cp's OEIS Frontend

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

%I A101986 #62 Dec 23 2024 14:53:42
%S A101986 0,2,9,23,46,80,127,189,268,366,485,627,794,988,1211,1465,1752,2074,
%T A101986 2433,2831,3270,3752,4279,4853,5476,6150,6877,7659,8498,9396,10355,
%U A101986 11377,12464,13618,14841,16135,17502,18944,20463,22061,23740,25502
%N A101986 Maximum sum of products of successive pairs in a permutation of order n+1.
%C A101986 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.
%C A101986 Comments from _Dmitry Kamenetsky_, Dec 15 2006: (Start)
%C A101986 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.
%C A101986 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)
%C A101986 Convolution of odd numbers and integers greater than 1. - _Reinhard Zumkeller_, Mar 30 2012
%C A101986 For n>0, a(n) is row 2 of the convolution array A213751. - _Clark Kimberling_, Jun 20 2012
%H A101986 Reinhard Zumkeller, <a href="/A101986/b101986.txt">Table of n, a(n) for n = 0..1000</a>
%H A101986 Leroy Quet, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/pipermail/seqfan/2005-January/004759.html">Max/Min Sums From Permutations</a> - a message on Jan 28 2005 to the sequence list, asking for someone to provide more values and to submit these to OEIS.
%H A101986 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A101986 a(n) = n*(2*n^2 + 9*n + 1)/6.
%F A101986 a(n+1) = a(n) + A008865(n+2); a(n) = A160805(n) - 4. [_Reinhard Zumkeller_, May 26 2009]
%F A101986 G.f.: x*(1+x)*(2-x)/(1-x)^4. - _L. Edson Jeffery_, Jan 17 2012
%F A101986 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
%F A101986 a(n) = A000330(n) + A005449(n) - A000217(n). - _Richard R. Forberg_, Aug 07 2013
%F A101986 a(n) = 1 + sum( A008865(i), i=1..n+1 ). [_Bruno Berselli_, Jan 13 2015]
%F A101986 a(n) = A000290(n) + A000330(n). - _J. M. Bergot_, Apr 26 2018
%e A101986 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.
%p A101986 a:=n->add((n+j^2),j=1..n): seq(a(n),n=0..41); # _Zerinvary Lajos_, Jul 27 2006
%t A101986 Table[(n + 9 n^2 + 2 n^3)/6, {n, 0, 41}] (* _Robert G. Wilson v_, Feb 04 2005 *)
%o A101986 (J) 0 1 9 2 & p. % 6 & p. (A) NB. the polynomial P such that P(n) is a(n).
%o A101986 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.
%o A101986 (PARI) a(n)=n*(2*n^2+9*n+1)/6 \\ _Charles R Greathouse IV_, Jan 17 2012
%o A101986 (Haskell)
%o A101986 a101986 n = sum $ zipWith (*) [1,3..] (reverse [2..n+1])
%o A101986 -- _Reinhard Zumkeller_, Mar 30 2012
%Y A101986 Pairwise sums of A005581.
%Y A101986 Cf. A000290, A000330, A008865, A026035, A160805.
%K A101986 nonn,easy
%O A101986 0,2
%A A101986 Eugene McDonnell (eemcd(AT)mac.com), Jan 29 2005
%E A101986 Edited by _Bruno Berselli_, Jan 13 2015
%E A101986 Name edited by _Alois P. Heinz_, Feb 02 2019