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%I #38 Feb 15 2023 10:08:38
%S 1,3,10,22,39,61,88,120,157,199,246,298,355,417,484,556,633,715,802,
%T 894,991,1093,1200,1312,1429,1551,1678,1810,1947,2089,2236,2388,2545,
%U 2707,2874,3046,3223,3405,3592,3784,3981,4183,4390,4602,4819,5041,5268,5500
%N a(n) = (5*n^2 - 11*n + 8)/2.
%C Binomial transform of [1, 2, 5, 0, 0, 0, ...] = A020821.
%H Harvey P. Dale, <a href="/A140066/b140066.txt">Table of n, a(n) for n = 1..1000</a>
%H Franck Ramaharo, <a href="https://arxiv.org/abs/1802.07701">Statistics on some classes of knot shadows</a>, arXiv:1802.07701 [math.CO], 2018.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A007318 * [1, 2, 5, 0, 0, 0, ...].
%F From _R. J. Mathar_, May 06 2008: (Start)
%F a(n) = A000217(n) + 4*A000217(n-2).
%F O.g.f.: x*(1+4*x^2)/(1-x)^3. (End)
%F a(n) = (8 - 11*n + 5*n^2)/2. - _Emeric Deutsch_, May 07 2008
%F a(n) = a(n-1) + 5*n - 8 (with a(1)=1). - _Vincenzo Librandi_, Nov 24 2010
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=3, a(3)=10. - _Harvey P. Dale_, Jan 28 2012
%e a(4) = 22 = (1, 3, 3, 1) dot (1, 2, 5, 0) = (1, + 6 + 15 + 0).
%p seq((8-11*n+5*n^2)*1/2,n=1..40); # _Emeric Deutsch_, May 07 2008
%t Table[(5n^2-11n+8)/2,{n,40}] (* or *) LinearRecurrence[{3,-3,1},{1,3,10},40] (* _Harvey P. Dale_, Jan 28 2012 *)
%o (PARI) a(n)=(5*n^2-11*n+8)/2 \\ _Charles R Greathouse IV_, Jun 17 2017
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, May 03 2008
%E More terms from _R. J. Mathar_ and _Emeric Deutsch_, May 06 2008
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