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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.

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).

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

For more information see A236104.

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. [An equivalent assertion is that the sum of row n of A237048 is the number of odd divisors of n, and this was proved by Hartmut F. W. Hoft in a comment in A237048. - N. J. A. Sloane, Dec 07 2020]

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)

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

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

Alternating row sums give A240542. - Omar E. Pol, Apr 16 2014

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, or in accordance with the triangles of the same family: the partial column sum of A237048. - Omar E. Pol, Apr 28 2017, Nov 24 2020

The above conjecture is true. The proof will be added soon (it uses the generating function for the columns). - N. J. A. Sloane, Nov 24 2020

T(n,k) is also the total length of all line segments between the k-th vertex and the central vertex of the largest Dyck path of the symmetric representation of sigma(n). In other words: T(n,k) is the sum of the last (A003056(n)-k+1) terms of the n-th row of A237591. - Omar E. Pol, Sep 07 2021

T(n,k) is also the Manhattan distance between the k-th vertex and the central vertex of the Dyck path described in the n-th row of the triangle A237593. - Omar E. Pol, Jan 11 2023

Gives an identity for sigma(n): alternating sum of row n equals the sum of divisors of n. For proof see Max Alekseyev link.

Row n has length A003056(n) hence column k starts in row A000217(k).

The number of positive terms in row n is A001227(n), the number of odd divisors of n.

If n = 2^j then the only positive integer in row n is T(n,1) = 2^(j+1) - 1.

If n is an odd prime then the only two positive integers in row n are T(n,1) = 2n - 1 and T(n,2) = n - 2.

If T(n,k) = 3 then T(n+1,k+1) = 1, the first element of the column k+1.

The partial sums of column k give the column k of A236104.

The connection with the symmetric representation of sigma is as follows: A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> A237270.

Alternating sum of row n equals the number of units cubes that protrude from the n-th level of the stepped pyramid described in A245092. - Omar E. Pol, Oct 28 2015

Conjecture: T(n,k) is the difference between the square of the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the square of the total number of partitions of all positive integers < n into exactly k consecutive parts. - Omar E. Pol, Feb 14 2018

From Omar E. Pol, Nov 24 2020: (Start)

T(n,k) is also the number of steps in the first n levels of the k-th double-staircase that has at least one step in the n-th level of the "Double- staircases" diagram, otherwise T(n,k) = 0, (see the Example section).

For the connection with A280851 see also the algorithm of A280850 and the conjecture of A296508. (End)

The number of zeros in the n-th row equals A238005(n). - Omar E. Pol, Sep 11 2021

Apart from the alternating row sums and the sum of divisors function A000203 another connection with Euler's pentagonal theorem is that in the irregular triangle of A238442 the k-th column starts in the row that is the k-th generalized pentagonal number A001318(k) while here the k-th column starts in the row that is the k-th generalized hexagonal number A000217(k). Both A001318 and A000217 are successive members of the same family: the generalized polygonal numbers. - Omar E. Pol, Sep 23 2021

Other triangle with the same row lengths and alternating row sums equals sigma(n) is A252117. - Omar E. Pol, May 03 2022

Gives an identity for the twice sigma function (A074400), the sum of the even divisors of 2n.

Alternating sum of row n equals A074400(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = 2*A000203(n) = A074400(n).

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

The number of positive terms in row n is A001227(n).

For more information see A196020.

Gives an identity for the sum of remainders of n mod k, for k = 1,2,3,...,n. Alternating sum of row n equals A004125(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A004125(n).

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

Gives an identity for the abundance of n. Alternating sum of row n equals the abundance of n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A033880(n).

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

Alternating sum of row n equals the sum of aliquot divisors of n, i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A001065(n).

Row n has length A003056(n).

Column k starts in row A000217(k).

The number of positive terms in row n is A001227(n), for n >= 2.

If n = 2^j then the only positive integer in row n is T(n,1) = n - 1, for j >= 1.

If n is an odd prime then the only two positive integers in row n are T(n,1) = n - 1 and T(n,2) = n - 2.

Gives an identity for the sum of all aliquot divisors of all positive integers <= n.

Alternating sum of row n equals A153485(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A153485(n).

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

Column 1 is A000217. Columns 2-3 are A008794, A211547, but without the zeros.

Column k lists the partial sums of the k-th column of triangle A231347 which gives an identity for the sum of aliquot divisors of n. - Omar E. Pol, Nov 11 2014

Alternating sum of row n equals A235796(n), i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A235796(n).

Row n has length A003056(n) hence column k starts in row A000217(k).

Column k starts with k+1 zeros and then lists the odd numbers interleaved with k zeros.

It appears that row n lists all zeros iff n is a power of 2.

Alternating row sums give the Chowla's function, i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A048050(n).

Row n has length A003056(n) hence column k starts in row A000217(k).

Column 1 gives 0 together with A001477.

Column 2 is A193356.

The number of positive terms in row n is A001227(n), if n >= 3. - Omar E. Pol, Apr 18 2016