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a(n) = n*2^n + 2*n^2 + 1.
0

%I #12 Mar 07 2020 07:53:30

%S 1,5,17,43,97,211,457,995,2177,4771,10441,22771,49441,106835,229769,

%T 491971,1049089,2228803,4719241,9962195,20972321,44041075,92275657,

%U 192939043,402654337,838862051,1744831817,3623880115,7516194337

%N a(n) = n*2^n + 2*n^2 + 1.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (7, -19, 25, -16, 4).

%F G.f.: (-1 - 10*x^4 + 6*x^3 - x^2 + 2*x)/((2*x-1)^2*(x-1)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 14 2009

%F a(n) = 7*a(n-1) - 19*a(n-2) + 25*a(n-3) - 16*a(n-4) + 4*a(n-5); a(0)=1, a(1)=5, a(2)=17, a(3)=43, a(4)=97. - _Harvey P. Dale_, Dec 06 2012

%t Table[n 2^n+2n^2+1,{n,0,30}] (* or *) LinearRecurrence[{7,-19,25,-16,4},{1,5,17,43,97},30] (* _Harvey P. Dale_, Dec 06 2012 *)

%K nonn

%O 0,2

%A _N. J. A. Sloane_