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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
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
1,2
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
LINKS
Problem solved on the Art of Problem Solving forum, Urinal-choice permutations. [From Joel B. Lewis, May 16 2009, corrected from pers. comm.]
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
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Rick L. Shepherd, Jul 06 2004
EXTENSIONS
More terms from Joel B. Lewis, May 16 2009
STATUS
approved