%I #27 Nov 06 2024 10:39:03
%S 2,3,9,74,8098,101114070,10080916639334518,
%T 234737156891222571756748160861129,
%U 104728182461244680288139397973895577148266725366426255244889745185
%N Denominators of r-Egyptian fraction expansion for sqrt(1/2), where r = (1,1/2,1/3,1/4,...)
%C Suppose that r is a sequence of rational numbers r(k) <= 1 for k >= 1, and that x is an irrational number in (0,1). Let f(0) = x, n(k) = floor(r(k)/f(k-1)), and f(k) = f(k-1) - r(k)/n(k). Then x = r(1)/n(1) + r(2)/n(2) + r(3)/n(3) + ... , the r-Egyptian fraction for x.
%C Guide to related sequences:
%C r(k) x denominators
%C 1 sqrt(1/2) A069139
%C 1 sqrt(1/3) A144983
%C 1 sqrt(2) - 1 A006487
%C 1 sqrt(3) - 1 A118325
%C 1 tau - 1 A117116
%C 1 1/Pi A006524
%C 1 Pi-3 A001466
%C 1 1/e A006526
%C 1 e - 2 A006525
%C 1 log(2) A118324
%C 1 Euler constant A110820
%C 1 (1/2)^(1/3) A269573
%C .
%C 1/k sqrt(1/2) A269993
%C 1/k sqrt(1/3) A269994
%C 1/k sqrt(2) - 1 A269995
%C 1/k sqrt(3) - 1 A269996
%C 1/k tau - 1 A269997
%C 1/k 1/Pi A269998
%C 1/k Pi-3 A269999
%C 1/k 1/e A270001
%C 1/k e - 2 A270002
%C 1/k log(2) A270314
%C 1/k Euler constant A270315
%C 1/k (1/2)^(1/3) A270316
%C .
%C Using the 12 choices for x shown above (that is, sqrt(1/2) to (1/2)^(1/3)), the denominator sequence of the r-Egyptian fraction for x appears for each of the following sequences (r(k)):
%C r(k) = 1 (see above)
%C r(k) = 1/k (see above)
%C r(k) = 2^(1-k): A270347-A270358
%C r(k) = 1/Fibonacci(k+1): A270394-A270405
%C r(k) = 1/prime(k): A270476-A270487
%C r(k) = 1/k!: A270517-A270527 (A000027 for x = e - 2)
%C r(k) = 1/(2k-1): A270546-A270557
%C r(k) = 1/(k+1): A270580-A270591
%H Clark Kimberling, <a href="/A269993/b269993.txt">Table of n, a(n) for n = 1..12</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>
%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%e sqrt(1/2) = 1/2 + 1/(2*3) + 1/(3*9) + ...
%t r[k_] := 1/k; f[x_, 0] = x; z = 10;
%t n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
%t f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
%t x = Sqrt[1/2]; Table[n[x, k], {k, 1, z}]
%o (PARI) r(k) = 1/k;
%o x = sqrt(1/2);
%o f(x, k) = if(k<1, x, f(x, k - 1) - r(k)/n(x, k));
%o n(x, k) = ceil(r(k)/f(x, k - 1));
%o for(k = 1, 10, print1(n(x, k),", ")) \\ _Indranil Ghosh_, Mar 27 2017, translated from Mathematica code
%Y Cf. A269573, A069139, A270347, A270394, A270476, A270517, A270546, A270580.
%K nonn,frac,easy
%O 1,1
%A _Clark Kimberling_, Mar 15 2016