

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

Sequence allows us to find X values of 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^29*X+5=(X+5)(X1)^2=Y^2, looking for perfect squares of the form X+5=n^2. Another example is X^3+X^25*X+3=(X+3)*(X1)^2=Y^2 with solutions generated from perfect squares of the form X+3=n^2.  R. J. Mathar, Nov 20 2007
a(n) = A170949(A002522(n1)) for n>0.  Reinhard Zumkeller, Mar 08 2010
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^22)^2(n^21)*(4*n)^2=4 can be written as A060626(n+1)^2a(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(n1), 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
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]^21] (* 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^21, 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

Simplified definition.  N. J. A. Sloane, Sep 05 2010


STATUS

approved



