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A288492 Indices of terms of A288349 that are powers of 2.

%I A288492 #43 Jul 19 2017 21:36:31
%S A288492 1,2,3,18,95,440,1897,7882,32139,129804,521741,2092046,8378383,
%T A288492 33533968,134176785,536789010,2147319827,8589606932,34359083029,
%U A288492 137437642774,549753192471,2199018012696,8796082536473,35184351117338,140737446412315,562949869535260
%N A288492 Indices of terms of A288349 that are powers of 2.
%C A288492 The sequence is derived from Chinese 2017 college entrance examination mathematics questions.
%H A288492 Zhining Yang, <a href="/A288492/b288492.txt">Table of n, a(n) for n = 1..200</a>
%H A288492 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8,-21,22,-8).
%F A288492 From _Colin Barker_, Jun 23 2017: (Start)
%F A288492 G.f.: x*(1 - 6*x + 8*x^2 + 14*x^3 - 22*x^4 + 8*x^5) / ((1 - x)^2*(1 - 2*x)*(1 - 4*x)).
%F A288492 a(n) = 2 - 5*2^(n-2) + 2^(2*n-3) + n for n>2.
%F A288492 a(n) = 8*a(n-1) - 21*a(n-2) + 22*a(n-3) - 8*a(n-4) for n>6.
%F A288492 (End)
%e A288492 a(4) = 18 means the 18th element of the sum of the concatenate subsequences [2^0, 2^1, ..., 2^k] = 1+1+2+1+2+4+1+2+4+8+1+2+4+8+16+1+2+4 = 64, and 64 is power of 2.
%t A288492 Position[Accumulate@ Flatten@ Array[2^Range[0, #] &, 2000, 0], k_ /; IntegerQ@ Log2@ k][[All, 1]] (* per Name, or *)
%t A288492 Table[2 - 5*2^(n - 2) + 2^(2 n - 3) + n + Boole[n == 2], {n, 26}] (* or *)
%t A288492 LinearRecurrence[{8, -21, 22, -8}, {1, 2, 3, 18, 95, 440}, 26] (* or *)
%t A288492 Rest@ CoefficientList[Series[x (1 - 6 x + 8 x^2 + 14 x^3 - 22 x^4 + 8 x^5)/((1 - x)^2*(1 - 2 x) (1 - 4 x)), {x, 0, 26}], x] (* _Michael De Vlieger_, Jun 19 2017 *)
%o A288492 (PARI) for(k=0,100,p=(2^k-3)*(2^k-2)/2+k; print1(p, ", "))
%o A288492 (PARI) ispower2(n) = (n==1) || (n==2) || (ispower(n,,&two) && (two==2));
%o A288492 lista(nn) = select(x->ispower2(x), vector(nn, n, t=floor(sqrt(2*n)+1/2); 2^t+2^(n-t*(t-1)/2)-t-2), 1); \\ _Michel Marcus_, Jun 20 2017
%o A288492 (PARI) Vec(x*(1 - 6*x + 8*x^2 + 14*x^3 - 22*x^4 + 8*x^5) / ((1 - x)^2*(1 - 2*x)*(1 - 4*x)) + O(x^30)) \\ _Colin Barker_, Jun 23 2017
%Y A288492 Cf. A059268, A288349.
%K A288492 nonn,easy
%O A288492 1,2
%A A288492 _Zhining Yang_, Jun 10 2017