A056155 Positive integer k, 1 <= k <= n, which maximizes k^(n+1-k).
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 20, 20, 20
Offset: 1
Examples
a(5) = 3 because 3^(5+1-3) = 27 is larger than k^(5+1-k) for any other k (1 <= k <= n) besides k = 3.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A003320.
Programs
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Mathematica
nn = 79; Monitor[a = Table[RankedMax[Table[k^(n + 1 - k), {k, 1, n}], 1], {n, 1, nn}];, n] Monitor[b = Flatten[Table[Position[Table[k^(n + 1 - k), {k, 1, n}], a[[n]]], {n, 1, nn}]], n] (* Mats Granvik, Jan 26 2017 *) a[n_] := MaximalBy[Range[n], #^(n + 1 - #)&][[1]]; Array[a, 100] (* Jean-François Alcover, Dec 11 2020 *)
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PARI
a(n) = my(v = vector(n, k, k^(n+1-k))); vecsort(v,,1)[#v]; \\ Michel Marcus, Jan 28 2017
Formula
a(n) ~ e^(LambertW(e*(n + 1)) - 1). - Mats Granvik, Jan 26 2017
Comments