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A367377
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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.
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7
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1, 6, 3, 72, 78, 9, 120, 10728, 1014, 21, 5184, 28920, 1598472, 12246, 81, 1440, 53752896, 6969720, 230297976, 171366, 147, 373248, 4157280
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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, #<b&&widthPattern[#]!=umw[n, k]&] (* computation of T(n, k) stops at bound b *)
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CROSSREFS
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Cf. A235791, A237048, A237270, A237591, A237593, A249223, A250071, A262045, A318843, A341969, A342592, A342594, A342595, A342596.
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KEYWORD
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AUTHOR
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STATUS
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approved
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