cp's OEIS Frontend

This sequence is the companion to A342595 in that a(n) is the smallest number k that has row n of the table in A342595 as its width pattern in the symmetric representation of sigma(k).

The number of possible width patterns of length n occurring up to the diagonal in symmetric representations of sigma is A001405(n). Those are realized for n <= 4. For larger n the actual number of width patterns is smaller. Only p symmetric patterns of length 2p-1 are realizable when a number has p odd divisors and p is prime. Patterns such as 1 0 1 2 3 ... k-1 k k-1 ... 3 2 1 0 1, k >= 4, i.e., numbers with at least 6 odd divisors, cannot be realized as width patterns in the symmetric representation of sigma. If n = 2^s * p * q^2, s >= 0, p < q odd primes, then 2^(s+1) < p and row(n) < 2^(s+1) * p must hold which leads to the contradiction q^2 < p^2; if n = 2^s * p^2 * q, s >= 0, p < q odd primes, then again 2^(s+1) < p and row(n) < 2^(s+1) * p must hold which leads to the contradiction p * q < p^2.

The numbers T(n, 1) instantiating a single unimodal pattern of width n form A250071(n). This first column is not increasing since T(5, 1) = 5184 > 1440 = T(6, 1).

The numbers T(1, k) instantiating the repeating unimodal patterns 1, 1, 0, 1, ..., 1, 0, 1, 0, ..., 0, 1, 0, 1, ... of width 1 form A318843(k). This first row is not increasing since T(1, 11) = 59049 > 29095 = T(1, 12).

The rows in the table are infinite since the numbers T(n, 1) * p^(k-1) >= T(n, k), with p the smallest prime greater than 2 * T(n, 1), instantiate the width pattern for T(n, k), though equality need not hold, as T(1, 4) = 21 = 3 * 7 < 1 * 3^3 = 27 demonstrates.

Conjecture 1: None of the rows and columns are increasing.

Conjecture 2: T(n, p) = T(n, 1) * A151800(2*T(n, 1))^(p-1) for n >= 1 and primes p.

Conjecture 3: T(p, q), p and q primes, is a record for its upper left hand rectangle in the table. Only one prime number index generally is not sufficient as the inequality 4157280 = T(6, 2) < 5 * 10^6 < T(5, 2) shows.

Those numbers in this sequence with only parts of width 1 in their symmetric representation of sigma form column 5 in the table of A357581. - Hartmut F. W. Hoft, Oct 04 2022

This sequence is a permutation of A174905. Numbers in the even numbered columns of the table form A241008 and those in the odd numbered columns form A241010. The first row of the table is A318843.

This sequence is a subsequence of A240062 and each column in this sequence is a subsequence in the respective column of A240062.

Each term has 4 odd divisors and has the form 2^k * p * q, k > 0, p and q prime, 2 < p < 2^(k+1) < 2^(k+1) * p < q. The inequalities ensure that the four 1's in row a(n) of triangle in A237048 are in positions 1, p, 2^(k+1), and 2^(k+1) * p <= floor( (sqrt(8*a(n)+1) - 1)/2 ) < q and establish width pattern 1210 in SRS(a(n)) up to the diagonal. Also since p < 2^(k+1), numbers of the form 2^k * p^3 force p^2 < 2^(k+1) * p which creates a width pattern of the form 1212121.

When a(n) satisfies q = 2^(k+1) * p + 1 it is the smallest number with prime factor p whose two parts of SRS(a(n)) meet at the diagonal since in this case 2^(k+1) * p = floor( (sqrt(8*a(n)+1) - 1)/2 ). The first 4 numbers with p = 3 are 2* 3 * 13 = 78, 2^4 * 3 * 97 = 4656, 2^5 * 3 * 193 = 18528 and 2^7 * 3 * 769 = 295296. The smallest number with prime factor p = 47 has 355 digits.

Conjecture: The subsequence of numbers m whose two parts of SRS(m) meet at the diagonal is infinite.

Those numbers in this sequence with only parts of width 1 in their symmetric representation of sigma form column 6 in the table of A357581. - Hartmut F. W. Hoft, Oct 04 2022

The sequence is not increasing with the maximum width of the symmetric representation just like A347979.

Observation: a(2)..a(14) ending in 5. - Omar E. Pol, Sep 23 2021

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.

For the 30 known terms the symmetric representation of sigma consists of a single part, i.e., this is a subsequence of A174973 = A238443.

The sequence is not increasing with the maximum width of the symmetric representation of sigma.

Also a(33) = 2162160 is the only further number in the sequence less than 2500000.

The first row of the table below is A318843 and the first column is A250070.

T(1,k+1) <= 3^k, for all k>=0, since for k=2j the (j+1)-st part in the symmetric representation of sigma(3^k) extends across the diagonal, and for k=2j+1 the (j+1)-st part is completed before the diagonal.

The data computed so far for a partially filled table of 15 rows and 15 columns, show that all rows, all columns (except column 4 for n <= 6 *10^7), and the diagonal are nonmonotonic.