A144410 - OEIS

%I #58 Dec 10 2022 10:46:53

%S 8,80,224,440,728,1088,1520,2024,2600,3248,3968,4760,5624,6560,7568,

%T 8648,9800,11024,12320,13688,15128,16640,18224,19880,21608,23408,

%U 25280,27224,29240,31328,33488,35720,38024,40400,42848,45368,47960,50624,53360,56168,59048,62000,65024,68120,71288,74528,77840,81224,84680,88208,91808,95480

%N a(n) = 4*(3*n+1)*(3*n+2).

%C The sequence lists all numbers k such that k+1 is a square and k+4 is divisible by 12. - _Bruno Berselli_, Sep 28 2017

%H Vincenzo Librandi, <a href="/A144410/b144410.txt">Table of n, a(n) for n = 0..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.: 8*(1 + 7*x + x^2)/(1 - x)^3. - _Michael De Vlieger_, Sep 29 2017

%F a(n) = 8*A060544(n+1).

%F a(n) = A136016(2*n+1).

%F a(n) = a(m) + 36*(n - m)*(n + m + 1). For m = n-1, a(n) = a(n-1) + 72*n. - _Bruno Berselli_, Sep 29 2017

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n >= 3. - _Klaus Purath_, Jul 05 2020

%F E.g.f.: 4*(2 +18*x +9*x^2)*exp(x). - _G. C. Greubel_, Mar 27 2021

%F From _Amiram Eldar_, Dec 10 2022: (Start)

%F Sum_{n>=0} 1/a(n) = Pi/(12*sqrt(3)) (A244977).

%F Sum_{n>=0} (-1)^n/a(n) = log(2)/6. (End)

%p A144410:= n-> 4*(3*n+1)*(3*n+2); seq(A144410(n), n=0..60); # _G. C. Greubel_, Mar 27 2021

%t Table[4 (3 n + 1) (3 n + 2), {n, 0, 51}] (* or *)

%t CoefficientList[Series[8 (1 + 7 x + x^2)/(1 - x)^3, {x, 0, 51}], x] (* _Michael De Vlieger_, Sep 29 2017 *)

%o (Magma) [4*(3*n+1)*(3*n+2): n in [0..40]]; // _Vincenzo Librandi_, Aug 07 2011

%o (PARI) a(n)=4*(3*n+1)*(3*n+2) \\ _Charles R Greathouse IV_, Jun 17 2017

%o (Sage) [4*(3*n+1)*(3*n+2) for n in (0..60)] # _G. C. Greubel_, Mar 27 2021

%Y Cf. A060544, A136016, A244977.

%K nonn,easy

%O 0,1

%A _Paul Curtz_, Sep 30 2008