its asymptotic behavior is lim_{k -> infinity} a(k)/k = log 2. The relative error is abs(a(k) - k*log(2))/(k*log(2)) <= 2/sqrt(k).

Conjecture 1: the sequence of first positions of the alternating runs of odd and even numbers in a(k) equals A028982. Example: the positions in (1),(2),2,(3),3,5,5,(6),(7),7,7,9,9,9,11,(12),12,(13),13,15,... are 1,2,4,8,9,16,18,... Checked with a Mathematica function through a(1000000).

See A235791, A237591 and A237593 for additional formulas and properties.

Conjecture 2: The sequence of differences a(n) - a(n-1), for n>=1, appears to be equal to A067742(n), the sequence of middle divisors of n; as an empty sum, a(0) = 0, (which was conjectured by Michel Marcus in the entry A237593). Checked with the two respective Mathematica functions up to n=100000. - Hartmut F. W. Hoft, Jul 17 2014

The number of occurrences of n is A259179(n). - Omar E. Pol, Dec 11 2016

Conjecture 3: a(n) is also the difference between the total number of partitions of all positive integers <= n into an odd number of consecutive parts, and the total number of partitions of all positive integers <= n into an even number of consecutive parts. - Omar E. Pol, Apr 28 2017

Conjecture 4: a(n) is also the total number of central subparts of all symmetric representations of sigma of all positive integers <= n. - Omar E. Pol, Apr 29 2017

a(n) is also the sum of the odd-indexed terms of the n-th row of the triangle A237591. - Omar E. Pol, Jun 20 2018

a(n) is the total number of middle divisors of all positive integers <= n (after Michel Marcus's conjecture in A237593). - Omar E. Pol, Aug 18 2021

Row n has length A001227(n), the number of odd divisors of n, and also the number of entries in row n of A341970.

Let 1 <= n, 1 <= d <= A001227(n) and k = A341970(A060831(n-1) + d). Expression s(n, k) = r*n - r*(r+1)*(r+2)/6 + k with r = floor((sqrt(8*n + 1) - 1)/2) translates position (row n, column k) in the triangle of A235791 to its position in sequence A235791.

The absolute values in row n are the smallest parts of the partitions of n into consecutive parts (with the partitions ordered by increasing number of parts). - Omar E. Pol, Dec 31 2024

The original name was: Number of "ON" cells at n-th stage in a cellular automaton (or pseudo cellular automaton) related to sigma (see Comments for precise definition).

The numbers in the n-th row of the triangle are the coordinates on the diagonal in the first quadrant of the polygons constructed by alternately adding and subtracting squares taken from the n-th row of A236104. The boundary from (0,n) to (n,0) of the final polygon is the Dyck path as defined in the n-th row of A237593. Therefore, using the arguments in A196020, A236104 and A071561, sigma(n) equals the area of its symmetric representation, for all n>=1.

The right border gives A240542.

For an image of the construction process of the Dyck path for sigma(15) see the image file in the Links section.

The length of the n-th row is A003056(n). - Omar E. Pol, Nov 03 2015

Decimal value of binary compactification of A235791.

Prime product compactification of A235791.

All terms are squarefree.

Row n is a palindromic composition of 2*n.

T(n,k) is also the length of the k-th segment in a Dyck path on the first quadrant of the square grid, connecting the x-axis with the y-axis, from (n, 0) to (0, n), starting with a segment in vertical direction, see example.

Conjecture 1: the area under the n-th Dyck path equals A024916(n), the sum of all divisors of all positive integers <= n.

If the conjecture is true then the n-th Dyck path represents the boundary segments after the alternating sum of the elements of the n-th row of A236104.

Conjecture 2: two adjacent Dyck paths never cross (checked by hand up to n = 128), hence the total area between the n-th Dyck path and the (n-1)-st Dyck path is equal to sigma(n) = A000203(n), the sum of divisors of n.

The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> this sequence --> A239660 --> A237270 --> A237271.

PARI scripts area(n) and chkcross(n) have been written to check the 2 properties and have been run up to n=10000. - Michel Marcus, Mar 27 2014

Mathematica functions have been written that verified the 2 properties through n=30000. - Hartmut F. W. Hoft, Apr 07 2014

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

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

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.

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).

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)

The symmetric representation of sigma, so defined, is row n of A237270. - Peter Munn, Jan 06 2025

It appears that, for the n-th set, the number of cells lying on the first diagonal is equal to A067742(n), the number of middle divisors of n. - Michel Marcus, Jun 21 2014

Checked Michel Marcus's conjecture with two Mathematica functions up to n=100000, for more information see A240542. - Hartmut F. W. Hoft, Jul 17 2014

A003056(n) is also the number of peaks of the Dyck path related to the n-th row of triangle. - Omar E. Pol, Nov 03 2015

The number of peaks of the Dyck path associated to the row A000396(n) of this triangle equals the n-th Mersenne prime A000668(n), hence Mersenne primes are visible in two ways at the pyramid described in A245092. - Omar E. Pol, Dec 19 2016

The limit as n approaches infinity (area under the Dyck path described in the n-th row of triangle divided by n^2) equals Pi^2/12 = zeta(2)/2. (Cf. A072691.) - Omar E. Pol, Dec 18 2021

The connection between the isosceles triangle and the stepped pyramid is due to the fact that this object can also be interpreted as a pop-up card. - Omar E. Pol, Nov 09 2022

T(n,k) is the number of cells in the k-th region of the n-th set of regions in a diagram of the symmetry of sigma(n), see example.

Row n is a palindromic composition of sigma(n).

Row sums give A000203.

Row n has length A237271(n).

In the row 2n-1 of triangle both the first term and the last term are equal to n.

If n is an odd prime then row n is [m, m], where m = (1 + n)/2.

The connection with A196020 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> this sequence.

For the boundary segments in an octant see A237591.

For the boundary segments in a quadrant see A237593.

For the boundary segments in the spiral see also A239660.

For the parts in every quadrant of the spiral see A239931, A239932, A239933, A239934.

We can find the spiral on the terraces of the stepped pyramid described in A244050. - Omar E. Pol, Dec 07 2016

T(n,k) is also the area of the k-th terrace, from left to right, at the n-th level, starting from the top, of the stepped pyramid described in A245092 (see Links section). - Omar E. Pol, Aug 14 2018

The original name was: Triangle read by rows: T(n,k) = A235791(n,k) - A235791(n,k+1), assuming that the virtual right border of triangle A235791 is A000004.

T(n,k) is also the length of the k-th segment in a zig-zag path on the first quadrant of the square grid, connecting the point (n, 0) with the point (m, m), starting with a segment in vertical direction, where m <= n.

Conjecture: the area of the polygon defined by the x-axis, this zig-zag path and the diagonal [(0, 0), (m, m)], is equal to A024916(n)/2, one half of the sum of all divisors of all positive integers <= n. Therefore the reflected polygon, which is adjacent to the y-axis, with the zig-zag path connecting the point (0, n) with the point (m, m), has the same property. And so on for each octant in the four quadrants.

For the representation of A024916 and A000203 we use two octants, for example: the first octant and the second octant, or the 6th octant and the 7th octant, etc., see A237593.

At least up to n = 128, two zig-zag paths never cross (checked by hand).

The finite sequence formed by the n-th row of triangle together with its mirror row gives the n-th row of triangle A237593.

The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> this sequence --> A237593 --> A239660 --> A237270 --> A237271.

Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014. (Start)

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

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.

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).

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)

From Hartmut F. W. Hoft, Apr 07 2014: (Start)

The row sum is A235791(n,1) - A235791(n,floor((sqrt(8n+1)-1)/2)+1) = n - 0.

Mathematica function has been written to check the conjecture as well as non-crossing zig-zag paths (Dyck paths rotated by 90 degrees) up through n=30000 (same applies to A237593). (End)

The n-th zig-zag path ending at the point (m, m), where m = A240542(n). - Omar E. Pol, Apr 16 2014

From Omar E. Pol, Aug 23 2015: (Start)

n is an odd prime if and only if T(n,2) = 1 + T(n-1,2) and T(n,k) = T(n-1,k) for the rest of the values of k.

The elements of the n-th row of triangle together with the elements of the n-th row of triangle A261350 give the n-th row of triangle A237593.

T(n,k) is also the area (or the number of cells) of the k-th vertical side at the n-th level (starting from the top) in the left hand part of the front view of the stepped pyramid described in A245092, see Example section.

(End)

From Omar E. Pol, Nov 19 2015: (Start)

T(n,k) is also the number of cells between the k-th and the (k+1)st line segments (from left to right) in the n-th row of the diagram as shown in Example section.

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

Conjecture: the values f(n,k) in the n-th row of the triangle are either 1 or 2 for all k with ceiling((sqrt(4*n+1)-1)/2) <= k <= floor((sqrt(8*n+1)-1)/2) = r(n), the length of the n-th row, though the lower bound need not be minimal; tested through 2500000. See also A285356. - Hartmut F. W. Hoft, Apr 17 2017

Conjecture: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts. - Omar E. Pol, Apr 30 2017

From Omar E. Pol, Aug 31 2021: (Start)

It appears that T(n,2)/T(n,1) converges to 1/3.

It appears that T(n,3)/T(n,2) converges to 1/2.

It appears that T(n,4)/T(n,3) converges to 3/5.

It appears that T(n,5)/T(n,4) converges to 2/3. (End)

In other words: T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(n). - Omar E. Pol, Sep 08 2021