cp's OEIS Frontend

A132411 a(0) = 0, a(1) = 1 and a(n) = n^2 - 1 with n >= 2.

%I A132411 #76 Jun 03 2025 15:09:54
%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 1368,1443,1520,1599,1680,1763,1848,1935,2024,2115,2208,2303,2400,2499,2600
%N A132411 a(0) = 0, a(1) = 1 and a(n) = n^2 - 1 with n >= 2.
%C A132411 X values of solutions to 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_, Nov 20 2007
%C A132411 Sum of possible divisors of a prime number up to its square root, with duplicate entries removed. - _Odimar Fabeny_, Aug 25 2010
%C A132411 a(0) = 0, a(1) = 1 and a(n) is the smallest k different from n such that n divides k and n+1 divides k+1. - _Michel Lagneau_, Apr 27 2013
%C A132411 The identity (4*n^2-2)^2 - (n^2-1)*(4*n)^2 = 4 can be written as A060626(n+1)^2 - a(n+2)*A008586(n+2)^2 = 4. - _Vincenzo Librandi_, Jun 16 2014
%C A132411 Conjecture: the sequence terms are the exponents in the expansion of the q-series Sum_{n >= 0} q^(2*n) * Product_{k >= 2*n+2} 1 - (-q)^k = 1 + q^3 + q^8 + q^15 + q^24 + .... - _Peter Bala_, May 10 2025
%H A132411 Vincenzo Librandi, <a href="/A132411/b132411.txt">Table of n, a(n) for n = 0..1000</a>
%H A132411 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A132411 a(n) = A005563(n-1), n > 1.
%F A132411 G.f.: x + x^2*(-3+x)/(-1+x)^3. - _R. J. Mathar_, Nov 20 2007
%F A132411 Starting (1, 3, 8, 15, 24, ...) = binomial transform of [1, 2, 3, -1, 1, -1, ...]. - _Gary W. Adamson_, May 12 2008
%F A132411 a(n) = A170949(A002522(n-1)) for n > 0. - _Reinhard Zumkeller_, Mar 08 2010
%F A132411 Sum_{n>0} 1/a(n) = 7/4. - _Enrique Pérez Herrero_, Dec 18 2015
%F A132411 Sum_{n>=1} (-1)^(n+1)/a(n) = 3/4. - _Amiram Eldar_, Sep 27 2022
%F A132411 From _Elmo R. Oliveira_, May 29 2025: (Start)
%F A132411 E.g.f.: exp(x)*(x^2 + x - 1) + x + 1.
%F A132411 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 5. (End)
%e A132411 0^3 - 1^2 + 2 = 1^2, 1^3 - 2^2 + 3 = 0^2, 3^3 - 4^2 + 5 = 4^2.
%e A132411 For P(n) = 29 we have sqrt(29) = 5.3851... so possible divisors are 3 and 5; for P(n) = 53 we have sqrt(53) = 7.2801... so possible divisors are 3, 5 and 7. - _Odimar Fabeny_, Aug 25 2010
%p A132411 a:= n-> `if`(n<2, n, n^2-1):
%p A132411 seq(a(n), n=0..55);  # _Alois P. Heinz_, Jan 24 2021
%t A132411 Join[{0, 1}, LinearRecurrence[{3, -3, 1}, {3, 8, 15}, 80]] (* and *) Table[If[n < 2, n, n^2 - 1], {n, 0, 80}] (* _Vladimir Joseph Stephan Orlovsky_, Feb 14 2012 *)
%t A132411 Join[{0,1},Range[2,50]^2-1] (* _Harvey P. Dale_, Feb 27 2013 *)
%t A132411 CoefficientList[Series[x + x^2 (-3 + x)/(-1 + x)^3, {x, 0, 60}], x] (* _Vincenzo Librandi_, May 01 2014 *)
%o A132411 (Magma) [0,1] cat [n^2 - 1: n in [2..60]]; // _Vincenzo Librandi_, May 01 2014
%o A132411 (PARI) concat(0, Vec(x+x^2*(-3+x)/(-1+x)^3 + O(x^100))) \\ _Altug Alkan_, Dec 18 2015
%o A132411 (PARI) a(n)=if(n>1,n^2-1,n) \\ _Charles R Greathouse IV_, Dec 18 2015
%Y A132411 Cf. A002522, A005563, A008586, A028560, A060626, A170949.
%K A132411 nonn,easy
%O A132411 0,3
%A A132411 _Mohamed Bouhamida_, Nov 12 2007
%E A132411 Definition simplified by _N. J. A. Sloane_, Sep 05 2010