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 A132411 a(0) = 0, a(1) = 1 and a(n) = n^2 - 1 with n>=2. 14
 0, 1, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS X values of solutions to the equation X^3 - (X + 1)^2 + X + 2 = Y^2. 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. 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 Sum of possible divisors of a prime number up to its square root, with duplicate entries removed. - Odimar Fabeny, Aug 25 2010 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 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 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (3, -3, 1). FORMULA a(n) = A005563(n-1), n>1. G.f.: x+x^2*(-3+x)/(-1+x)^3. - R. J. Mathar, Nov 20 2007 Starting (1, 3, 8, 15, 24,...) = binomial transform of [1, 2, 3, -1, 1, -1,...]. - Gary W. Adamson, May 12 2008 a(n) = A170949(A002522(n-1)) for n>0. - Reinhard Zumkeller, Mar 08 2010 Sum(n>0, 1/a(n)) = 7/4. - Enrique Pérez Herrero, Dec 18 2015 EXAMPLE 0^3 - 1^2 + 2 = 1^2, 1^3 - 2^2 + 3 = 0^2, 3^3 - 4^2 + 5 = 4^2. For P(n) = 29 we have sqrt(29) = 5.3851 then possible divisors are 3 and 5; for P(n) = 53 we have sqrt(53) = 7.2801 then possible divisors are 3, 5 and 7. [Odimar Fabeny, Aug 25 2010] MAPLE for n from 1 to 100 do :ii :=0 :for k from 1 to 20000 while(ii=0) do: if n<>k and irem(k, n)=0 and irem(k+1, n+1)=0 then printf(`%d, `, k):ii:=1:else fi:od:od: # Michel Lagneau, Apr 27 2013 MATHEMATICA 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 *) Join[{0, 1}, Range[2, 50]^2-1] (* Harvey P. Dale, Feb 27 2013 *) CoefficientList[Series[x + x^2 (-3 + x)/(-1 + x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, May 01 2014 *) PROG (MAGMA) [0, 1] cat [n^2 - 1: n in [2..60]]; // Vincenzo Librandi, May 01 2014 (PARI) concat(0, Vec(x+x^2*(-3+x)/(-1+x)^3 + O(x^100))) \\ Altug Alkan, Dec 18 2015 (PARI) a(n)=if(n>1, n^2-1, n) \\ Charles R Greathouse IV, Dec 18 2015 CROSSREFS Cf. A028560, A005563. Sequence in context: A013648 A258837 A131386 * A005563 A067998 A066079 Adjacent sequences:  A132408 A132409 A132410 * A132412 A132413 A132414 KEYWORD nonn,easy AUTHOR Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 12 2007 EXTENSIONS Definition simplified by N. J. A. Sloane, Sep 05 2010 STATUS approved

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Last modified October 17 06:08 EDT 2019. Contains 328106 sequences. (Running on oeis4.)