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2-adic continued fraction of zero, where a(n) = 1 if n is odd, otherwise -2*A006519(n/2).
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%I #24 Oct 28 2023 04:04:39

%S 1,-2,1,-4,1,-2,1,-8,1,-2,1,-4,1,-2,1,-16,1,-2,1,-4,1,-2,1,-8,1,-2,1,

%T -4,1,-2,1,-32,1,-2,1,-4,1,-2,1,-8,1,-2,1,-4,1,-2,1,-16,1,-2,1,-4,1,

%U -2,1,-8,1,-2,1,-4,1,-2,1,-64,1,-2,1,-4,1,-2,1,-8,1,-2,1,-4,1,-2,1,-16,1,-2,1,-4,1,-2,1,-8,1,-2,1,-4,1,-2,1,-32,1,-2,1

%N 2-adic continued fraction of zero, where a(n) = 1 if n is odd, otherwise -2*A006519(n/2).

%C Limit of convergents equals zero; only the 6th convergent is indeterminate. Other 2-adic continued fractions of zero are: A118821, A118824, A118830. A006519(n) is the highest power of 2 dividing n; A080277 = partial sums of A038712, where A038712(n) = 2*A006519(n) - 1.

%C Multiplicative because both A006519 and A165326 are. - _Andrew Howroyd_, Aug 01 2018

%H Antti Karttunen, <a href="/A118827/b118827.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = A165326(n) * A006519(n). - _Andrew Howroyd_, Aug 01 2018

%F From _Amiram Eldar_, Oct 28 2023: (Start)

%F Multiplicative with a(2^e) = -2^e, and a(p^e) = 1 for an odd prime p.

%F Dirichlet g.f.: zeta(s) * (1 - 2^(1-s) + 1/(2-2^s)).

%F Sum_{k=1..n} a(k) ~ (-1/(2*log(2))) * n *(log(n) + gamma - log(2)/2 - 1), where gamma is Euler's constant (A001620). (End)

%e For n >= 1, convergents A118828(k)/A118829(k):

%e at k = 4*n: -1/(2*A080277(n));

%e at k = 4*n+1: -1/(2*A080277(n)-1);

%e at k = 4*n+2: -1/(2*A080277(n)-2);

%e at k = 4*n-1: 0.

%e Convergents begin:

%e 1/1, -1/-2, 0/-1, -1/2, -1/1, 1/0, 0/1, 1/-8,

%e 1/-7, -1/6, 0/-1, -1/10, -1/9, 1/-8, 0/1, 1/-24,

%e 1/-23, -1/22, 0/-1, -1/26, -1/25, 1/-24, 0/1, 1/-32,

%e 1/-31, -1/30, 0/-1, -1/34, -1/33, 1/-32, 0/1, 1/-64, ...

%t Array[If[OddQ@ #, 1, -2*2^(IntegerExponent[#, 2] - 1)] &, 99] (* _Michael De Vlieger_, Nov 06 2018 *)

%o (PARI) a(n)=local(p=+1,q=-2);if(n%2==1,p,q*2^valuation(n/2,2))

%Y Cf. A001620, A006519, A080277; convergents: A118828/A118829; variants: A118821, A118824, A118830; A100338, A165326.

%K cofr,sign,mult

%O 1,2

%A _Paul D. Hanna_, May 01 2006