OFFSET
1,1
COMMENTS
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.
Guide to related sequences:
r(k) x denominators
1 sqrt(1/2) A069139
1 sqrt(1/3) A144983
1 sqrt(2) - 1 A006487
1 sqrt(3) - 1 A118325
1 tau - 1 A117116
1 1/Pi A006524
1 Pi-3 A001466
1 1/e A006526
1 e - 2 A006525
1 log(2) A118324
1 Euler constant A110820
1 (1/2)^(1/3) A269573
.
1/k sqrt(1/2) A269993
1/k sqrt(1/3) A269994
1/k sqrt(2) - 1 A269995
1/k sqrt(3) - 1 A269996
1/k tau - 1 A269997
1/k 1/Pi A269998
1/k Pi-3 A269999
1/k 1/e A270001
1/k e - 2 A270002
1/k log(2) A270314
1/k Euler constant A270315
1/k (1/2)^(1/3) A270316
.
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)):
r(k) = 1 (see above)
r(k) = 1/k (see above)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..12
Eric Weisstein's World of Mathematics, Egyptian Fraction
EXAMPLE
sqrt(1/2) = 1/2 + 1/(2*3) + 1/(3*9) + ...
MATHEMATICA
r[k_] := 1/k; f[x_, 0] = x; z = 10;
n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
x = Sqrt[1/2]; Table[n[x, k], {k, 1, z}]
PROG
(PARI) r(k) = 1/k;
x = sqrt(1/2);
f(x, k) = if(k<1, x, f(x, k - 1) - r(k)/n(x, k));
n(x, k) = ceil(r(k)/f(x, k - 1));
for(k = 1, 10, print1(n(x, k), ", ")) \\ Indranil Ghosh, Mar 27 2017, translated from Mathematica code
KEYWORD
nonn,frac,easy
AUTHOR
Clark Kimberling, Mar 15 2016
STATUS
approved