A186053 Smallest perimeter among all sets of nonnegative integers whose volume (sum) is n.

0, 1, 2, 2, 4, 5, 3, 6, 7, 6, 4, 8, 8, 9, 7, 5, 10, 11, 9, 10, 8, 6, 11, 12, 13, 10, 11, 9, 7, 14, 12, 13, 14, 11, 12, 10, 8, 16, 16, 13, 14, 15, 12, 13, 11, 9, 16, 17, 17, 14, 15, 16, 13, 14, 12, 10, 16, 17, 18, 18, 15, 16, 17, 14, 15, 13, 11, 22, 17, 18, 19, 19, 16, 17

Offset: 0

Author

John W. Layman, Feb 11 2011

Keywords

• nonn

Comments

The volume and perimeter of a set S of nonnegative integers are introduced in the reference. The volume is defined simply as the sum of the elements of S, and the perimeter is defined as the sum of the elements of S whose predecessor and successor are not both in S.

For all partitions into distinct parts (with first part 0 implied), the perimeter of a set is the sum of parts p such that not both of p-1 and p+1 are in the partition. The partition with smallest perimeter is sometimes, but not often unique. For example, there are three partitions of volume 37 achieving the minimal perimeter 16, namely [ 0 1 2 3 4 5 6 7 x 9 ], [ x 2 x 5 6 7 8 9 ], and [0 x 2 x 5 6 7 8 9 ] (where x is for one or more skipped parts), but there is only one partition of 36 with minimal perimeter 8, namely [ 0 1 2 3 4 5 6 7 8 ]. [Joerg Arndt, Jun 03 2013]

Examples

			For n=8, the set S={0,1,2,5} has sum 8 and the perimeter 7 (the sum of 2 and 5).  No other set of volume 8 has a smaller perimeter, so a(8)=7.
		

Links

• Joerg Arndt and Alois P. Heinz, Table of n, a(n) for n = 0..2000 (terms n = 0..201 from Joerg Arndt)
• Patrick Devlin, Sets with High Volume and Low Perimeter, arXiv:1107.2954 [math.CO], 2011.
• Patrick Devlin, Integer Subsets with High Volume and Low Perimeter, arXiv:1202.1331 [math.CO], 2012.
• Patrick Devlin, Integer Subsets with High Volume and Low Perimeter, INTEGERS, Vol. 12, #A32.
• J. Miller, F. Morgan, E. Newkirk, L. Pedersen and D. Seferis, Isoperimetric Sets of Integers, Math. Mag. 84 (2011) 37-42.

Crossrefs

Cf. A227538.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n=0 and i<>0, `if`(t>1, i+1, 0),
      `if`(i<0, infinity, min(`if`(t>1, i+1, 0)+b(n, i-1, iquo(t, 2)),
      `if`(i>n, NULL, `if`(t=2, i+1, 0)+b(n-i, i-1, iquo(t, 2)+2)))))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 23 2013

Mathematica

notBoth[lst_List] := Select[lst, !MemberQ[lst, #-1] || !MemberQ[lst, #+1]&]; Table[s=Select[IntegerPartitions[n], Length[#]==Length[Union[#]]&]; s=Append[#,0]&/@s; Min[Table[Plus@@notBoth[i], {i,s}]], {n,40}]
(* Second program: *)
b[n_, i_, t_] := b[n, i, t] = If[n == 0 && i != 0, If[t > 1, i+1, 0], If[ i<0, Infinity, Min[If[t>1, i+1, 0] + b[n, i-1, Quotient[t, 2]], If[i > n, Infinity, If[t == 2, i+1, 0] + b[n-i, i-1, Quotient[t, 2]+2]]]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Aug 29 2016, after Alois P. Heinz *)

Formula

If t(n) is a triangular number t(n)=n*(n+1)/2, then a(t(n))=n.

a(n) = A002024(n) + A182298(A025581(n)) unless n is one of the 114 known counterexamples listed in A182246. This allows the n-th term to be calculated in order log(log(n)) time using constant memory. The first n terms can be calculated in order n time (using order n memory). [Result and details in above listed paper by Patrick Devlin (2012).]