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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 October 22 12:30 EDT 2019. Contains 328318 sequences. (Running on oeis4.)