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A080663 Numbers of the form 3*n^2 - 1. 15

%I

%S 2,11,26,47,74,107,146,191,242,299,362,431,506,587,674,767,866,971,

%T 1082,1199,1322,1451,1586,1727,1874,2027,2186,2351,2522,2699,2882,

%U 3071,3266,3467,3674,3887,4106,4331,4562,4799,5042,5291,5546,5807,6074,6347,6626

%N Numbers of the form 3*n^2 - 1.

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

%C 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

%C a(n) = unsigned real term in (1 + ni)^3. E.g. (1 + 4i)^3 = (-47 - 52i); where 52 = A121670(4). Note that (4 + i)^3 = (52 + 47i) = (A121670(4) + A080663(4)i). - _Gary W. Adamson_, Aug 14 2006

%C 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

%C 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

%C 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

%C 2*a(n+1) = surface area of a rectangular prism with consecutive integer sides: n, n+1, and n+2, (n>0). - _Wesley Ivan Hurt_, Sep 06 2014

%D Ethan D. Bolker, Elementary Number Theory: An Algebraic Approach. Mineola, New York: Dover Publications (1969, reprinted 2007): p. 7, Problem 6.6.

%D E. Grosswald, Topics from the Theory of Numbers, 1966 p 64 problem 11

%H Nathaniel Johnston, <a href="/A080663/b080663.txt">Table of n, a(n) for n = 1..10000</a>

%H Cino Hilliard, <a href="http://web.archive.org/web/20080411092534/http://groups.msn.com/BC2LCC/3n21isnotsquare.msnw">3n^2-1 not square</a>. [Archived copy as of Apr 11 2008 from web.archive.org]

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SymmetricPolynomial.html">Symmetric Polynomial</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 3*n^2 - 1. - _Stephen Crowley_, Jul 06 2009

%F a(n) = a(n-1) - 3*a(n-2) + 3*a(n-3). - _Johannes W. Meijer_, Oct 16 2009

%F G.f.: x*(2+5*x-x^2)/(1-x)^3. - _Joerg Arndt_, Sep 06 2014

%F a(n) = 6*n + a(n-1) - 3 for n > 1. - _Vincenzo Librandi_, Aug 08 2010

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

%t Table[Inner[Times, {n, n + 1, n + 2}, {n + 2, n + 3, n + 4}, Plus], {n, 47}] (* _Vladimir Joseph Stephan Orlovsky_, Feb 08 2010 *)

%t 3Range[47]^2 - 1 (* _Alonso del Arte_, Oct 19 2013 *)

%o (PARI) nosquare(n) = { for(x=1,n, y = 3*x*x-1; print1(y" ") ) } checkit(n) = { for(x=1,n, y = 3*x*x-1; if(!issquare(y),print1(y" ")) ) }

%o (PARI) Vec(x*(2+5*x-x^2)/(1-x)^3+O(x^66)) \\ _Joerg Arndt_, Sep 06 2014

%o (MAGMA) [3*n^2-1 : n in [1..50]]; // _Wesley Ivan Hurt_, Sep 04 2014

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

%K nonn,easy

%O 1,1

%A _Cino Hilliard_, Mar 01 2003

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Last modified January 21 21:30 EST 2020. Contains 331128 sequences. (Running on oeis4.)