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A006271 Numerators of a continued fraction for 1 + sqrt(2).
(Formerly M1555)
2

%I M1555 #28 Jan 20 2022 10:18:18

%S 2,5,197,7761797,467613464999866416197,

%T 102249460387306384473056172738577521087843948916391508591105797

%N Numerators of a continued fraction for 1 + sqrt(2).

%C With b(n) = floor((1+sqrt(2))^n) (cf. A080039) the terms appear to be b(2*3^n). - _Joerg Arndt_, Apr 29 2013

%C Note that 1 + sqrt(2) = (c + sqrt(c^2+4))/2 and has regular continued fraction [c, c, ...] with c = 2. With b(n) = A006266(n), it can be expanded into an irregular continued fraction f(1) = b(1) and f(n) = (b[n-1]^2+1)/(b[n]-b[n-1]), and numerator(f(n)) = a(n) (cf. Shallit). - _Michel Marcus_, Apr 29 2013

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H F. L. Bauer, <a href="http://www.ega-math.narod.ru/Nquant/Bauer.htm">Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence</a>, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.

%H N. J. Fine, <a href="http://www.jstor.org/stable/2321014">Infinite products for k-th roots</a>, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.

%H Jeffrey Shallit, <a href="http://archive.org/details/jresv80Bn2p285">Predictable regular continued cotangent expansions</a>, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

%F From _Peter Bala_, Jan 18 2022: (Start)

%F a(n) = (3 + 2*sqrt(2))^3^(n-1) + (3 - 2*sqrt(2))^3^(n-1) - 1 for n >= 1.

%F a(n) = A006266(n)^2 + 1 for n >= 1.

%F a(1) = 5 and a(n) = a(n-1)^3 + 3*a(n-1)^2 - 3 for n >= 2.

%F a(1) = 5 and a(n) = 8*(Product_{k = 1..n-1} a(k))^2 - 3 for n >= 2.

%F 2 - Product_{n = 1..N} (1 + 2/a(n))^2 = 8/(a(N+1) + 3). Therefore

%F sqrt(2) = (1 + 2/5) * (1 + 2/197) * (1 + 2/7761797) * (1 + 2/ 467613464999866416197) * ... - see Bauer.

%F The convergence is cubic - see Fine. The first six factors of the product give sqrt(2) correct to more than 500 decimal places. (End)

%p a := proc (n) option remember; if n = 1 then 5 else a(n-1)^3 + 3*a(n-1)^2 - 3 end if; end proc:

%p seq(a(n), n = 1 .. 5); # _Peter Bala_, Jan 19 2022

%Y For denominators see A006272. Cf. A002814, A006266, A006273, A006275, A006276.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_.

%E Previous values for a(3) and a(4) were 776 and 1797. They have been merged into 7761797 to reflect the 2nd continued fraction on page 6 of Shallit paper by _Michel Marcus_, Apr 29 2013

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Last modified April 25 13:27 EDT 2024. Contains 371971 sequences. (Running on oeis4.)