A183110 Period-length of the ultimate periodic behavior of the orbit of a list [1,1,1,...,1] of n 1's under the mapping defined in the comments.

1, 2, 1, 3, 3, 1, 4, 4, 4, 1, 5, 5, 5, 5, 1, 6, 6, 6, 6, 6, 1, 7, 7, 7, 7, 7, 7, 1, 8, 8, 8, 8, 8, 8, 8, 1, 9, 9, 9, 9, 9, 9, 9, 9, 1, 10, 10, 10, 10, 10, 10, 10, 10, 10, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 1, 13, 13

Offset: 1

Author

John W. Layman, Feb 01 2011

Keywords

• nonn
• easy

Comments

We use the list mapping introduced in A092964, whereby one removes the first term of the list, z(1), and adds 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list.

This is also conjectured to be the length of the longest cycle of pebble-moves among the partitions of n (cf. A201144). - Andrew V. Sutherland

Examples

			Under the indicated mapping the list [1,1,1,1,1,1,1] of seven 1's results in the orbit [1,1,1,1,1,1,1], [2,1,1,1,1,1], [2,2,1,1,1], [3,2,1,1], [3,2,2], [3,3,1], [4,2,1], [3,2,1,1], ... which is clearly periodic with period-length 4, so a(7) = 4.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A037306, A092964, A178572, A178574.

Programs

Mathematica

f[p_] := Module[{pp, x, lpp, m, i}, pp = p; x = pp[[1]]; pp = Delete[pp,1]; lpp = Length[pp]; m = Min[x, lpp]; For[i = 1, i ≤ m, i++, pp[[i]]++]; For[i = 1, i ≤ x - lpp, i++, AppendTo[pp, 1]]; pp]; orb[p_] := Module[{s, v}, v = p; s = {v}; While[! MemberQ[s, v = f[v]], AppendTo[s, v]]; s]; attractor[p_] := Module[{orbp, pos, len, per}, orbp = orb[p]; pos = Flatten[Position[orbp, f[orbp[[-1]]]]][[1]] - 1; (*pos = steps to enter period*) len = Length[orbp] - pos; per = Take[orbp, -len]; Sort[per]]; a = {}; For[n = 1, n ≤ 80, n++, {rn = Table[1, {k, 1, n}]; orbn = orb[rn]; lenorb = Length[orbn]; lenattr = Length[attractor[rn]]; AppendTo[a, lenattr]}]; Print[a];

Formula

It appears, but has not yet been proved, that a(n)=1 if n=t(k) and a(n)=k if t(k-1) < n < t(k) where t(k) is the k-th triangular number t(k) = k*(k+1)/2.