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For the construction of the n-th row of this triangle start with a copy of the n-th row of the triangle A196020.

Then replace each element of the m-th pair of positive integers (x, y) with the value (x - y)/2, where "y" is the m-th even-indexed term of the row, and "x" is its previous nearest odd-indexed term not used in another pair in the same row, if such a pair exist. Otherwise T(n,k) = A196020(n,k). (See example).

Observation 1: at least for the first 28 rows of the triangle the nonzero terms in the n-th row are also the subparts of the symmetric representation of sigma(n), assuming the ordering of the subparts in the same row does not matter.

Question 1: are always the nonzero terms of the n-th row the same as all the subparts of the symmetric representation of sigma(n)? If not, what is the index of the row in which appears the first counterexample?

Note that the "subparts" are the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.

For more information about "subparts" see A279387 and A237593.

About the question 1, it appears that the n-th row of the triangle A280851 and the n-th row of this triangle contain the same nonzero numbers, though in different order; checked through n = 250000. - Hartmut F. W. Hoft, Jan 31 2018

From Omar E. Pol, Feb 02 2018: (Start)

Observation 2: at least for the first 28 rows of the triangle we have that in the n-th row the odd-indexed terms, from left to right, together with the even-indexed terms, from right to left, form a finite sequence in which the nonzero terms are the same as the n-th row of triangle A280851, which lists the subparts of the symmetric representation of sigma(n).

Question 2: Are always the same for all rows? If not, what is the index of the row in which appears the first counterexample? (End)

Conjecture: the odd-indexed terms of the n-th row together with the even-indexed terms of the same row but listed in reverse order give the n-th row of triangle A296508 (this is the same conjecture from A296508). - Omar E. Pol, Apr 20 2018

Gives an identity for A000593. Alternating sum of row n equals the sum of odd divisors of n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A000593(n).

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

Since the odd-indexed rows of the triangle A266537 contain all zeros then odd-indexed rows of this triangle are the same as the odd-indexed rows of the triangle A196020.

If T(n,k) is the second odd number in the column k then T(n+1,k+1) = 1 is the first element in the column k+1.

Alternating row sums of A196020 give A000203.

Alternating row sums of A266537 give A146076.

Alternating sum of row n in the triangle A196020 equals A000203(n). This is the standard row sums of triangle A196020.

a(n) is the total number of steps in all odd-indexed double-staircases of the diagram of A196020 with n levels (see the example).

a(n) is also the total number of steps in all odd-indexed double-staircases of the diagram described in A335616 with n levels that have at least one step in the bottom level of the diagram.

Sigma(n) <= a(n).

The graph of the sum-of-divisors function A000203 is intermediate between the graph of this sequence and the graph of A353154 (see link). - Omar E. Pol, May 13 2022

Conjecture: indices of zeros give A082662.

a(n) is the total number of steps in all even-indexed double-staircases of the diagram of A196020 with n levels.

a(n) is also the total number of steps in all even-indexed double-staircases of the diagram described in A335616 with n levels that have at least one step in the bottom level of the diagram.

The graph of the sum-of-divisors function A000203 is intermediate between the graph of A353149 and the graph of this sequence (see the Links section). - Omar E. Pol, May 13 2022

Every column of this square array is associated to an isosceles triangle and to a stepped pyramid in the same way as the sequence A196020 is associated to the isosceles triangle of A237593 and to the pyramid described in A245092. Hence there are infinitely many isosceles triangles and infinitely many pyramids that are associated to this sequence.

In the Example section appears the triangles and the top views of the pyramids associated to the columns 1 and 2.

The sequence A196020 is associated to the isosceles triangle of A237593 as follows: A196020 --> A236104 --> A235791 --> A237591 --> A237593. Then the structure of the pyramid described in A245092 arises after the 90-degree-zig-zag folding of every row of the isosceles triangle of A237593.

Note that the first m terms of column k are also the first m terms of A000203, where m = A000217(k) + k = A000217(k+1) - 1 = A000096(k).

Multiplicative: If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).

Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (this sequence) (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001

A number n is abundant if sigma(n) > 2n (cf. A005101), perfect if sigma(n) = 2n (cf. A000396), deficient if sigma(n) < 2n (cf. A005100).

a(n) is the number of sublattices of index n in a generic 2-dimensional lattice. - Avi Peretz (njk(AT)netvision.net.il), Jan 29 2001 [In the language of group theory, a(n) is the number of index-n subgroups of Z x Z. - Jianing Song, Nov 05 2022]

The sublattices of index n are in one-to-one correspondence with matrices [a b; 0 d] with a>0, ad=n, b in [0..d-1]. The number of these is Sum_{d|n} d = sigma(n), which is a(n). A sublattice is primitive if gcd(a,b,d) = 1; the number of these is n * Product_{p|n} (1+1/p), which is A001615. [Cf. Grady reference.]

Sum of number of common divisors of n and m, where m runs from 1 to n. - Naohiro Nomoto, Jan 10 2004

a(n) is the cardinality of all extensions over Q_p with degree n in the algebraic closure of Q_p, where p>n. - Volker Schmitt (clamsi(AT)gmx.net), Nov 24 2004. Cf. A100976, A100977, A100978 (p-adic extensions).

Let s(n) = a(n-1) + a(n-2) - a(n-5) - a(n-7) + a(n-12) + a(n-15) - a(n-22) - a(n-26) + ..., then a(n) = s(n) if n is not pentagonal, i.e., n != (3 j^2 +- j)/2 (cf. A001318), and a(n) is instead s(n) - ((-1)^j)*n if n is pentagonal. - Gary W. Adamson, Oct 05 2008 [corrected Apr 27 2012 by William J. Keith based on Ewell and by Andrey Zabolotskiy, Apr 08 2022]

Write n as 2^k * d, where d is odd. Then a(n) is odd if and only if d is a square. - Jon Perry, Nov 08 2012

Also total number of parts in the partitions of n into equal parts. - Omar E. Pol, Jan 16 2013

Note that sigma(3^4) = 11^2. On the other hand, Kanold (1947) shows that the equation sigma(q^(p-1)) = b^p has no solutions b > 2, q prime, p odd prime. - N. J. A. Sloane, Dec 21 2013, based on postings to the Number Theory Mailing List by Vladimir Letsko and Luis H. Gallardo

Limit_{m->infinity} (Sum_{n=1..prime(m)} a(n)) / prime(m)^2 = zeta(2)/2 = Pi^2/12 (A072691). See more at A244583. - Richard R. Forberg, Jan 04 2015

a(n) + A000005(n) is an odd number iff n = 2m^2, m>=1. - Richard R. Forberg, Jan 15 2015

a(n) = a(n+1) for n = 14, 206, 957, 1334, 1364 (A002961). - Zak Seidov, May 03 2016

Equivalent to the Riemann hypothesis: a(n) < H(n) + exp(H(n))*log(H(n)), for all n>1, where H(n) is the n-th harmonic number (Jeffrey Lagarias). See A057641 for more details. - Ilya Gutkovskiy, Jul 05 2016

a(n) is the total number of even parts in the partitions of 2*n into equal parts. More generally, a(n) is the total number of parts congruent to 0 mod k in the partitions of k*n into equal parts (the comment dated Jan 16 2013 is the case for k = 1). - Omar E. Pol, Nov 18 2019

From Jianing Song, Nov 05 2022: (Start)

a(n) is also the number of order-n subgroups of C_n X C_n, where C_n is the cyclic group of order n. Proof: by the correspondence theorem in the group theory, there is a one-to-one correspondence between the order-n subgroups of C_n X C_n = (Z x Z)/(nZ x nZ) and the index-n subgroups of Z x Z containing nZ x nZ. But an index-n normal subgroup of a (multiplicative) group G contains {g^n : n in G} automatically. The desired result follows from the comment from Naohiro Nomoto above.

The number of subgroups of C_n X C_n that are isomorphic to C_n is A001615(n). (End)

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