A367377 Square array T(n, k), n >= 1, k >= 1, read by antidiagonals, of the least numbers whose symmetric representation of sigma instantiate the unimodal width pattern 1, 2, ..., n, ..., 2, 1 repeated k times separated by instances of width 0.

1, 6, 3, 72, 78, 9, 120, 10728, 1014, 21, 5184, 28920, 1598472, 12246, 81, 1440, 53752896, 6969720, 230297976, 171366, 147, 373248, 4157280

Offset: 1

Author

Hartmut F. W. Hoft, Nov 15 2023

Keywords

• nonn
• tabl
• more

Comments

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.

Examples

			The corner of the table begins:
  --------------------------------------------------------------------
     Pattern   |   once    twice  3 times   4 times  5 times  6 times
  --------------------------------------------------------------------
        1      |      1        3        9        21       81      147
       121     |      6       78     1014     12246   171366  1922622
      12321    |     72    10728  1598472 230297976
     1234321   |    120    28920  6969720
    123454321  |   5184 53752896
   12345654321 |   1440  4157280
  1234567654321| 373248
  ...
T(3, 4) must have 12 odd divisors and as least number must have 2^3 * 3^2 as a factor in order to create the initial width pattern 1 2 3 2 1 0. Therefore, since the next smallest prime larger than 16 * 9 is 149, T(3, 4) is 2^3 * 3^2 * 149^3 or 2^3 * 3^2 * 149 * p for suitable prime p which leads to p = 21467 < 22201 = 149^2.
All other numbers in the table were found by exhaustive computations.
		

Crossrefs

Cf. A235791, A237048, A237270, A237591, A237593, A249223, A250071, A262045, A318843, A341969, A342592, A342594, A342595, A342596.

Programs

Mathematica

t249223[n_] := FoldList[#1+(-1)^(#2+1)KroneckerDelta[Mod[n-#2 (#2+1)/2, #2]]&, 1, Range[2, Floor[(Sqrt[8n+1]-1)/2]]] (* row n in triangle of A249223 *)
t262045[n_] := Join[t249223[n], Reverse[t249223[n]]] (* row n in triangle of A262045 *)
widthPattern[n_] := Map[First, Split[t262045[n]]]
umw[n_, k_] := Most[Flatten[Table[Join[Range[n], Range[n-1, 0, -1]], k]]]
a367377[{n_, k_}, b_] := NestWhile[#+1&, 1, #