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 A343015 Decimal expansion of the probability that at least 2 of 23 randomly selected people share a birthday, considering leap years. 0
 5, 0, 6, 8, 7, 6, 0, 9, 3, 1, 6, 5, 2, 7, 8, 4, 5, 5, 2, 2, 2, 4, 3, 9, 3, 1, 3, 1, 6, 0, 5, 1, 1, 2, 3, 7, 7, 7, 3, 5, 2, 6, 9, 9, 8, 2, 5, 4, 8, 5, 2, 6, 1, 0, 5, 6, 1, 9, 4, 1, 2, 1, 4, 3, 8, 1, 4, 1, 3, 7, 2, 5, 8, 4, 6, 7, 8, 6, 3, 3, 5, 4, 8, 4, 9, 5, 1 (list; constant; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS The usual solution of the Birthday Problem, 1 - ((365!)/((365 - 23)! * 365^23)) = 0.507297... (A333507), is based on the assumption that all the years have 365 days. The solution given by Nandor (2004) includes leap years, i.e., 97 years of 366 days in each cycle of 400 years of the Gregorian calendar. With the addition of leap-year days, i.e., the possibility of having a birthday on February 29, the probability is reduced to 0.506876... This constant is a rational number: its numerator and denominator have 111 and 112 digits, respectively. The sequence has a period of 7.983424...*10^108. LINKS M. J. Nandor, Including Leap Year in the Canonical Birthday Problem, The Mathematics Teacher, Vol. 97, No. 2 (2004), pp. 87-89. Eric Weisstein's World of Mathematics, Birthday Problem. Wikipedia, Birthday problem. Wikipedia, Gregorian calendar. FORMULA Equals 1 - (365!/((365 - 23)! * 365^23)) * (146000/146097)^23 * (1 + 97 * 365 * 23/146000/(366 - 23)). EXAMPLE 0.50687609316527845522243931316051123777352699825485... MATHEMATICA RealDigits[1 - (365!/((365 - 23)! * 365^23)) * (146000/146097)^23 * (1 + 97 * 365 * 23/146000/(366 - 23)), 10, 100][[1]] CROSSREFS Cf. A011763, A014088, A333507. Sequence in context: A320375 A200419 A271522 * A069206 A291800 A091685 Adjacent sequences:  A343012 A343013 A343014 * A343016 A343017 A343018 KEYWORD nonn,cons AUTHOR Amiram Eldar, Apr 02 2021 STATUS approved

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Last modified September 28 06:57 EDT 2021. Contains 347703 sequences. (Running on oeis4.)