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A320521
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a(n) is the smallest even number k such that the symmetric representation of sigma(k) has n parts.
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3
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2, 10, 50, 230, 1150, 5050, 22310, 106030, 510050, 2065450, 10236350
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OFFSET
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1,1
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COMMENTS
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It appears that a(n) = 2 * q where q is odd and that the symmetric representation of sigma(a(n)/2) has the same number of parts as that for a(n). Number a(12) > 15000000. - Hartmut F. W. Hoft, Sep 22 2021
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LINKS
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EXAMPLE
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a(1) = 2 because the second row of A237593 is [2, 2], and the first row of the same triangle is [1, 1], therefore between both symmetric Dyck paths there is only one part: [3], equaling the sum of the divisors of 2: 1 + 2 = 3. See below:
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a(2) = 10 because the 10th row of A237593 is [6, 2, 1, 1, 1, 1, 2, 6], and the 9th row of the same triangle is [5, 2, 2, 2, 2, 5], therefore between both symmetric Dyck paths there are two parts: [9, 9]. Also there are no even numbers k < 10 whose symmetric representation of sigma(k) has two parts. Note that the sum of these parts is 9 + 9 = 18, equaling the sum of the divisors of 10: 1 + 2 + 5 + 10 = 18. See below:
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a(3) = 50 because the 50th row of A237593 is [26, 9, 4, 3, 3, 1, 2, 1, 1, 1, 1, 2, 1, 3, 3, 4, 9, 26], and the 49th row of the same triangle is [25, 9, 5, 3, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 3, 5, 9, 25], therefore between both symmetric Dyck paths there are three parts: [39, 15, 39]. Also there are no even numbers k < 50 whose symmetric representation of sigma(k) has three parts. Note that the sum of these parts is 39 + 15 + 39 = 93, equaling the sum of the divisors of 50: 1 + 2 + 5 + 10 + 25 + 50 = 93. (The diagram of the symmetric representation of sigma(50) = 93 is too large to include.)
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MATHEMATICA
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a320521[n_, len_] := Module[{list=Table[0, len], i, v}, For[i=2, i<=n, i+=2, v=Count[a341969[i], 0]+1; If[list[[v]]==0, list[[v]]=i]]; list]
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CROSSREFS
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Cf. A000203, A018262, A005843, A196020, A235791, A236104, A237048, A237591, A237593, A239663, A239665, A240062, A245092, A262626, A296508.
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KEYWORD
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nonn,more,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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