A080663 a(n) = 3*n^2 - 1.

2, 11, 26, 47, 74, 107, 146, 191, 242, 299, 362, 431, 506, 587, 674, 767, 866, 971, 1082, 1199, 1322, 1451, 1586, 1727, 1874, 2027, 2186, 2351, 2522, 2699, 2882, 3071, 3266, 3467, 3674, 3887, 4106, 4331, 4562, 4799, 5042, 5291, 5546, 5807, 6074, 6347, 6626

Offset: 1

Author

Cino Hilliard, Mar 01 2003

Keywords

• nonn
• easy

Comments

These numbers cannot be perfect squares. See the Hilliard link for a proof.

2nd elementary symmetric polynomial of n, n + 1 and n + 2: n(n+1) + n(n+2) + (n+1)(n+2). - Zak Seidov, Mar 23 2005

This sequence equals for n >= 2 the third right hand column of triangle A165674. Its recurrence relation leads to Pascal's triangle A007318. Crowley's formula for A080663(n-1) leads to Wiggen's triangle A028421 and the o.g.f. of this sequence, without the first term, leads to Wood's polynomials A126671. See also A165676, A165677, A165678 and A165679. - Johannes W. Meijer, Oct 16 2009

The Diophantine equation x(x+1) + (x+2)(x+3) = (x+y)^2 + (x-y)^2 has solutions x = a(n), y = 3n. - Bruno Berselli, Mar 29 2013

A simpler proof that these numbers can't be perfect squares can easily be constructed using congruences: If the equation x^2 = 3y^2 - 1 has a solution in positive integers, then x^2 = 2 mod 3. Obviously we can't have x = 0 mod 3, and x = 1 mod 3 doesn't work either because then x^2 = 1 mod 3 also. That leaves x = 2 mod 3, but then x^2 = 1 mod 3. - Alonso del Arte, Oct 19 2013

2*a(n+1) is surface area of a rectangular prism with consecutive integer sides: n, n+1, and n+2, (n>0). - Wesley Ivan Hurt, Sep 06 2014

Numbers m such that 3*m+3 is a square. So, these are the numbers m such that the system of equations x=sqrt(m-2yz), y=sqrt(m+1-2xz), z=sqrt(m+2-2xy) admits 3 real positive solutions whose sum is an integer. See the Rechtman link. - Michel Marcus, Jun 06 2020

References

• Ethan D. Bolker, Elementary Number Theory: An Algebraic Approach. Mineola, New York: Dover Publications (1969, reprinted 2007): p. 7, Problem 6.6.
• E. Grosswald, Topics from the Theory of Numbers, 1966 p 64 problem 11

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..10000
• Cino Hilliard, 3n^2 - 1 not square. [Archived copy as of Apr 11 2008 from web.archive.org]
• Ana Rechtman, Juin 2020, 1er défi, Images des Mathématiques, CNRS, 2020 (in French).
• Leo Tavares, Illustration: Conjoined Trapezoids
• Eric Weisstein's World of Mathematics, Symmetric Polynomial.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A007318, A028421, A080663, A121670, A126671, A165674, A165676, A165677, A165678, A165679.

Cf. A005449, A115067.

Programs

Magma

[3*n^2-1 : n in [1..50]]; // Wesley Ivan Hurt, Sep 04 2014

Maple

A080663 := proc(n) return 3*n^2-1: end proc: seq(A080663(n), n=1..50); # Nathaniel Johnston, Oct 16 2013

Mathematica

3*Range[47]^2 - 1 (* Alonso del Arte, Oct 19 2013 *)

PARI

list(n) = { for(x=1,n, y = 3*x*x-1; print1(y, ", ") ) } \\ edited by Michel Marcus, Feb 01 2020

PARI

Vec(x*(2+5*x-x^2)/(1-x)^3+O(x^66)) \\ Joerg Arndt, Sep 06 2014

Formula

a(n) = -Re((1 + n*i)^3) where i=sqrt(-1). - Gary W. Adamson, Aug 14 2006

a(n) = 3*n^2 - 1. - Stephen Crowley, Jul 06 2009

a(n) = a(n-1) - 3*a(n-2) + 3*a(n-3). - Johannes W. Meijer, Oct 16 2009

G.f.: x*(2 + 5*x - x^2)/(1-x)^3. - Joerg Arndt, Sep 06 2014

a(n) = a(n-1) + 6*n - 3 for n > 1. - Vincenzo Librandi, Aug 08 2010

E.g.f.: 1 + exp(x)*(3*x^2 + 3*x - 1). - Stefano Spezia, Feb 01 2020

From Amiram Eldar, Feb 04 2021: (Start)

Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(3))*cot(Pi/sqrt(3)))/2.

Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(3))*csc(Pi/sqrt(3)) - 1)/2.

Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(3))*csc(Pi/sqrt(3)).

Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(3))*sin(sqrt(2/3)*Pi)/sqrt(2). (End)

a(n) = A005449(n) + A115067(n). - Leo Tavares, May 25 2022

a(n) = (n-1)*n + (n-1)*(n+1) + n*(n+1), for n >= 1. See the Zak Seidov comment above. - Wolfdieter Lang, Aug 15 2024