A047997 Triangle of numbers a(n,k) = number of balance positions when k equal weights are placed at a k-subset of the points {-n, -(n-1), ..., n-1, n} on a centrally pivoted rod.
1, 1, 2, 1, 3, 5, 1, 4, 8, 12, 1, 5, 13, 24, 32, 1, 6, 18, 43, 73, 94, 1, 7, 25, 69, 141, 227, 289, 1, 8, 32, 104, 252, 480, 734, 910, 1, 9, 41, 150, 414, 920, 1656, 2430, 2934, 1, 10, 50, 207, 649, 1636, 3370, 5744, 8150, 9686, 1, 11, 61, 277, 967
Offset: 1
Examples
From _Gus Wiseman_, Apr 18 2023: (Start) Triangle begins: 1 1 2 1 3 5 1 4 8 12 1 5 13 24 32 1 6 18 43 73 94 1 7 25 69 141 227 289 1 8 32 104 252 480 734 910 1 9 41 150 414 920 1656 2430 2934 Row n = 4 counts the following balanced subsets: {0} {-1,1} {-1,0,1} {-3,0,1,2} {-2,2} {-2,0,2} {-4,0,1,3} {-3,3} {-3,0,3} {-2,-1,0,3} {-4,4} {-3,1,2} {-2,-1,1,2} {-4,0,4} {-3,-1,0,4} {-4,1,3} {-3,-1,1,3} {-2,-1,3} {-3,-2,1,4} {-3,-1,4} {-3,-2,2,3} {-4,-1,1,4} {-4,-1,2,3} {-4,-2,2,4} {-4,-3,3,4} (End)
Links
- R. E. Odeh and E. J. Cockayne, Balancing weights on the integer line, J. Combin. Theory, 7 (1969), 130-135.
Crossrefs
Programs
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Mathematica
a[n_, k_] := Length[ IntegerPartitions[ n*(2k - n + 1)/2, n, Range[2k - n + 1]]]; Flatten[ Table[ a[n, k], {k, 1, 11}, {n, 1, k}]] (* Jean-François Alcover, Jan 02 2012 *) Table[Length[Select[Subsets[Range[-n,n]],Length[#]==k&&Total[#]==0&]],{n,8},{k,n}] (* Gus Wiseman, Apr 16 2023 *)
Formula
Equivalent to number of partitions of n(2k-n+1)/2 into up to n parts each no more than 2k-n+1 so a(n, k)=A067059(n, n(2k-n+1)/2); row sums are A047653(n)-1 = A212352(n). - Henry Bottomley, Aug 11 2001
Comments