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

Author

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

Keywords

• sign
• easy

Comments

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

Examples

			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 + ...
		

Links

• 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).

Crossrefs

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]

Programs

Haskell

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

Magma

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

Mathematica

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

PARI

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

Formula

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)