A160805 -id:A160805 - cp's OEIS frontend

A008865 a(n) = n^2 - 2.

-1, 2, 7, 14, 23, 34, 47, 62, 79, 98, 119, 142, 167, 194, 223, 254, 287, 322, 359, 398, 439, 482, 527, 574, 623, 674, 727, 782, 839, 898, 959, 1022, 1087, 1154, 1223, 1294, 1367, 1442, 1519, 1598, 1679, 1762, 1847, 1934, 2023, 2114, 2207, 2302, 2399, 2498

Offset: 1

N. J. A. Sloane, R. K. Guy

For n >= 2, least m >= 1 such that f(m, n) = 0 where f(m,n) = Sum_{i=0..m} Sum_{k= 0..i} (-1)^k*(floor(i/n^k) - n*floor(i/n^(k+1))). - Benoit Cloitre, May 02 2004
For n >= 3, the a(n)-th row of Pascal's triangle always contains a triple forming an arithmetic progression. - Lekraj Beedassy, Jun 03 2004
Let C = 1 + sqrt(2) = 2.414213...; and 1/C = 0.414213... Then a(n) = (n + 1 + 1/C) * (n + 1 - C). Example: a(6) = 34 = (7 + 0.414...) * (7 - 2.414...). - Gary W. Adamson, Jul 29 2009
The sequence (n-4)^2-2, n = 7, 8, ... enumerates the number of non-isomorphic sequences of length n, with entries from {1, 2, 3} and no two adjacent entries the same, that minimally contain each of the thirteen rankings of three players (111, 121, 112, 211, 122, 212, 221, 123, 132, 213, 231, 312, 321) as embedded order isomorphic subsequences. By "minimally", we mean that the n-th symbol is necessary for complete inclusion of all thirteen words. See the arXiv paper below for proof. If n = 7, these sequences are 1213121, 1213212, 1231213, 1231231, 1231321, 1232123, and 1232132, and for each case, there are 3! = 6 isomorphs. - Anant Godbole, Feb 20 2013
a(n), n >= 0, with a(0) = -2, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 8 for b = 2*n. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
With a different offset, this is 2*n^2 - (n + 1)^2, which arises in one explanation of why Bertrand's postulate does not automatically prove Legendre's conjecture: as n gets larger, so does the range of numbers that can have primes that satisfy Bertrand's postulate yet do nothing for Legendre's conjecture. - Alonso del Arte, Nov 06 2013
x*(x + r*y)^2 + y*(y + r*x)^2 can be written as (x + y)*(x^2 + s*x*y + y^2). For r >= 0, the sequence gives the values of s: in fact, s = (r + 1)^2 - 2. - Bruno Berselli, Feb 20 2019
For n >= 2, the continued fraction expansion of sqrt(a(n)) is [n-1; {1, n-2, 1, 2n-2}]. For n=2, this collapses to [1; {2}]. - Magus K. Chu, Sep 06 2022

			G.f. = -x + 2*x^2 + 7*x^3 + 14*x^4 + 23*x^5 + 34*x^6 + 47*x^7 + 62*x^8 + 79*x^9 + ...

T. D. Noe, Table of n, a(n) for n = 1..1000
Anant Godbole and Martha Liendo, Waiting time distribution for the emergence of superpatterns, arxiv 1302.4668 [math.PR], 2013.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A145067 (Zero followed by partial sums of A008865).
Cf. A000027, A013648.
Cf. A028871 (primes).
Cf. A263766 (partial products).
Cf. A270109. [Bruno Berselli, Mar 17 2016]

a008865 = (subtract 2) . (^ 2) :: Integral t => t -> t
a008865_list = scanl (+) (-1) [3, 5 ..]
-- Reinhard Zumkeller, May 06 2013

[n^2 - 2: n in [1..60]]; // Vincenzo Librandi, May 01 2014

Range[50]^2 - 2 (* Harvey P. Dale, Mar 14 2011 *)

{for(n=1, 47, print1(n^2-2, ","))} \\ Klaus Brockhaus, Oct 17 2008

For n > 1: a(n) = A143053(A000290(n)), A143054(a(n)) = A000290(n). - Reinhard Zumkeller, Jul 20 2008
G.f.: (x-5*x^2+2*x^3)/(-1+3*x-3*x^2+x^3). - Klaus Brockhaus, Oct 17 2008
E.g.f.: (x^2 + x -2)*exp(x) + 2. - G. C. Greubel, Aug 19 2017
a(n+1) = A101986(n) - A101986(n-1) = A160805(n) - A160805(n-1). - Reinhard Zumkeller, May 26 2009
For n > 1, a(n) = floor(n^5/(n^3 + n + 1)). - Gary Detlefs, Feb 10 2010
a(n) = a(n-1) + 2*n - 1 for n > 1, a(1) = -1. - Vincenzo Librandi, Nov 18 2010
Right edge of the triangle in A195437: a(n) = A195437(n-2, n-2). - Reinhard Zumkeller, Nov 23 2011
a(n)*a(n-1) + 2 = (a(n) - n)^2 = A028552(n-2)^2. - Bruno Berselli, Dec 07 2011
a(n+1) = A000096(n) + A000096(n-1) for all n in Z. - Michael Somos, Nov 11 2015
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=1} 1/a(n) = (1 - sqrt(2)*Pi*cot(sqrt(2)*Pi))/4.
Sum_{n>=1} (-1)^n/a(n) = (1 - sqrt(2)*Pi*cosec(sqrt(2)*Pi))/4. (End)
Assume offset 0. Then a(n) = 2*LaguerreL(2, 1 - n). - Peter Luschny, May 09 2021
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = sqrt(2/3)*sin(sqrt(3)*Pi)/sin(sqrt(2)*Pi).
Product_{n>=2} (1 + 1/a(n)) = -Pi/(sqrt(2)*sin(sqrt(2)*Pi)). (End)

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

Original entry on oeis.org

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

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

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

			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.

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).

Pairwise sums of A005581.

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

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

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.

a:=n->add((n+j^2),j=1..n): seq(a(n),n=0..41); # Zerinvary Lajos, Jul 27 2006

Table[(n + 9 n^2 + 2 n^3)/6, {n, 0, 41}] (* Robert G. Wilson v, Feb 04 2005 *)

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

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

Edited by Bruno Berselli, Jan 13 2015

Name edited by Alois P. Heinz, Feb 02 2019

cp's OEIS Frontend

A008865 a(n) = n^2 - 2.

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