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a(n) = (9n^4 - 18n^3 + 18n^2 - 9n + 2)/2.
1

%I #27 Sep 08 2022 08:45:08

%S 1,28,190,703,1891,4186,8128,14365,23653,36856,54946,79003,110215,

%T 149878,199396,260281,334153,422740,527878,651511,795691,962578,

%U 1154440,1373653,1622701,1904176,2220778,2575315,2970703,3409966,3896236,4432753,5022865

%N a(n) = (9n^4 - 18n^3 + 18n^2 - 9n + 2)/2.

%D E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982; p. 810.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F From _Harvey P. Dale_, Jun 10 2011: (Start)

%F G.f.: (x*(x*(x+3)*(x+20)+23)+1)/(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), a(0) = 1, a(1) = 28, a(2) = 190, a(3) = 703, a(4) = 1891. (End)

%F a(n) = (3*n^2 - 3*n + 1)*(3*n^2 - 3*n + 2)/2. - _Bruno Berselli_, Jan 28 2017

%t Table[(9n^4 + 18n^2 + 5)/32, {n, 1, 71, 2}] (* or *) LinearRecurrence[ {5, -10, 10, -5, 1}, {1, 28, 190, 703, 1891}, 36] (* _Harvey P. Dale_, Jun 10 2011 *)

%o (PARI) a(n)=(((9*n-18)*n+18)*n-9)*n/2+1 \\ _Charles R Greathouse IV_, Jun 10 2011

%o (Magma) [(9*n^4 - 18*n^3 + 18*n^2 - 9*n + 2)/2 : n in [1..40]]; // _Wesley Ivan Hurt_, Jan 27 2017

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_, Feb 21 2003