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%I #26 Jul 02 2023 18:42:51
%S 11,15,27,31,43,57,73,91,111,133,157,183,211,241,273,307,343,381,421,
%T 463,507,553,601,651,703,757,813,871,931,993,1057,1123,1191,1261,1333,
%U 1407,1483,1561,1641,1723,1807,1893,1981,2071,2163,2257,2353,2451,2551,2653
%N Trajectory of 11 under repeated application of the map n -> n + 2*square excess of n (see A094765).
%C The trajectory of 3 gives A002061 and 5 gives essentially the same trajectory as 3.
%H Robert Israel, <a href="/A094766/b094766.txt">Table of n, a(n) for n = 0..10000</a>
%H S. H. Weintraub, <a href="http://www.jstor.org/stable/4145074">An interesting recursion</a>, Amer. Math. Monthly, 111 (No. 6, 2004), 528-530.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1).
%F Numbers given satisfy a(n) = n^2 + 5n + 7, for n>2. - _Ralf Stephan_, Dec 04 2004
%F From _Robert Israel_, Oct 23 2015: (Start)
%F This is because for x = m^2 + 5*m + 7, (m+2)^2 < x < (m+3)^2 so A094765(x) = x + 2*(x-(m+2)^2) = m^2 + 7*m + 13 = (m+1)^2 + 5*(m+1) + 7.
%F Similarly, for any positive integer k, the trajectory of k^2 + k + 1 is n^2 + (2k+1) n + k^2 + k + 1 for n >= 0.
%F G.f.: 4 + 2*x + 6*x^2 + (7-8*x+3*x^2)/(1-x)^3. (End)
%p f:= n -> 3*n - 2*floor(sqrt(n))^2:
%p g:= proc(n) option remember; f(procname(n-1)) end proc:
%p g(0):= 11:
%p seq(g(n),n=0..100); # _Robert Israel_, Oct 23 2015
%t NestList[3*#-2*Floor[Sqrt[#]]^2&,11,50] (* _Harvey P. Dale_, Feb 26 2022 *)
%o (PARI) lista(nn) = {print1(n=11, ", "); for (k=2, nn, m = 3*n - 2*sqrtint(n)^2; print1(m, ", "); n = m;);} \\ _Michel Marcus_, Oct 23 2015
%o (PARI) Vec(4+2*x+6*x^2+(7-8*x+3*x^2)/(1-x)^3 + O(x^100)) \\ _Altug Alkan_, Oct 23 2015
%Y Cf. A002061, A053186, A094765.
%K nonn
%O 0,1
%A _N. J. A. Sloane_, Jun 10 2004
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