A165806 - OEIS

%I #26 Sep 08 2022 08:45:48

%S 19,67,145,253,391,559,757,985,1243,1531,1849,2197,2575,2983,3421,

%T 3889,4387,4915,5473,6061,6679,7327,8005,8713,9451,10219,11017,11845,

%U 12703,13591,14509,15457,16435,17443,18481,19549,20647,21775,22933,24121,25339

%N a(n) = 15n^2 + 3n + 1.

%C Polynomials f(x) have the following property: f(x + n*f(x)) is congruent to f(x); here n is an integer.

%C This can be proved by Taylor's theorem.

%C After rationalization of the denominator, the quotient q(n,x) = f(x + n*f(x))/f(x) consists of two parts:

%C a) a rational integer and b) an irrational part.

%C 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).

%H Vincenzo Librandi, <a href="/A165806/b165806.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: x*(19 + 10*x + x^2)/(1-x)^3. - _R. J. Mathar_, Sep 29 2009

%F From _G. C. Greubel_, Apr 08 2016: (Start)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F E.g.f.: (15*x^2 + 18*x + 1)*exp(x). (End)

%e When we substitute sqrt(2) for x in the quadratic expression x^2 + 3x + 13 we get 15 + 3*sqrt(2).

%e sqrt(2) + (15 + 3*sqrt(2)) = (15 + 4*sqrt(2)). When this is substituted in f(x) we get 270 + 132*sqrt(2).

%e (270 + 132*sqrt(2))/(15+3*sqrt(2)) = 19 + 5*sqrt(2).

%t LinearRecurrence[{3,-3,1},{19,67,145}, 100] (* _G. C. Greubel_, Apr 08 2016 *)

%t Table[15n^2+3n+1,{n,50}] (* _Harvey P. Dale_, Mar 14 2020 *)

%o (PARI) a(n)=15*n^2+3*n+1 \\ _Charles R Greathouse IV_, Sep 28 2011

%o (Magma) [15*n^2 + 3*n + 1: n in [1..50]]; // _Vincenzo Librandi_, Sep 29 2011

%Y Cf. A005563, A045944.

%K nonn,easy

%O 1,1

%A _A.K. Devaraj_, Sep 28 2009

%E Definition simplified, sequence extended by _R. J. Mathar_, Sep 29 2009