1 hour ago · Science · hide · 0 comments

The previous post looked at how many digits are in the reduced fraction for the nth harmonic number. I was curious about how long the cycle of digits in a harmonic number might be. I wrote about the period length for the digits of fractions almost a decade ago. This post includes code so I can apply it to harmonic demoninators. from sympy import lcm, factorint, n_order def period(n): factors = factorint(n) exp2 = factors.get(2, 0) exp5 = factors.get(5, 0) r = max(exp2, exp5) d = n // (2**exp2 * 5**exp5) s = 1 if d == 1 else n_order(10, d) return (r, s) This function returns two numbers: r is the number of non-repeating digits at the beginning and s is the length of the repeating part. The following code from functools import reduce def lcm_range(n): return reduce(lcm, range(1, n + 1)) print( period( lcm_range(50) ) ) prints (5, 1275120) meaning that 1/lcm(1, 2, 3, …, 49, 50) has five non-repeating digits following by 1,275,120 digits that repeat ad infinitum. And so the decimals in…

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