Ed-- Either this is not mysterious or I'm missing something. First of all, 5/2 is the average of 2 and 3. If the sum went from k=2 to oo, instead of 1 to oo, so that the p[k] were just the odd primes, then the 5/2 crossing would be left out, and I don't think the other crossings would be changed much. Second, it looks like that formula will have zeroes near the averages of adjacent elements of *any* sorted sequence. Have you tried substituting (e.g.) the Fibonacci sequence starting at the second 1? You should get crossings at... 3/2, 5/2, 8/2, 13/2, 21/2 ... (kinda interesting itself). I would guess the accuracy (-log of error) goes up like the distance between neighbors? Or maybe distance squared? Have you plotted that? (I don't have Mathematica.) --Steve
From: Ed Pegg Jr <ed@mathpuzzle.com>
Daniel Tisdale (daniel6874@gmail.com) sent MathWorld an interesting item.
"The sum $f(x)=sum_(k=1)^infty(-1)^ke^(-(x-p_k)^2)$ has as zeros precisely the interprimes, with the single additional point 5/2"
The plot agreed precisely, so I added it.?? http://mathworld.wolfram.com/Interprime.html
John Pardon (jpardon@princeton.edu)?? pointed out that it's wrong... that it gives a series of highly accurate approximations instead.
How accurate? f[x_] := N[Sum[(-1)^k*Exp[-(x-Prime[k])^2],{k,1,2500}],1000] Table[Floor[Log[10, Abs[1/f[Prime[n]/2 + Prime[n + 1]/2]]]], {n, 1, 400}]
2, 1, 3, 15, 35, 43, 21, 6, 10, 156, 10, 6, 73, 6, 21, 10, 52, 10, 6, 10, 10, 43, 21, 27, 6, 21, 111, 21, 6, 73, 27, 10, 21, 125, 21, 10, 21, 444 ...
The accuracies to the 400th Interprime have the following distribution: {{1, 1}, {2, 1}, {3, 1}, {6, 44}, {10, 61}, {15, 1}, {21, 70}, {27, 47}, {35, 23}, {43, 24}, {52, 32}, {62, 6}, {73, 23}, {85,13}, {111, 14}, {125, 19}, {156, 10}, {173, 1}, {210, 1}, {229, 2}, {271, 2}, {340, 1}, {417, 1}, {444, 2}}
Just the accuracies, in numbers of digits {1, 2, 3, 6, 10, 15, 21, 27, 35, 43, 52, 62, 73, 85, 111, 125, 156, 173, 210, 229, 271, 340, 417, 444}
Differences: 1, 1, 3, 4, 5, 6, 6, 8, 8, 9, 10, 11, 12, 26, 14, 31, 17, 37, 19, 42, 69, 77, 27
For starters, 444 digits of accuracy has occurred more than once.? 52 digits of accuracy has occurred 32 times. Something seems to be happening here.?
Ed Pegg Jr