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A209082
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Least power separator of the partitions of n.
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1
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1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8
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OFFSET
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1,2
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COMMENTS
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The least power separator of the partitions of n is introduced here as the least positive integer m for which the sums x(1)^m + x(2)^m + ... + x(k)^m, as {x(1),x(2),...,x(k)} ranges through the partitions of n (as in A000041), are distinct.
In the following table, d(n,h)=[number of partitions x(1),x(2),...,x(k) of n]-[number of distinct sums x(1)^m + x(2)^m + ... + x(k)^m], so that a(n) is the least h for which d(n,h)=0.
n.....d(n,1)..d(n,2)..d(n,3)..d(n,4)..d(n,5)..d(n,6)
1.....0.......0.......0.......0.......0.......0
2.....1.......0.......0.......0.......0.......0
3.....2.......0.......0.......0.......0.......0
4.....4.......0.......0.......0.......0.......0
5.....6.......0.......0.......0.......0.......0
6.....10......2.......0.......0.......0.......0
7.....14......2.......0.......0.......0.......0
8.....21......4.......2.......0.......0.......0
9.....29......9.......3.......0.......0.......0
10....41......15......6.......0.......0.......0
11....55......24......1.......0.......0.......0
12....76......38......16......0.......0.......0
13....100.....55......24......1.......0.......0
14....134.....81......39......1.......0.......0
15....175.....115.....61......2.......0.......0
16....230.....159.....91......3.......4.......0
17....296.....214.....130.....5.......7.......0
18....384.....293.....186.....7.......12......0
19....489.....384.....254.....12......20......0
20....626.....509.....349.....16......33......0...1...0
21....791.....662.....467.....27......48......0...1...0
22....1001....857.....625.....40......79......0...2...0
For 0<n<20, it appears that for d(n,q)=0 for all q>m but not for 19<n<32. Only the first sixteen row sequences (excluding column 1) are monotonic.
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LINKS
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EXAMPLE
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The partitions of 6: {6},{5,1},{4,2},{4,1,1},{3,3},{3,2,1},{3,1,1,1},{2,2,2},{2,2,1,1},{2,1,1,1,1},{1,1,1,1,1,1}. The 11 sums x(1)+...x(k) all = 6. The 11 sums x(1)^2+...x(k)^2 are, respectively, 36,26,20,18,18,14,12,12,10,8,6, which are not distinct. The 11 sums x(1)^3+...x(k)^3 are, respectively, 216,126,72,66,54,36,30,24,18,12,6, and these are distinct, so that a(6)=3.
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MATHEMATICA
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p[n_] := IntegerPartitions[n]
p[n_, k_] := p[n]^k
s[n_, k_] := Map[Plus @@ # &, p[n, k]]
d[n_, m_] := Length[p[n]] - Length[Union[s[n, m]]]
t = Table[d[n, m], {n, 1, 40}, {m, 1, 25}]
a[n_] := First[Position[t[[n]], 0]]
Flatten[Table[a[n], {n, 1, 40}]] (* A209082 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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