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%I #28 Apr 04 2016 11:52:22
%S 5,4,6,9,14,23,40,73,138,267,524,1037,2062,4111,8208,16401,32786,
%T 65555,131092,262165,524310,1048599,2097176,4194329,8388634,16777243,
%U 33554460,67108893,134217758,268435487,536870944,1073741857,2147483682,4294967331,8589934628
%N a(1) = 5; a(n) is the sum of |a(m) - m| for m < n.
%H Colin Barker, <a href="/A270841/b270841.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,2).
%F For n > 2, a(n) = a(n - 2) + |a(n - 1) - (n - 1)|
%F For n > 2, a(n) = 2^(n - 2) + n + 1. - _Peter Kagey_, Mar 25 2016
%F From _Colin Barker_, Mar 28 2016: (Start)
%F a(n) = 4*a(n-1)-5*a(n-2)+2*a(n-3) for n>4.
%F G.f.: x*(5-16*x+15*x^2-5*x^3) / ((1-x)^2*(1-2*x)).
%F (End)
%e a(1) = 5;
%e a(2) = 4 because |a(1) - 1| = 4
%e a(3) = 6 because |a(2) - 2| + |a(1) - 1| = 6
%e a(4) = 9 because |a(3) - 3| + |a(2) - 2| + |a(1) - 1| = 9
%e (...)
%t a[1] := 5
%t a[n_]:= 2^(n - 2) + n + 1 (* _Peter Kagey_, Mar 25 2016 *)
%o (Java)
%o public static int[] a(int n) {
%o int[] terms = new int[n];
%o terms[0] = 0;
%o terms[1] = 5;
%o for (int i = 2; i < n; i++) {
%o int count = 0;
%o for (int j = 0; j < i; j++) {
%o count = count + abs(j - terms[j]);
%o }
%o terms[i] = count;
%o }
%o return terms;
%o }
%o (Ruby)
%o def a270841(n); n == 1 ? 5 : 2**(n - 2) + n + 1 end
%o # _Peter Kagey_, Mar 25 2016
%o (PARI) Vec(x*(5-16*x+15*x^2-5*x^3)/((1-x)^2*(1-2*x)) + O(x^50)) \\ _Colin Barker_, Mar 28 2016
%o (PARI) a(n)=if(n>1, 2^(n-2)+n+1, 5) \\ _Charles R Greathouse IV_, Mar 28 2016
%K easy,nonn
%O 1,1
%A _Alec Jones_, Mar 23 2016
%E More terms from _Colin Barker_, Mar 28 2016
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