On Sun, Oct 24, 2010 at 1:24 AM, Bill Gosper <billgosper@gmail.com> wrote:
Bill Gosper:
Has anyone listed the largest terms in Hans's data?
Hans Havermann:
I have. :)
For his 13th birthday, funster Neil Bickford has just computed and verified 485M partial quotients (http://www.neilbickford.com/picf.htm). The longest previous CF we can find remains Hans Havermann's 180M, wherein Neil found two glitches--a "q" instead of a "1" at position 4,311,037 (a picked 2^6 bit) and a semicolon following term 100,000,000.
The probably biggest news is the survival of 878783625 as the largest known partial quotient, seeming to signal the end of a pattern of "premature" high water marks, assuming the lg 1+t distribution for the tails of almost all reals. If pi is to maintain its reputation, the next high should be a doozy. (I dimly recall deriving that lg 1+t implies that, in an n term burst, the largest should be about e*n. Can anyone help here?)
The Lord heps they that heps theyselves. The probability that a generic tail of a CF is > k is
(c43) p(k) = logb(2,1+1/k) 1 log(- + 1) k (d43) p(k) = ---------- log(2) Therefore, the probability that the max of n independent terms exceeds k is (c44) 1-(1-p(k))^n n (d44) 1 - (1 - p(k)) E.g., for three terms, (c45) makelist(float(1-(1-log(1/k+1)/log(2))^3),k,1,9) (d45) [1.0, 0.92851, 0.79984, 0.68824, 0.59974, 0.5298, 0.47375, 0.42806, 0.39021] The probability that the max of three *consecutive* terms exceeds k is (c46) p(k)+sum(p(j)-p(1/k+j),j,1,k-1)+sum(sum(p(1/(1/k+j)+i)-p(1/j+i),i,1,k-1),j,1,k-1) k - 1 ==== \ 1 (d46) p(k) + > (p(j) - p(- + j)) / k ==== j = 1 k - 1 k - 1 ==== ==== \ \ 1 1 + > > (p(----- + i) - p(- + i)) / / 1 j ==== ==== - + j j = 1 i = 1 k E.g., for 1<=k<=9: (c47) makelist(float(apply_nouns(apply_nouns(log(1/k+1)/log(2)+sum(log(1/j+1)/log(2)-log(1/(1/k+j)+1)/log(2),j,1,k-1)+sum(sum(log(1/(1/(1/k+j)+i)+1)/log(2)-log(1/(1/j+i)+1)/log(2),i,1,k-1),j,1,k-1)))),k,1,9) (d47) [1.0, 0.94111, 0.81825, 0.70548, 0.6144, 0.54199, 0.48389, 0.43657, 0.39741] So the (false) independence assumption seems to be within about 1%. This says that the prob. that the max of n terms exceeds k*n is about (c48) 1-%e^-(1/(log(2)*k)) 1 - -------- log(2) k (d48) 1 - %e For various landmarks in pi, (c49) makelist(sfloat(1-%e^-(1/(log(2)*k))),k,[3/1,7/2,15/3,292/5,20776/432,878783625/11505030,878783625/458585379]) (d49) [0.38177, 0.33781, 0.25064, 0.0244, 0.02955, 0.0187, 0.52898] so Neil had only a 53% chance to beat my record. (Which I had only a 1.9% chance of setting.) --rwg