A095698 Number of permutations of {1,2,3,...,n} where, for 1 < i <= n, the i-th number has maximized sum of the i-1 absolute differences from all previous numbers of the permutation.

1, 2, 4, 6, 14, 18, 46, 54, 146, 162, 454, 486, 1394, 1458, 4246, 4374, 12866, 13122, 38854, 39366, 117074, 118098, 352246, 354294, 1058786, 1062882, 3180454, 3188646, 9549554, 9565938, 28665046, 28697814, 86027906, 86093442, 258149254

Offset: 1

Author

Rick L. Shepherd, Jul 06 2004

Keywords

• nonn
• easy

Comments

Another variant of A095236: Here each phone after the first selected (which can still be any) is chosen such that the total distance in the normal sense from the chosen phone to all previously-chosen phones in the row is maximized. (Equivalently, the average distance is maximized.) Another space- or privacy-conscious selection strategy. Are there any applications of this sequence to phyllotaxy? Gregarious (or eavesdropping) strategy: If, instead, the total (average) distance is minimized, the sequence generated is 1,2,4,8,16,32,64,128,256,512,..., apparently the nonnegative powers of 2.

In the gregarious case (suggested by the above comment), the permutations that result are exactly those that avoid the permutation patterns 132 and 312. See link to Art of Problem Solving Forums for proof of formula below. - Joel B. Lewis, May 16 2009

Taking every other term gives A008776 (even-indexed terms) and A027649 (odd-indexed terms). - Joel B. Lewis, May 16 2009

With Lewis's formulas, the addition of the g.f.s for a(2*n) and a(2*n+1) yields the conjectures below: 2*x/(-3*x^2+1) - (-x^2+1)/(-6*x^4+5*x^2-1) = (-4*x^3-x^2+2*x+1)/(6*x^4-5*x^2+1). - Georg Fischer, Nov 19 2022

Examples

			a(4)=6 as these six permutations of {1,2,3,4} are counted (as in A095236(4)): (1,4,2,3), (1,4,3,2), (2,4,1,3), (3,1,4,2), (4,1,2,3) and (4,1,3,2).
In particular, (2,4,3,1) and (3,1,2,4), counted in A095236(4), are not counted here.
		

Links

• Problem solved on the Art of Problem Solving forum, Urinal-choice permutations. [From _Joel B. Lewis_, May 16 2009, corrected from pers. comm.]
• Index entries for linear recurrences with constant coefficients, signature (0, 5, 0, -6).

Crossrefs

Cf. A008776, A027649, A095236.

Programs

Mathematica

CoefficientList[Series[(-4*x^3-x^2+2*x+1)/(6*x^4-5*x^2+1), {x,0,34}], x] (* Georg Fischer, Nov 19 2022 *)

Formula

a(1) = 1; Conjectured: For k >= 1, a(2k) = a(2k-1) + 2^(k-1) and a(2k+1) = 2*a(2k-1) + a(2k) (needs proof or a reference).

a(2n) = 2 * 3^(n - 1) for n >= 1. a(2n + 1) = 2 * 3^n - 2^n for n >= 0. - Joel B. Lewis, May 16 2009

Conjecture: a(n) = 5*a(n-2) - 6*a(n-4); g.f.: x*(1+2*x-x^2-4*x^3)/((1-2*x^2)*(1-3*x^2)). - Colin Barker, Jul 27 2012

Conjecture: a(n) = 2^(((-1)^n + 2*n - 5)/4)*((-1)^n-1) - 2*3^(((-1)^n + 2*n - 5)/4)*((-1)^n-2). - Luce ETIENNE, Dec 20 2014

Extensions

More terms from Joel B. Lewis, May 16 2009