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A048058 a(n) = n^2 + n + 11.

%I A048058 #41 Oct 28 2024 19:40:05
%S A048058 11,13,17,23,31,41,53,67,83,101,121,143,167,193,221,251,283,317,353,
%T A048058 391,431,473,517,563,611,661,713,767,823,881,941,1003,1067,1133,1201,
%U A048058 1271,1343,1417,1493,1571,1651,1733,1817,1903,1991,2081,2173,2267,2363,2461,2561
%N A048058 a(n) = n^2 + n + 11.
%C A048058 Fontebasso lists this as a prime-generating polynomial due to Legendre, but doesn't give a reference. - _Charles R Greathouse IV_, Jun 30 2021
%H A048058 Seiichi Manyama, <a href="/A048058/b048058.txt">Table of n, a(n) for n = 0..10000</a>
%H A048058 P. A. Fontebasso, <a href="https://books.google.com.bo/books?id=-gwMAAAAYAAJ&amp;pg=RA3-PA130#v=onepage&amp;q">Un'altra formula che dà una serie limitata di numeri primi</a>, Supplemento al Periodico di matematica (1899), p. 130.
%H A048058 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A048058 For n > 4: a(n) = A176271(n+1,6). - _Reinhard Zumkeller_, Apr 13 2010
%F A048058 a(n) = 2*n + a(n-1) (with a(0)=11). - _Vincenzo Librandi_, Aug 06 2010
%F A048058 Sum_{n>=0} 1/a(n) = Pi*tanh(Pi*sqrt(43)/2)/sqrt(43). - _Amiram Eldar_, Jan 17 2021
%F A048058 From _Elmo R. Oliveira_, Oct 28 2024: (Start)
%F A048058 G.f.: (11 - 20*x + 11*x^2)/(1 - x)^3.
%F A048058 E.g.f.: (11 + 2*x + x^2)*exp(x).
%F A048058 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)
%p A048058 with(combinat):seq(fibonacci(3, n)+n+10, n=0..45); # _Zerinvary Lajos_, Jun 07 2008
%t A048058 f[n_]:=n^2+n+11;f[Range[0,100]] (* _Vladimir Joseph Stephan Orlovsky_, Mar 12 2011*)
%o A048058 (PARI) a(n)=n^2+n+11 \\ _Charles R Greathouse IV_, Jun 29 2015
%Y A048058 Cf. A002522, A048059, A048097, A176271.
%K A048058 nonn,easy
%O A048058 0,1
%A A048058 _N. J. A. Sloane_