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A288492
Indices of terms of A288349 that are powers of 2.
1
1, 2, 3, 18, 95, 440, 1897, 7882, 32139, 129804, 521741, 2092046, 8378383, 33533968, 134176785, 536789010, 2147319827, 8589606932, 34359083029, 137437642774, 549753192471, 2199018012696, 8796082536473, 35184351117338, 140737446412315, 562949869535260
OFFSET
1,2
COMMENTS
The sequence is derived from Chinese 2017 college entrance examination mathematics questions.
FORMULA
From Colin Barker, Jun 23 2017: (Start)
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)).
a(n) = 2 - 5*2^(n-2) + 2^(2*n-3) + n for n>2.
a(n) = 8*a(n-1) - 21*a(n-2) + 22*a(n-3) - 8*a(n-4) for n>6.
(End)
EXAMPLE
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.
MATHEMATICA
Position[Accumulate@ Flatten@ Array[2^Range[0, #] &, 2000, 0], k_ /; IntegerQ@ Log2@ k][[All, 1]] (* per Name, or *)
Table[2 - 5*2^(n - 2) + 2^(2 n - 3) + n + Boole[n == 2], {n, 26}] (* or *)
LinearRecurrence[{8, -21, 22, -8}, {1, 2, 3, 18, 95, 440}, 26] (* or *)
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 *)
PROG
(PARI) for(k=0, 100, p=(2^k-3)*(2^k-2)/2+k; print1(p, ", "))
(PARI) ispower2(n) = (n==1) || (n==2) || (ispower(n, , &two) && (two==2));
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
(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
CROSSREFS
Sequence in context: A048047 A114165 A166510 * A184719 A076016 A226961
KEYWORD
nonn,easy
AUTHOR
Zhining Yang, Jun 10 2017
STATUS
approved