**Continued Fractions**which served as

*my*introduction to the topic around 1967.

The sequence [3; 7, 15, 1, 292, 1, 1, ...] is of course the start of the continued-fraction expansion for π. I wish C.D. Olds had used (and

*insisted on*) that semicolon (instead of a comma) after the initial 3. The book shows us how to calculate from these numbers

*convergents*to π, numerators and denominators of the convergents being derived separately. For numerators we pretend things start with [0, 1, ...] and for denominators, [1, 0, ...]. The numerator- and denominator-sequences are then calculated by taking an entry from the continued-fraction sequence (starting with 3),

*multiplying*it by the previous number, and

*adding*the previous-to-previous number.

So, for numerators: [0, 1, 3*1+0 = 3; 7*3+1 = 22, 15*22+3 = 333, ...]

For denominators: [1, 0, 3*0+1 = 1; 7*1+0 = 7, 15*7+1 = 106, ...]

Dropping our pretend prepends and dividing: [3/1; 22/7, 333/106, 355/113, ...]. These are the well-known convergents to π. Note that the convergents alternate in their relation to π: [<π; >π, <π, >π, ...]. Subtracting π: [<0; >0, <0, >0, ...]. Or, if you will, their

*sign*: [-1; 1, -1, 1, ...].

One can do the same thing for τ (pronounced

*tau*), π's less well-known parent (more commonly referred to as 2π). The continued-fraction expansion for τ is [6; 3, 1, 1, 7, 2, 146, ...]. The convergents are: [6; 19/3, 25/4, 44/7, ...]. Like π, these convergents alternate in their relation to τ: [<τ; >τ, <τ, >τ, ...], their sign likewise.

What happens when one divides the convergents to τ by the convergents to π? The resulting sequence is [2; 133/66, 1325/666, 4972/2485, ...], numbers very close or equal to 2. To generate their sign, subtract 2: [0; 1/66, -7/666, 2/2485, ...], hence [0; 1, -1, 1, ...]. The twist is that the sign sequence is now littered with zeros (in seemingly

*random*— but decidedly

*clustered*— locations). How do those zeros affect the sign alternation? The surprising (to me) answer is: not at all. Remove all zeros from Convergents[τ]/Convergents[π]-2 and what remains will alternate sign, regardless of the number or placement of the previously intervening, single-or-consecutive zeros.

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