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A056469
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Number of elements in the continued fraction for Sum_{k=0..n} 1/2^2^k.
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5
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2, 3, 4, 6, 10, 18, 34, 66, 130, 258, 514, 1026, 2050, 4098, 8194, 16386, 32770, 65538, 131074, 262146, 524290, 1048578, 2097154, 4194306, 8388610, 16777218, 33554434, 67108866, 134217730, 268435458, 536870914, 1073741826, 2147483650
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OFFSET
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0,1
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COMMENTS
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Let f_1(x) := 1 - sqrt(1 - x^2) = 2*x^2 + 2*x^4 + 4*x^6 + ... and for n>1 let f_n(x) := f_{n-1}(f_1(x)) = x^(2^n)*(2 + 2^n*x^2 + 2^n*a(n-1)*x^4 + ...). - Michael Somos, Jun 29 2023
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LINKS
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FORMULA
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a(0)=2; for n > 0, a(n) = 2^(n-1) + 2 = A052548(n-1) + 2.
a(n) = 3*a(n-1) - 2*a(n-2) for n > 2.
G.f.: -(x^2+3*x-2) / ((x-1)*(2*x-1)). (End)
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EXAMPLE
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G.f. = 2 + 3*x + 4*x^2 + 6*x^3 + 10*x^4 + 18*x^5 + 34*x^6 + ... - Michael Somos, Jun 29 2023
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MATHEMATICA
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LinearRecurrence[{3, -2}, {2, 3, 4}, 40] (* Harvey P. Dale, Apr 23 2015 *)
a[ n_] := If[n < 0, 0, Floor[2^n/2] + 2]; (* Michael Somos, Jun 29 2023 *)
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PROG
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(Sage) [floor(gaussian_binomial(n, 1, 2)+3) for n in range(-1, 32)] # Zerinvary Lajos, May 31 2009
(PARI) {a(n) = if(n<0, 0, 2^n\2 + 2)}; /* Michael Somos, Jun 29 2023 */
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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