A101986 Maximum sum of products of successive pairs in a permutation of order n+1.

0, 2, 9, 23, 46, 80, 127, 189, 268, 366, 485, 627, 794, 988, 1211, 1465, 1752, 2074, 2433, 2831, 3270, 3752, 4279, 4853, 5476, 6150, 6877, 7659, 8498, 9396, 10355, 11377, 12464, 13618, 14841, 16135, 17502, 18944, 20463, 22061, 23740, 25502

Offset: 0

Author

Eugene McDonnell (eemcd(AT)mac.com), Jan 29 2005

Keywords

• nonn
• easy

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

Examples

			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.
		

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).

Crossrefs

Pairwise sums of A005581.

Cf. A000290, A000330, A008865, A026035, A160805.

Programs

Haskell

a101986 n = sum $ zipWith (*) [1,3..] (reverse [2..n+1])
-- Reinhard Zumkeller, Mar 30 2012

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.

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 *)

PARI

a(n)=n*(2*n^2+9*n+1)/6 \\ Charles R Greathouse IV, Jan 17 2012

Formula

a(n) = n*(2*n^2 + 9*n + 1)/6.

a(n+1) = a(n) + A008865(n+2); a(n) = A160805(n) - 4. [Reinhard Zumkeller, May 26 2009]

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) = A000330(n) + A005449(n) - A000217(n). - Richard R. Forberg, Aug 07 2013

a(n) = 1 + sum( A008865(i), i=1..n+1 ). [Bruno Berselli, Jan 13 2015]

a(n) = A000290(n) + A000330(n). - J. M. Bergot, Apr 26 2018

Extensions

Edited by Bruno Berselli, Jan 13 2015

Name edited by Alois P. Heinz, Feb 02 2019