cp's OEIS Frontend

T(i, j) = -1 for i >= 1 odd, nonprime, j even with 1 < j < i; also for i prime and all j with 1 < j < i.

The single value T(10, 4) = -1 has been verified; see the conjecture below.

T(i, i) <= 3^(i-1) for all i >=1 . Equality holds for all primes i. T(i, i) = A318843(i), for all i >= 1.

A038547(i) is the smallest number with exactly i odd divisors. Thus odd number A038547(i) occurs in row i of triangle T(i, j) so that A038547 is a subsequence of this sequence. For i prime, A038547(i) = T(i, i). For 4 <= i <= 10^9 nonprime, A038547(i) is in the third column, T(i, 3), except for i=8; furthermore, the first part of SRS(A038547(i)) has width 1 and size (A038547(i)+1)/2.

T(i, 1) <= 2 * 3^(i-1) and it is even for all i >1. Equality holds for all primes i.

T(i, 2) <= 2 * 3^(i/2-1) * p for all even i where p is the smallest prime greater than 4 * 3^(i/2-1). Equality holds when i = 2 * h where h is prime.

The positive numbers in columns 1..6 are subsequences of A174973, A239929, A279102, A280107, A320066, A320511, respectively.

Conjectures:

All entries T(i, j) in columns j >= 3 are odd.

T(i, 1)/2 is odd for all i > 1.

T(i, 1) = 2 * T(i, 3) for all nonprime i > 3, for i = 3, but not for i = 8.

T(i, 2)/2 is odd for all even i > 2.

T(i, 3) = A038547(i) for all nonprime i > 3, except i = 8.

T(2*i, 2*j) = -1 for j >= 2 and all prime i satisfying i >= prime(j+1).

From Omar E. Pol, Jun 08 2025: (Start)

T(i,j) is also the smallest number k whose symmetric representation of sigma(k) has i subparts and j parts, or -1 if no such k exists.

Observations:

At least for i < 12 if i is prime then T(i,1) = 2*T(i,i).

At least for i < 12 if i is prime then all terms in row i are -1's except the first and the last term. (End)

Numbers m = 2^k * q, k >= 0 and q > 1 odd, without odd prime factors p < 2^(k+1).

This sequence is a proper subsequence of A238524. Numbers 78 = A370206(1) = A238524(55) and 102 = A237287(72) are not in this sequence since their width pattern (A341969) is 1210121.

A000079 is not a subsequence since SRS(2^k), k>=0, consists of a single part of width 1.

Let m = 2^k * q, k >= 0 and q > 1 odd, be a number in this sequence and s the size of the first part of SRS(m) which has width 1 and consists of 2^(k+1) - 1 legs of width 1. Therefore, s = Sum_{i=1..2^(k+1)-1} a237591(m, i) = a235791(m, 1) - a235791(m, 2^(k+1)) = ceiling((m+1)/1 - (1+1)/2) - ceiling((m+1)/2^(k+1) - (2^(k+1) + 1)/2) = (2^(k+1) - 1)(q+1)/2. In other words, point (m, s) is on the line s(m) = (2^(k+1) - 1)/2^(k+1) * m + (2^(k+1) - 1)/2.

For every odd number m in this sequence, the first part of SRS(m) has size (m+1)/2.

Let u = 2^k * Product_{i=1..PrimePi(2^(k+1)} p_i, where p_i is the i-th prime, and let v be the number of elements in this sequence that are in the set V = {m = 2^k * q | 1 < m <= u } then T(j + t*v, k) = T(j, k) + t*u, 1 <= j and 1 <= t, holds for the elements in column k.

Let the nonincreasing multiset cL = { c_1, ... , c_s } be a factorization of n, let dL = { d_1, ... , d_s } be any set of s distinct odd primes, let q = dL^(cL - 1) = d_1^(c_1 - 1) * ... * d_s^(c_s - 1), and let k satisfy 2^k < q < 2^(k+1). Then SRS(2^k * q) is unimodal of maximum height n, 2^k * q has 2n odd divisors and its width pattern has 2n-1 entries. The smallest possible choice for 2^k * q is with the increasing sequence of odd primes d_i = p_(i+1), 1 <= i <= s. The overall smallest 2^k * q is the minimum among all factorizations of n. The smallest number m for which SRS(m) has two unimodal parts of maximum width n requires the additional prime factor r > 2^(k+1) * q which yields m = 2^k * q * r.

This sequence is column 2 in the array of A367377 and a(2) = A370206(1).

Maximum width 3 can occur an odd number of times in the width pattern of SRS(m) only for numbers m in this sequence for which SRS(m) has an odd number of parts. In that case width 3 must occur at the diagonal of SRS(m). However, the center part of SRS(m) need not be unimodal.

When m is odd and SRS(m) has maximum width 3 then SRS(m) has at least 3 parts because the first and last parts of SRS(m) consist of a single leg of width 1. Therefore, the first two rows of the table contain only even numbers. The numbers in the third row appear to be odd and divisible by 15.

Since for numbers m^2 in the sequence the width at the diagonal of SRS(m^2) is 1, the area m of its center part is odd so that this sequence is a proper subsequence of A016754 and since SRS(m^2) has an odd number of parts it is a proper subsequence of A319529. The smallest odd square not in this sequence is 225 = 15^2. SRS(225) is {113, 177, 113}, its center part has maximum width 2, its width at the diagonal is 1.

The k+1 parts of SRS(p^(2k)), p an odd prime and k >= 0, through the diagonal including the center part have areas (p^(2k-i) + p^i)/2 for 0 <= i <= k. They form a strictly decreasing sequence. Since p^(2k) has 2k+1 divisors and SRS(p^(2k)) has 2k+1 parts, all of width 1 (A357581), the even powers of odd primes form a proper subsequence of A244579. For the subsequence of squares of odd primes p, SRS(p^2) consists of the 3 parts { (p^2 + 1)/2, p, (p^2 + 1)/2 } see A001248, A247687 and A357581.

The areas of the parts of SRS(m^2) need not be in descending order through the diagonal as a(112) = 275^2 = 75625 with SRS(75625) = (37813, 7565, 3443, 1525, 715, 738, 275, 738, 715, 1525, 3443, 7565, 37813) demonstrates.

An equivalent description of the sequence is: The center part of SRS(m^2) has width 1, m is odd, and A249223(m^2, m-1) = 0.

Conjectures (true for all a(n) <= 10^8):

(1) The central part of SRS(a(n)) is the minimum of all parts of SRS(a(n)), 1 <= n.

(2) The terms in this sequence are the squares of the terms in A244579.

All terms m in this sequence for which SRS(m) consists of 1 or 2 parts are even.

Let m = 2^k * q, k >= 0 and q > 2 odd, be a number in this sequence. Let c be the number of divisors r <= A003056(m) of q for which there is at most one pair of divisors s and t of q satisfying r < s < t <= min( 2^(k+1) * r, A003056(m)). Call such triples (r, s, t) good triples. Then at least one good triple exists for number m.

Let w be the number of times that width 3 occurs in the width pattern of m (row m in the triangle of A341969). Then c = (w + 1)/2 when the width at the diagonal is equal to 3 and c = w/2 otherwise.

Sequence a(n) is the subsequence of A375611 for which the symmetric representation of sigma(a(n)) has at least two parts. The width at the diagonal can be any of the 3 widths.

Let m = 2^k * q, k >= 0 and q odd, be a number in this sequence. Let c be the number of divisors s <= A003056(m) of q for which there is at most one divisor t of q satisfying s < t <= min( 2^(k+1) * s, A003056(m). Let w be the number of times width 2 occurs in the width pattern of m (row m in the triangle of A341960). Then c = (w + 1)/2 when the width at the diagonal is equal to 2 and c = w/2 otherwise.

Number m = 2^k * q, k >= 0 and q odd, is in this sequence precisely when for any divisor s <= A003056(m) of q there is at most one divisor t of q satisfying s < t <= min(2^(k+1) * s, A003056(m)), and at least one such pair s < t of successive odd divisors exists. Equivalently, row m of the triangle in A249223 contains at least one 2, but no number larger than 2.

It appears that all repeated terms in A351903 occur in pairs only, and either are odd or multiples of 10; true through a(4543) = 999999.

It is not known whether this sequence is infinite. A351903 is infinite, but not increasing.