%I A132411
%S A132411 0,1,3,8,15,24,35,48,63,80,99,120,143,168,195,224,255,288,323,360,399,
%T A132411 440,483,528,575,624,675,728,783,840,899,960,1023,1088,1155,1224,1295,
%U A132411 1368,1443,1520,1599,1680,1763
%N A132411 Sequence allows us to find X values of the equation: X^3 - (X + 1)^2 
               + X + 2 = Y^2.
%C A132411 To prove that X = 1 or X = n^2 - 1: Y^2 = X^3 - (X + 1)^2 + X + 2 = X^3 
               - X^2 - X + 1 = (X + 1)(X^2 - 2X + 1) = (X + 1)*(X - 1)^2 it means: 
               X = 1 or (X + 1) must be a perfect square, so X = 1 or X = n^2 - 
               1 with n>=1. which gives: (X, Y) = (0, 1) or (X, Y) = (1, 0) or (X, 
               Y) = (n^2 - 1, n*(n^2 - 2))with n>=2.
%C A132411 An equivalent technique of integer factorization would work for example 
               for the equation X^3+3*X^2-9*X+5=(X+5)(X-1)^2=Y^2, looking for perfect 
               squares of the form X+5=n^2. Another example is X^3+X^2-5*X+3=(X+3)*(X-1)^2=Y^2 
               with solutions generated from perfect squares of the form X+3=n^2. 
               - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 20 2007
%F A132411 a(o) = 0, a(1) = 1 and a(n) = n^2 - 1 with n>=2.
%F A132411 G.f.: x+x^2*(-3+x)/(-1+x)^3 . - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), 
               Nov 20 2007
%F A132411 Starting (1, 3, 8, 15, 24,...) = binomial transfor of [1, 2, 3, -1, 1, 
               -1,...] - Gary W. Adamson (qntmpkt(AT)yahoo.com), May 12 2008
%e A132411 0^3 - 1^2 + 2 = 1^2, 1^3 - 2^2 + 3 = 0^2, 3^3 - 4^2 + 5 = 4^2.
%Y A132411 Cf. A028560, A005563.
%Y A132411 Sequence in context: A083656 A013648 A005563 this_sequence A147998 A164003 
               A067998
%Y A132411 Adjacent sequences: A132408 A132409 A132410 this_sequence A132412 A132413 
               A132414
%K A132411 nonn
%O A132411 0,3
%A A132411 Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 12 2007