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A145910 a(n) = (1 + 3*n)*(4 + 3*n)/2.

%I A145910 #53 May 31 2025 05:17:55
%S A145910 2,14,35,65,104,152,209,275,350,434,527,629,740,860,989,1127,1274,
%T A145910 1430,1595,1769,1952,2144,2345,2555,2774,3002,3239,3485,3740,4004,
%U A145910 4277,4559,4850,5150,5459,5777,6104,6440,6785,7139,7502,7874
%N A145910 a(n) = (1 + 3*n)*(4 + 3*n)/2.
%H A145910 R. C. Alperin, <a href="https://www.fq.math.ca/Papers/58-2/alperin09212019.pdf">A nonlinear recurrence and its relations to Chebyshev polynomials</a>, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
%H A145910 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A145910 a(n) = a(n-1) + 3*(3*n+1) = a(n-1) + A017197(n+1).
%F A145910 G.f.: (-2 - 8*x + x^2)/(x-1)^3. - _R. J. Mathar_, Jan 06 2011
%F A145910 a(n) = A144449(n)/8.
%F A145910 a(n) = 2*a(n-1) - a(n-2) + 9.
%F A145910 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A145910 From _Amiram Eldar_, Mar 11 2022: (Start)
%F A145910 Sum_{n>=0} 1/a(n) = 2/3.
%F A145910 Sum_{n>=0} (-1)^n/a(n) = 4*Pi/(9*sqrt(3)) + 4*log(2)/9 - 2/3. (End)
%F A145910 From _Elmo R. Oliveira_, Nov 15 2024: (Start)
%F A145910 E.g.f.: exp(x)*(4 + 24*x + 9*x^2)/2.
%F A145910 a(n) = A085001(n)/2. (End)
%p A145910 A145910:=n->(1+3*n)*(4+3*n)/2: seq(A145910(n), n=0..100); # _Wesley Ivan Hurt_, Jul 25 2017
%t A145910 Table[(1+3n)(4+3n)/2, {n,0,50}] (* _Harvey P. Dale_, Feb 23 2011 *)
%o A145910 (PARI) a(n)=(1+3*n)*(4+3*n)/2 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y A145910 Cf. A017197, A085001, A144449.
%K A145910 nonn,easy
%O A145910 0,1
%A A145910 _Paul Curtz_, Oct 24 2008
%E A145910 Terms a(11)-a(42) from _Vincenzo Librandi_, Nov 17 2009