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A054404
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Number of daughters to wait before picking in sultan's dowry problem.
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1
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0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 26, 26, 26, 27, 27, 27
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OFFSET
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1,5
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COMMENTS
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The correct rule can be found in the Gardner reference (p. 60) and in the Wikipedia article (see link): if the number of candidates is n, then the optimal r (the number of candidates to skip) is the r that maximizes (r/n)(1/r+1/(r+1)+...+1/(n-1)). - Zvi Mendlowitz (zvi113(AT)zahav.net.il), Jul 12 2007
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REFERENCES
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M. Gardner, My Best Mathematical and Logic Puzzles, Dover, 1994
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LINKS
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Table of n, a(n) for n=1..75.
Eric Weisstein's World of Mathematics, Sultans Dowry Problem.
Wikipedia, Secretary problem.
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FORMULA
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a(n) = the integer r that maximizes (r/n)(1/r+1/(r+1)+...+1/(n-1)) - Zvi Mendlowitz (zvi113(AT)zahav.net.il), Jul 12 2007
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MATHEMATICA
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a[n_] := r /. Last[ Maximize[ {(r/n)*Sum[1/k, {k, r, n - 1}], 0 <= r < n/2}, r, Integers]]; a[1] = 0; a[2] = 1; Table[a[n], {n, 1, 75}] (* Jean-François Alcover, Dec 13 2011, after Zvi Mendlowitz *)
(* The code above may not work in Mma 8 *)
PR[n_, r_] := (r/n)*Sum[1/k, {k, r, n - 1}];
maxi[li_] := {Do[If[li[[n + 1]] <
li[[n]], aux = n; Break[]], {n, 1, Length[li] - 1}], aux}[[2]];
SEQ[1] = 0; SEQ[2] = 1; SEQ[n_] := maxi[Table[PR[n, i], {i, 1, n - 1}]];
Table[SEQ[n], {n, 1, 133}] (* José María Grau Ribas, May 11 2013 *)
a[1]=0; a[2]=1; a[n_] := Block[{r}, r /. Last@ Maximize[{(r/n) * (PolyGamma[0, n] - PolyGamma[0, r]), 1 <= r < n/2}, r, Integers]]; Array[a, 75] (* Giovanni Resta, May 11 2013 *)
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CROSSREFS
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Sequence in context: A026414 A077219 A026405 * A008671 A199017 A189709
Adjacent sequences: A054401 A054402 A054403 * A054405 A054406 A054407
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KEYWORD
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nonn,changed
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AUTHOR
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Eric W. Weisstein
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EXTENSIONS
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Corrected by Zvi Mendlowitz (zvi113(AT)zahav.net.il), Jul 12 2007
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STATUS
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approved
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