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A226049 Denominators of signed Egyptian fractions with sums converging to the golden ratio. 12
3, 4, 28, 986, 1759060, 4906125653220, 26098452883705403280615128, 1041141217110191230944018432385108773652321818272607 (list; graph; refs; listen; history; text; internal format)
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

REFERENCES

Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3, Fall 1998, p. 176.  Solution published in Vol. 12, No. 1, Winter 2000, pp. 61-62.

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

Cf. A226050.

Sequence in context: A140896 A005326 A298561 * A100600 A288868 A076001

Adjacent sequences:  A226046 A226047 A226048 * A226050 A226051 A226052

KEYWORD

nonn,frac

AUTHOR

Clark Kimberling, May 24 2013

STATUS

approved

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Last modified December 13 17:24 EST 2018. Contains 318086 sequences. (Running on oeis4.)