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%I #30 Feb 19 2023 03:28:53
%S 15,143,399,783,1295,1935,2703,3599,4623,5775,7055,8463,9999,11663,
%T 13455,15375,17423,19599,21903,24335,26895,29583,32399,35343,38415,
%U 41615,44943,48399,51983,55695,59535,63503,67599,71823,76175,80655,85263,89999,94863,99855
%N a(n) = (8*n+3)*(8*n+5).
%C Cf. comment of _Reinhard Zumkeller_ in A177059: in general, (h*n+h-k)*(h*n+k) = h^2*A002061(n+1)+(h-k)*k-h^2; therefore a(n) = 64*A002061(n+1)-49. - _Bruno Berselli_, Aug 24 2010
%H Vincenzo Librandi, <a href="/A177065/b177065.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 128*n+a(n-1) with n>0, a(0)=15.
%F a(n) = A125169(A016754(n) - 1). - _Reinhard Zumkeller_, Jul 05 2010
%F a(0)=15, a(1)=143, a(2)=399, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). - _Harvey P. Dale_, Mar 13 2013
%F G.f.: (15+98*x+15*x^2)/(1-x)^3. - _Vincenzo Librandi_, Apr 08 2013
%F From _Amiram Eldar_, Feb 19 2023: (Start)
%F a(n) = A017101(n)*A004770(n).
%F Sum_{n>=0} 1/a(n) = (sqrt(2)-1)*Pi/16.
%F Sum_{n>=0} (-1)^n/a(n) = (cos(Pi/8) * log(tan(3*Pi/16)) + sin(Pi/8) * log(cot(Pi/16)))/4.
%F Product_{n>=0} (1 - 1/a(n)) = sec(Pi/8)*cos(Pi/(4*sqrt(2))).
%F Product_{n>=0} (1 + 1/a(n)) = sec(Pi/8). (End)
%p A177065:=n->(8*n+3)*(8*n+5): seq(A177065(n), n=0..100); # _Wesley Ivan Hurt_, Apr 24 2017
%t Table[(8n+3)(8n+5),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{15,143,399},40] (* _Harvey P. Dale_, Mar 13 2013 *)
%t CoefficientList[Series[(15 + 98 x + 15 x^2)/(1-x)^3, {x, 0, 50}], x] (* _Vincenzo Librandi_, Apr 08 2013 *)
%o (Magma) [(8*n+3)*(8*n+5): n in [0..50]]; // _Vincenzo Librandi_, Apr 08 2013
%o (PARI) a(n)=(8*n+3)*(8*n+5) \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A002061, A004770, A016754, A017101, A125169, A177059.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, May 31 2010
%E Edited by _N. J. A. Sloane_, Jun 22 2010
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