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%I #41 Feb 29 2020 16:17:32
%S 0,4,44,174,472,1040,2004,3514,5744,8892,13180,18854,26184,35464,
%T 47012,61170,78304,98804,123084,151582,184760,223104,267124,317354,
%U 374352,438700,511004,591894,682024,782072,892740,1014754,1148864,1295844
%N a(n) = n^4 + 3*n^3 + 2*n^2 - 2*n.
%C a(n)/A269657(n) gives unforgeable word approximations (A003000) with increasing accuracy, as follows: 4/15, 44/79, 174/253, ... ~ 0.26 (A242430), 0.5569 (A019308), 0.68774 (A019309), 0.8055770, 0.83674321, 0.85937882, 0.87654509, 0.89000100, 0.9008270111, ....
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F From _Colin Barker_, Jan 15 2020: (Start)
%F G.f.: 2*x*(2 + 12*x - 3*x^2 + x^3) / (1 - x)^5.
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
%F (End)
%F E.g.f.: exp(x)*x*(4 + 18*x + 9*x^2 + x^3). - _Stefano Spezia_, Feb 03 2020
%p A330651 := n -> (((n+3)*n+2)*n-2)*n; # _M. F. Hasler_, Feb 29 2020
%t Numerator/@Table[(-2 n+2 n^2+3 n^3+n^4)/(1+3 n+6 n^2+4 n^3+n^4),{n,0,33}] (* _Ed Pegg Jr_, Jan 15 2020 *)
%o (PARI) Vec(2*x*(2 + 12*x - 3*x^2 + x^3) / (1 - x)^5 + O(x^40),-40) \\ _Colin Barker_, Jan 15 2020
%o (PARI) apply( {A330651(n)=(((n+3)*n+2)*n-2)*n}, [0..44]) \\ _M. F. Hasler_, Feb 29 2020
%Y Cf. A269657, A242430, A003000, A019308, A019309.
%K nonn,easy
%O 0,2
%A _Ed Pegg Jr_, Jan 15 2020
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