OFFSET
0,1
COMMENTS
With b(n) = floor((1+sqrt(2))^n) (cf. A080039) the terms appear to be b(2*3^n). - Joerg Arndt, Apr 29 2013
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
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
F. L. Bauer, Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.
Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.
FORMULA
From Peter Bala, Jan 18 2022: (Start)
a(n) = (3 + 2*sqrt(2))^3^(n-1) + (3 - 2*sqrt(2))^3^(n-1) - 1 for n >= 1.
a(n) = A006266(n)^2 + 1 for n >= 1.
a(1) = 5 and a(n) = a(n-1)^3 + 3*a(n-1)^2 - 3 for n >= 2.
a(1) = 5 and a(n) = 8*(Product_{k = 1..n-1} a(k))^2 - 3 for n >= 2.
2 - Product_{n = 1..N} (1 + 2/a(n))^2 = 8/(a(N+1) + 3). Therefore
sqrt(2) = (1 + 2/5) * (1 + 2/197) * (1 + 2/7761797) * (1 + 2/ 467613464999866416197) * ... - see Bauer.
The convergence is cubic - see Fine. The first six factors of the product give sqrt(2) correct to more than 500 decimal places. (End)
MAPLE
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:
seq(a(n), n = 1 .. 5); # Peter Bala, Jan 19 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
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
STATUS
approved