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a(n) = 16*n^4 + 32*n^3 + 36*n^2 + 20*n + 3.
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%I #24 Jul 13 2020 02:48:48

%S 3,107,699,2547,6803,15003,29067,51299,84387,131403,195803,281427,

%T 392499,533627,709803,926403,1189187,1504299,1878267,2318003,2830803,

%U 3424347,4106699,4886307,5772003,6773003,7898907,9159699,10565747

%N a(n) = 16*n^4 + 32*n^3 + 36*n^2 + 20*n + 3.

%C Let x(n) = (1/2)*(-(2*n + 1) + sqrt((2*n + 1)^2 + 4)) and f(n,k) = (-1)*Sum_{i=1..k} Sum_{j=1..i} (-1)^floor(j*x(n)). Then a(n) = k is the least integer k > 0 such that f(n, k) = 0. In other words, f(n, a(n)) = 0, and if f(n,k) = 0, then a(n) <= k. [Edited by _Petros Hadjicostas_, Jul 12 2020]

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

%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. - _Harvey P. Dale_, Jul 23 2013

%t Table[16n^4+32n^3+36n^2+20n+3,{n,0,30}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{3,107,699,2547,6803},30] (* _Harvey P. Dale_, Jul 23 2013 *)

%o (PARI) a(n) = 16*n^4+32*n^3+36*n^2+20*n+3

%Y Cf. A085005, A094201.

%K nonn,easy

%O 0,1

%A _Benoit Cloitre_, May 25 2004