The Millin series is defined as:

$$\sum_{n \mathop = 0}^\infty \frac 1 {F_{2^n} }$$

where $F_n$ denotes the $n$th Fibonacci number.

It can be shown to equal $\dfrac {7 - \sqrt 5} 2$.

But who was the D.A. Millin who it is named after?

EDIT: User https://hsm.stackexchange.com/users/16591/michael has located the issue of FQ in which Millin's name originally appears, where he was identified as a Pennsylvanian high-school student in 1974.

Further to this, I have found that the solution appeared in FQ issue Vol. 14 no. 2 (1976), but in this case his name appears as D.A. Miller.

The question arises as to whether Millin might have been a misprint. If his name truly is "Miller", then his precise identity may be very difficult to track down. There is a professor in Virginia with that name, but he appears a couple of decades too young.

Whoever he is, he may well be currently active.


2 Answers 2


I am the author of the Advanced Problem H-237 to the FQ issue Vol. 14 no. 2 (1976). The series in that Problem was later named the Millin series. At that time, I was a senior at the Annville-Cleona High School in Pennsylvania. This series was also part of the paper titled "Observations in Pure Mathematics" that I submitted to the 1974 Westinghouse Science Talent Search (now the Regeneron Science Talent Search). In the days before LaTeX, I wrote my submission to FQ using a combination of handwriting and a typewriter. Apparently, my handwritten signature was not clear and was misread at the time.

Frankly, I like the name "the Millin Series," and I hope no one makes an effort to change it. For my proof of this identity, see the last entry on this webpage

Dale A Miller, Research Scientist, Inria, France


The furthest back I came is this 1974 article from The Fibonacci Quarterly 12, No. 3.

Extract: enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.