OFFSET
1,1
COMMENTS
Suppose that f(n) is a strictly decreasing sequence of positive numbers having limit 0. Let g = (inverse of f). An algorithm for signed Egyptian fractions for r follows:
Let a(1) be the least positive integer k such that f(1)+...+f(k) >= r. Let p(r) = f(1)+...+f(a(1)). If p(r) = r the algorithm terminates.
Otherwise, let a(2) be the greatest k such that p(r)-f(k) <= r, so that a(2) = floor[g(p(r)-r)]. If p(r)-f(a(2)) = r, the algorithm terminates.
Otherwise, let a(3) be the least k such that p(r)-f(a(2))+f(k) >= r, so that a(3) = floor[g(r-p(r)+f(a(2)))]. If p(r)-f(a(2))+f(a(3)) = r, the algorithm terminates.
Otherwise, the algorithm continues inductively:
If n is odd, let a(n+1) be the greatest k such that p(r)-f(a(2))+f(a(3))-...+f(a(n))-f(k) <= r, so that a(n+1) = floor[g(p(r)-r-f(a(2))+f(a(3))-...-f(a(n)))]. If p(r)-f(a(2))+ f(a(3))- ...-f(a(n)) = r, the algorithm terminates.
If n is even, let a(n+1) be the least k such that
p(r)- f(a(2))+f(a(3))-...-f(a(n))+f(k) >= r, so that a(n+1) = floor[g(p(r)-r-f(a(2))+ f(a(3))-...+f(a(n)))]. If p(r)-f(a(2))+f(a(3))-...+f(a(n)) = r, the algorithm terminates.
Taking r = (1+sqrt(5))/2 and f(n) = 1/n gives the present sequence, and
r = 1/1 + 1/2 + 1/3 - 1/4 + 1/28 - 1/986 + 1/a(5) - 1/a(6) + ...
The 14th partial sum differs from r by less than 10^(-1000).
In the following guide to related sequences, EM represents the Euler-Mascheroni constant.:
r ............... f(n) ... a(n)
(1+sqrt(5))/2 ... 1/n .... A226049
e ............... 1/n .... A226050
pi .............. 1/n .... A226051
sqrt(2) ......... 1/n .... A226052
sqrt(1/2) ....... 1/n .... A226053
EM ...............1/n .... A226058
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..12
EXAMPLE
1 + 1/2 < r < 1 + 1/2 + 1/3, where r = golden ratio = (1 + sqrt(5))/2; so a(1) = 3.
1 + 1/2 + 1/3 - 1/4 < r, so a(2) = 4.
1 + 1/2 + 1/3 - 1/4 + 1/27 < r < 1 + 1/2 + 1/3 - 1/4 + 1/28, so a(3) = 28.
MATHEMATICA
$MaxExtraPrecision = Infinity;
nn = 10; f[n_] := 1/n; r = GoldenRatio; s = 0; b[1] = NestWhile[# + 1 &, 1, ! (s += f[#]) > r &]; u[1] = Sum[f[n], {n, 1, b[1]}]; c[1] = Floor[1/(u[1] - r)]; v[1] = u[1] - 1/c[1]; n = 1; While[n < nn/2, n++; b[n] = Floor[1/(r - v[n - 1])]; u[n] = v[n - 1] + 1/b[n]; c[n] = Floor[1/(u[n] - r)]; v[n] = u[n] - 1/c[n]]; a = Riffle[Table[b[i], {i, 1, nn/2}], Table[c[i], {i, 1, nn/2}]]
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Clark Kimberling, May 24 2013
STATUS
approved