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Search: id:A114604
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| 1, 5, 43, 709, 23003, 1481957, 190305691, 48796386661, 25003673060507, 25613941912987493, 52467767892904362139, 214929296497738201165669, 1760788099067877263041671323, 28849467307107603960961499533157
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OFFSET
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0,2
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COMMENT
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To win a game, you must flip n+1 heads in a row, where n is the total number of tails flipped so far. The probability of having won before n+1 tails (that is, winning by flipping n+1 or fewer heads in a row) is A114604 / A006125 The probability of winning for the first time after n tails (that is, by flipping n+1 heads in a row) is A005329 / A006125.
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FORMULA
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a(n) = numerator of sum from k = 0 to n of A005329/A006125. a(n) = a(n-1) * 2^(n+1) + A005329(n)
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EXAMPLE
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a(3) = 43 because 1/2 + 1/8 + 3/64 = 43/64, or because a(2) * 2^(2+1) + A005329(2) = 5 * 8 + 3 = 43.
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CROSSREFS
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Cf. A005329, A006125.
Sequence in context: A162695 A161635 A005989 this_sequence A085098 A099794 A142726
Adjacent sequences: A114601 A114602 A114603 this_sequence A114605 A114606 A114607
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KEYWORD
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easy,frac,nonn
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AUTHOR
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Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), Dec 14 2005
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