OFFSET
0,2
COMMENTS
1 3 5 4 2 is the 11th permutation, in lexical order. of order 5. Its reverse 2 4 5 3 1 is the 41st. The earliest permutation of order 6 is the 41st, 1 3 5 6 4 2. This pattern continues as far as I have looked, so its reversal 2 4 6 5 3 1 is the 191st and the earliest permutation of order 7 is the 191st, et cetera.
Comments from Dmitry Kamenetsky, Dec 15 2006: (Start)
This sequence is related to A026035, except here we take the maximum sum of products of successive pairs. Here is a method for generating such permutations. Start with two lists, the first has numbers 1 to n, while the second is empty.
Repeat the following operations until the first list is empty: 1. Move the smallest number of the first list to the leftmost available position in the second list. The move operation removes the original number from the first list. 2. Move the smallest number of the first list to the rightmost available position in the second list. For example when n=8, the permutation is 1, 3, 5, 7, 8, 6, 4, 2. (End)
Convolution of odd numbers and integers greater than 1. - Reinhard Zumkeller, Mar 30 2012
For n>0, a(n) is row 2 of the convolution array A213751. - Clark Kimberling, Jun 20 2012
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Leroy Quet, Max/Min Sums From Permutations - a message on Jan 28 2005 to the sequence list, asking for someone to provide more values and to submit these to OEIS.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
a(n) = n*(2*n^2 + 9*n + 1)/6.
G.f.: x*(1+x)*(2-x)/(1-x)^4. - L. Edson Jeffery, Jan 17 2012
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3, a(0)=0, a(1)=2, a(2)=9, a(3)=23. - L. Edson Jeffery, Jan 17 2012
a(n) = 1 + sum( A008865(i), i=1..n+1 ). [Bruno Berselli, Jan 13 2015]
EXAMPLE
The permutations of order 5 with maximum sum of products is 1 3 5 4 2 and its reverse, since (1*3)+(3*5)+(5*4)+(4*2) is 46. All others are empirically less than 46. So a(4) = 46.
MAPLE
a:=n->add((n+j^2), j=1..n): seq(a(n), n=0..41); # Zerinvary Lajos, Jul 27 2006
MATHEMATICA
Table[(n + 9 n^2 + 2 n^3)/6, {n, 0, 41}] (* Robert G. Wilson v, Feb 04 2005 *)
PROG
(J) 0 1 9 2 & p. % 6 & p. (A) NB. the polynomial P such that P(n) is a(n).
NB. where 0 1 9 2 are the coefficients in ascending order of the numerator of a rational polynomial and 6 is the (constant) coefficient of its denominator. J's primitive function p. produces a polynomial with these coefficients. Division is indicated by % . Thus the J expression (A) is equivalent to the formula above.
(PARI) a(n)=n*(2*n^2+9*n+1)/6 \\ Charles R Greathouse IV, Jan 17 2012
(Haskell)
a101986 n = sum $ zipWith (*) [1, 3..] (reverse [2..n+1])
-- Reinhard Zumkeller, Mar 30 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Eugene McDonnell (eemcd(AT)mac.com), Jan 29 2005
EXTENSIONS
Edited by Bruno Berselli, Jan 13 2015
Name edited by Alois P. Heinz, Feb 02 2019
STATUS
approved