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 A165806 a(n) = 15n^2 + 3n + 1. 10
 19, 67, 145, 253, 391, 559, 757, 985, 1243, 1531, 1849, 2197, 2575, 2983, 3421, 3889, 4387, 4915, 5473, 6061, 6679, 7327, 8005, 8713, 9451, 10219, 11017, 11845, 12703, 13591, 14509, 15457, 16435, 17443, 18481, 19549, 20647, 21775, 22933, 24121, 25339 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Polynomials f(x) have the following property: f(x + n*f(x)) is congruent to f(x); here n is an integer. This can be proved by Taylor's theorem. After rationalization of the denominator, the quotient q(n,x) = f(x + n*f(x))/f(x) consists of two parts: a) a rational integer and b) an irrational part. The present sequence is the integer part for f(x) = x^2 + 3x + 13 and x = sqrt(2), i.e., q(n,x) = a(n) + sqrt(2)*A045944(n). LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA G.f.: x*(19 + 10*x + x^2)/(1-x)^3. - R. J. Mathar, Sep 29 2009 From G. C. Greubel, Apr 08 2016: (Start) a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). E.g.f.: (15*x^2 + 18*x + 1)*exp(x). (End) EXAMPLE When we substitute sqrt(2) for x in the quadratic expression x^2 + 3x + 13 we get 15 + 3*sqrt(2). sqrt(2) + (15 + 3*sqrt(2)) = (15 + 4*sqrt(2)). When this is substituted in f(x) we get 270 + 132*sqrt(2). (270 + 132*sqrt(2))/(15+3*sqrt(2)) = 19 + 5*sqrt(2). MATHEMATICA LinearRecurrence[{3, -3, 1}, {19, 67, 145}, 100] (* G. C. Greubel, Apr 08 2016 *) PROG (PARI) a(n)=15*n^2+3*n+1 \\ Charles R Greathouse IV, Sep 28 2011 (MAGMA) [15*n^2 + 3*n + 1: n in [1..50]]; // Vincenzo Librandi, Sep 29 2011 CROSSREFS Cf. A005563, A045944. Sequence in context: A104047 A239714 A201781 * A142404 A139927 A228670 Adjacent sequences:  A165803 A165804 A165805 * A165807 A165808 A165809 KEYWORD nonn,easy AUTHOR A.K. Devaraj, Sep 28 2009 EXTENSIONS Definition simplified, sequence extended by R. J. Mathar, Sep 29 2009 STATUS approved

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Last modified January 29 10:34 EST 2020. Contains 331337 sequences. (Running on oeis4.)