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a(n) = 7*n^2 + 2*n - 15.
1

%I #56 Sep 08 2022 08:46:07

%S -6,17,54,105,170,249,342,449,570,705,854,1017,1194,1385,1590,1809,

%T 2042,2289,2550,2825,3114,3417,3734,4065,4410,4769,5142,5529,5930,

%U 6345,6774,7217,7674,8145,8630,9129,9642,10169,10710,11265,11834,12417,13014,13625,14250,14889,15542,16209,16890

%N a(n) = 7*n^2 + 2*n - 15.

%C Follows the integer values from 1 on the parabola: 7*n^2 + 2*n - 15.

%C Real roots: (-1 +- sqrt(106))/7. - _Wesley Ivan Hurt_, Mar 26 2014

%C The first in the family of parabolas of the form: prime(k+3)*n^2 + prime(k)*n - prime(k+1)*prime(k+2), where k >= 1 (k=1 gives a(n)). - _Wesley Ivan Hurt_, Mar 26 2014

%H Vincenzo Librandi, <a href="/A239796/b239796.txt">Table of n, a(n) for n = 1..1000</a>

%H Vi Hart, <a href="https://www.youtube.com/watch?v=v-pyuaThp-c">Doodling in Math Class: Connecting Dots</a> (2012) [Video]

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

%F a(n) = n * A017005(n) - 15. - _Wesley Ivan Hurt_, Mar 26 2014

%F G.f.: -x*(6 - 35*x + 15*x^2)/(1 - x)^3. - _Bruno Berselli_, Mar 27 2014

%e For n=3, a(3) = 7*3^2 + 2*3 - 15 = 54; for n=6, a(6) = 7*6^2 + 2*6 - 15 = 249.

%p A239796:=n->7*n^2 +2*n - 15; seq(A239796(n), n=1..50); # _Wesley Ivan Hurt_, Mar 26 2014

%t Table[7 n^2 + 2 n - 15, {n, 50}] (* _Wesley Ivan Hurt_, Mar 26 2014 *)

%t CoefficientList[Series[(6 - 35 x + 15 x^2)/(x - 1)^3, {x, 0, 50}], x] (* _Vincenzo Librandi_, Mar 29 2014 *)

%o (Magma) [7*n^2+2*n-15: n in [1..50]]; // _Bruno Berselli_, Mar 27 2014

%o (PARI) a(n)=7*n^2 + 2*n - 15 \\ _Charles R Greathouse IV_, Jan 21 2016

%K sign,easy

%O 1,1

%A _Katherine Guo_, Mar 26 2014