|
Search: id:A101986
|
|
|
| A101986 |
|
a(n) is the maximum sum of products of successive pairs in a permutation of order n. |
|
+0 3
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
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 (Dmitry.Kamenetsky(AT)rsise.anu.edu.au), 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)
|
|
REFERENCES
|
Message from Leroy Quet on January 28, 2005 6:45:43 PM PST to the sequence list, asking for someone to provide more values and to submit these to OEIS.
|
|
FORMULA
|
a(n) = (n+9n^2+2n^3)/6.
|
|
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.
|
|
MAPLE
|
a:=n->sum((n+j^2), j=1..n): seq(a(n), n=0..41); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 27 2006
|
|
MATHEMATICA
|
Table[(n + 9n^2 + 2n^3)/6, {n, 0, 41}] (from Robert G. Wilson v Feb 04 2005)
|
|
PROGRAM
|
(J language): the polynomial P such that P(n) is a(n) is: 0 1 9 2 & p. % 6 & p. (A)
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.
|
|
CROSSREFS
|
Pairwise sums of A005581.
Cf. A026035.
Sequence in context: A027702 A051897 A032636 this_sequence A023542 A023654 A062445
Adjacent sequences: A101983 A101984 A101985 this_sequence A101987 A101988 A101989
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Eugene McDonnell (eemcd(AT)mac.com), Jan 29 2005
|
|
|
Search completed in 0.002 seconds
|