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A177060 a(n) = (7*n+2)*(7*n+5) = 49*n^2 + 49*n + 10.

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%I A177060 #19 Oct 25 2024 06:34:04
%S A177060 10,108,304,598,990,1480,2068,2754,3538,4420,5400,6478,7654,8928,
%T A177060 10300,11770,13338,15004,16768,18630,20590,22648,24804,27058,29410,
%U A177060 31860,34408,37054,39798,42640,45580,48618,51754,54988,58320,61750,65278,68904,72628,76450
%N A177060 a(n) = (7*n+2)*(7*n+5) = 49*n^2 + 49*n + 10.
%C A177060 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) = 49*A002061(n+1) - 39. - _Bruno Berselli_, Aug 24 2010
%H A177060 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A177060 a(n) = 98*n + a(n-1) with a(0) = 10.
%F A177060 From _Amiram Eldar_, Feb 19 2023: (Start)
%F A177060 a(n) = A017005(n)*A017041(n).
%F A177060 Sum_{n>=0} 1/a(n) = tan(3*Pi/14)*Pi/21.
%F A177060 Product_{n>=0} (1 - 1/a(n)) = sec(3*Pi/14)*cos(sqrt(13)*Pi/14).
%F A177060 Product_{n>=0} (1 + 1/a(n)) = sec(3*Pi/14)*cos(sqrt(5)*Pi/14). (End)
%F A177060 From _Elmo R. Oliveira_, Oct 24 2024: (Start)
%F A177060 G.f.: 2*(5 + 39*x + 5*x^2)/(1 - x)^3.
%F A177060 E.g.f.: exp(x)*(10 + 49*x*(2 + x)).
%F A177060 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)
%e A177060 For n=1, a(1) = 98 + 10 = 108.
%e A177060 For n=2, a(2) = 98*2 + 108 = 304.
%e A177060 For n=3, a(3) = 98*3 + 304 = 598.
%t A177060 a[n_] := (7*n + 2)*(7*n + 5); Array[a, 40, 0] (* _Amiram Eldar_, Feb 19 2023 *)
%o A177060 (PARI) a(n)=49*n*(n+1)+10 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y A177060 Cf. A002061, A017005, A017041, A177059.
%K A177060 nonn,easy
%O A177060 0,1
%A A177060 _Vincenzo Librandi_, May 31 2010