cp's OEIS Frontend

All entries in the triangle are nonnegative since the number of 1's in odd-numbered columns of A237048 prior to column j, 1 <= j <= row(n), is at least as large as the number of 1's in even-numbered columns through column j. As a consequence:

(a) The two adjacent symmetric Dyck paths whose legs are defined by adjacent rows of triangle A237593 never cross each other (see also A235791 and A237591) and the rows in this triangle describe the widths between the legs.

(b) Let legs(n) denote the n-th row of triangle A237591, widths(n) the n-th row of this triangle, and c(n) the rightmost entry in the n-th row of this triangle (center of the Dyck path). Then area(n) = 2 * legs(n) . width(n) - c(n), where "." is the inner product, is the area between the two adjacent symmetric Dyck paths.

(c) For certain sequences of integers, it is known that area(n) = sigma(n); see A238443, A245685, A246955, A246956 and A247687.

Right border gives A067742. - Omar E. Pol, Jan 21 2017

For a proof that T(n, k) = |{ d : d|n and k/2 < d <= k }|, for 1 <= k <= row(n), where row(n) is the length of row n, an identity suggested by Peter Munn, see the link. A corollary to it is that the number of divisors of n in the half-open interval (row(n)/2, row(n)] equals the width of the symmetric representation of n at the diagonal: T(n, row(n)) = | { d : d|n and row(n)/2 < d <= row(n) } |. See also the comments and conjectures of Michel Marcus in A067742 and A237593. - Hartmut F. W. Hoft, Jun 24 2024

From Omar E. Pol, Jul 24 2024: (Start)

Conjecture 1: Every column is a periodic sequence.

Conjecture 2: The periods of the columns 1..8 are respectively: 1, 2, 6, 12, 60, 60, 420, 840.

Question 1: Is the period of the column k equal to A003418(k)? (End).

From Omar E. Pol, Jul 26 2024: (Start)

Column 1 gives A000012.

Column 2 gives A000035.

Conjecture 3: Column 3 gives [2, 0] together with A115357, hence column 3 gives 2 together with A171182.

Question 2: Except the first nine terms of A337976, is the column 4 the same as A337976?

Question 3: Except the first 14 terms of A366981, is the column 5 the same as A366981? (End)

From Hartmut F. W. Hoft, Aug 01 2024: (Start)

Conjectures 1 and 2 are true and the answer to question 1 is affirmative.

By definition, each column k in triangle T237048(n, k) of sequence A237048 is a periodic sequence of period k. Since the k-th term in row n of the triangle T(n, k) = Sum_{i=1..k} (-1)^(i+1) * T237048(n, i), with 1 <= k <= A003056(n), each initial subsequence T(n, 1) .. T(n, k) of row n in this triangle is periodic of period lcm(1, .. , k) = A003418(k). This implies that each column k in this sequence has period A003418(k).

Conjecture 3 and Question 2 are true. Since T237048(n, 1) = 1, T237208(n, 2) = 1 if n odd and 0 if n even, T237048(n, 3) = 1 if 3|n and 0 otherwise, and T237048(n, 4) = 1 if 4|(n-2) and 0 otherwise, equations T249223(n, 3) = 1 - (n mod 2) + delta( n mod 3) and T249223(n, 4) = 1 - (n mod 2) + delta( n mod 3) - delta( (n-2) mod 4) hold where delta(k) = 1 if k = 0 and 0 otherwise. With the 3rd column starting at n = A000217(3) = 6, each period starting in a row that is a multiple of 6 is [ 2 0 1 1 1 0 ], and appropriate shifts yield A115357 and A171182. With the 4th column starting at n = A000217(4) = 10, each period starting in a row n with 12|(n+2) is [ 0 0 2 0 0 1 1 0 1 0 1 1 ], and with a shift of 9 yields the apparently periodic A337976(10), A337976(11), ... (End)

Here T(n,k) is defined to be the "k-th width" of the symmetric representation of sigma(n), with n>=1 and 1<=k<=2n-1. Explanation: consider the diagram of the symmetric representation of sigma(n) described in A236104, A237593 and other related sequences. Imagine that the diagram for sigma(n) contains 2n-1 equidistant segments which are parallel to the main diagonal [(0,0),(n,n)] of the quadrant. The segments are located on the diagonal of the cells. The distance between two parallel segment is equal to sqrt(2)/2. T(n,k) is the length of the k-th segment divided by sqrt(2). Note that the triangle contains nonnegative terms because for some n the value of some widths is equal to zero. For an illustration of some widths see Hartmut F. W. Hoft's contribution in the Links section of A237270.

Row n has length 2*n-1.

Row sums give A000203.

If n is a power of 2 then all terms of row n are 1's.

If n is an even perfect number then all terms of row n are 1's except the middle term which is 2.

If n is an odd prime then row n lists (n+1)/2 1's, n-2 zeros, (n+1)/2 1's.

The number of blocks of positive terms in row n gives A237271(n).

The sum of the k-th block of positive terms in row n gives A237270(n,k).

It appears that the middle diagonal is also A067742 (which was conjectured by Michel Marcus in the entry A237593 and checked with two Mathematica functions up to n = 100000 by Hartmut F. W. Hoft).

It appears that the trapezoidal numbers (A165513) are also the numbers k > 1 with the property that some of the noncentral widths of the symmetric representation of sigma(k) are not equal to 1. - Omar E. Pol, Mar 04 2023

Here T(n,k) is defined to be the "k-th width" of the symmetric representation of A024916(n), with n>=1 and 1<=k<=2n-1.

If both A249351 and this sequence are written as isosceles triangles then the partial sums of the columns of A249351 give the columns of this isosceles triangle (see the second triangle in Example section).

For the definition of the k-th width of the symmetric representation of sigma(n) see A249351.

Note that for the geometric representation of the n-th row of the triangle we need the x-axis, the y-axis, and only a Dyck path which is given by the elements of the n-th row of the triangle A237593.

Row n has length 2*n-1.

Row sums give A024916.

The middle diagonal is A240542.

For the definition of k-th width of the symmetric representation of sigma(n) see A249351.

Row n list the first n terms of the n-th row of A249351.

It appears that the leading diagonal is also A067742 (which was conjectured by Michel Marcus in the entry A237593 and checked with two Mathematica functions up to n = 100000 by Hartmut F. W. Hoft).

For more information see A237591, A237593.

The equation y^2 = n^3 + A*n^2 + B*n + C, where A = 1, B = 2*r, C = r^2 is a minimal model of an elliptic curve with integral coefficients, for details see the Links section.

For a prime number p >= 5, the equation y^2 = p^3 + (p + r)^2 has 2 solutions, r_1 = p*(p - 3)/2 and r_2 = (p + 1)*(p^2 - p - 1)/2.

Factoring the equation y^2 = n^3 + n^2 + 2*r*n + r^2 yields (y+n+r)*(y-n-r) = n^3, which implies y+n+r = d and y-n-r = n^3/d for some divisor d of n^3. Thus a(n) is the number of divisors d of n^3 such that (d-n^3/d)/2 - n is a nonnegative integer. This resolves some of Thomas Scheuerle's conjectures. - Robin Visser, Sep 30 2023