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3 Oct 2014: 2g(V ) 2 = n(2g(W) 2) PV. (eP 1). 20. This result enables us to interpret the genus topologically (non-examinable).

17 Mar 2014: 314 32 185 44 91 21 52 17. 556 24 158 38 140 20 72 15. ... NA NA 150 20 59 5 90 8. NA NA 148 19 87 4 50 5.

13 Oct 2014: log d, with the speculation then being phrasedin a simple form (though see Footnote 32, and compare [20]). ... 5 10 15 20 25 30. Figure 2. Comparison of F̂ (t) and B̂10(t) for larger values of t.

18 Jul 2014: of degrees 20 and 30. By Theorem 2.1 it only remains to show they are scaled as. ... The rest is[15, Lemma 5.2.31 and Theorem 5.3.20]. 2. Theorem 4.5.

14 Oct 2014: 2. Between 0 and 10 there are four primes. Another example of two consecutive multiples often, between which there are four primes, is 10 and 20.

18 Sep 2014: all i. So f(g) / f(A). 20.

18 Sep 2014: So all leaves in C(Y ) havecyclic fundamental group. 20. Now let L be a closed vertical loop in an annulus C(Y ). Then the image ofπ1(L) in G is

20 Jan 2014: Only a few minutes ago six shells fell about 20 yards frommy dug-out amongst our waggons and last night a shell fuse came into it. ... Letter dated April 27th 15,1/5h York and Lancaster Rgt. Colver Letters, typed by P.M.E.Altham 20.

19 Dec 2014: M[H] such that M[H] |= (20 = λ and there is a Suslin tree).(iv) Suppose that M |= (P is a c.c.c. ... MM requires a large cardinal axiomfor its consistency. 20 Optional. Normal Functions and Mahlo Cardinals.

18 Sep 2014: 20. 1. 1 Introduction. Let M be a compact surface with a geometric structure. ... 20. 2. If χ(M) = 0 then M is a torus and admits a Euclidean structure.