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Sum of first n terms of the simple continued fraction of Sum_{k>=0} 1/2^(2^k) (cf. A007400).

2

`%I #10 Oct 13 2019 18:07:50
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`%S 0,1,5,7,11,15,21,25,27,31,37,39,43,49,53,57,59,63,69,71,75,79,85,89,
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`%T 91,97,101,103,107,113,117,121,123,127,133,135,139,143,149,153,155,
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`%U 159,165,167,171,177,181,185,187,193,197,199,203,207,213,217,219,225,229
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`%N Sum of first n terms of the simple continued fraction of Sum_{k>=0} 1/2^(2^k) (cf. A007400).
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`%F a(n) ~ 4n and a(n) < 4n.
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`%e The first 9 terms are 0, 1, 4, 2, 4, 4, 6, 4, 2, hence a(9)= 0 + 1 + 4 + 2 + 4 + 4 + 6 + 4 + 2 = 27.
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`%Y Cf. A007400.
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`%K easy,nonn
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`%O 1,3
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`%A _Benoit Cloitre_, Aug 18 2002
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