On 07/01/2017 16:21, Simon Plouffe wrote:
Hello,
each image of rank k is : the k'th position of n*n*phi n is from 1 to 16000000 disregarding the decimal point.
Oh! So, just to be absolutely clear: the pixel at (p,q) in image k contains the k'th decimal digit of (4000q+p)^2 phi. If that's right then I take it he "breakpoints" in the pattern happen where the number of digits before the decimal point changes. It would surely make more sense to number the digits relative to the decimal point rather than the first nonzero digit, in which case I bet the sudden changes in pattern would go away. So what we have then is a plot of, in effect, frac(Kn^2 phi) where, for a given plot, K is fixed; and we are seeing kinda-periodic dependence on n. The plots have K = 10^k, but I bet you would see something similar with K not a power of 10. And they plot "first digit after the decimal point of Kn^2 phi" rather than frac(Kn^2 phi) but obviously those plots look very similar. So the question (or at least *a* question) is: why should there be this kinda-periodic dependence of frac(Kn^2 phi) on n? It seems we have something a little like K[(n+t)^2-n^2]phi ~= integer for "many" n and fixed t, though I think this is too crude a description of what we see. That's the same as K[2tn+t^2]phi ~= integer. Perhaps what we actually have is K[(n+t)^2-n^2]phi ~= integer+const in which case the constant would come from Kt^2 phi, and the relevant fact would be that 2Knt phi ~= integer for "many" n, which would e.g. happen if 2Kt phi ~= integer. Of course phi is famously not very well approximated by rational numbers, but *any* number has rational approximations; the best ones for phi are ratios of consecutive Fibonacci numbers; do we have 2K.period a multiple of a "respectably-sized" Fibonacci number in each case? (If you play the same game with cubes instead of squares then the differences will be quadratic instead of linear and so you won't get periodicity.) -- g