MATHEMATICS ON SEQUENCES OF INTEGERS GENERATED BY A SIEVING PROCESS BY PAUL ERDŐS ANa ERI JABOTINSKY (Communicated by Prof. J. POPKEN at the meeting o

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1 MATHEMATICS ON SEQUENCES OF INTEGERS GENERATED BY A SIEVING PROCESS BY PAUL ERDŐS ANa ERI JABOTINSKY (Commuicated by Prof. J. POPKEN at the meetig of Jue 29, 1957) (29) PART II 4. The secod term o f the asymptotic expasio for ak (for bk = ak ad ay A > 1) Usig formula (4) ad (26), we shall prove that : (27) 7, _ 1 a a2 1 O(1). a, <k i Ideed, the umber Q i (4) is defied as the smallest iteger for which ak_q<q+1, whece, by (26) : (28) k-q= [I+0(1)] k log k' (28') Formula (4) ow becomes : akk -A[ T, (a 1)] C1 log k~ a2 with 0 < B <2. ai S Q a g But it ca be see by usig (26) ad (28) that : a2 0r < k ai - 1)=O[ k ( + r log r)] + o ((log k)z)' t k--<rlosr<k log k ad further from (21) ad (26) : (29') a, < a, =(1+0(1)) log k. a <Qai-1 ai< kai-1 Thus (27) follows from (28'), (29) ad (29'). Now we wat to prove that (30) a,,= log +(2+o(1)) (log log ) 2. We will omit some of the details. Put : (31) a = log + 2 (log log )2+ /() log log.

2 To prove (30) we must prove that : 125 (32) /()=o (log log ). First we show that for every e > 0 ad > o (e) (33) /() < e log log. The proof of /() > - e log log would be similar. If (33) would ot hold, a simple argumet shows that there would exist two ifiite sequeces k ad m k satisfyig : (34) mk' < k < Mk, l(mk) > J (k) +e, f(mk) > AU), 1 < u < mk, /(k) < Ak+v), 0 < V < Mk -k. By (26) ad (27) we have : (35) a ji 1 (1-á) +0(1)=[ 11 (1+~)+0()]+0(1)= ~a;,< m ~ 5ai < m<m ~ = lj (1+ ) } 0(1). -<ai<m a Hece from (31) by puttig m=mk ad =k i (35), for some c>e : log m+ 21 (log log m) 2 + (/() +c) log log m= (36) _ [log + 2(log log ) 2 +/() log log ] fj (1 + á2)+0(1). <_a i <m Now we show that for < ai< m (37) a i/i>log i+2(log log i)2 + (/()+o(1)) log log i. For i > this follows from the defiitio of i. For the ai satisfyig < ai < a it follows from (35) ad ai = (1 + o (l)) i log i by a simple computatio. Suppose ow that (33) does ot hold. The, from (35) ad (36), we have /(m)=/()+c ad : log m -{- 2 (log log m) 2 I [/ () -G- c] log log m < (38) < log +'(log log ) 2 +l() -log log - IT (1 + g( ~~ ) ), _<ak <m where g(k)=k[log k+2(log log k) 2 +If(k)-i-o(1)1 log log k]. Put m=1 +s. I the computatio which follows we will eglect terms which are o(log log ), or i estimatig the product o the right side of (38) we ca eglect terms which are o(log log /log ). We have (the equality sig is to be uderstood to mea that terms which are o(log log /log ) have bee eglected) (39) 1 [1 1 _ ~ 1,I _<ak<l+ő `1+g(k» ~klogk<m1+ő+ g(kj exp `'_<kloak<mg(]c~,

3 1 2 6 Heceforth it is to be uderstood that i all the products < k log k < l+s. We have : ad sums (log log k) 2 log log k g(k) - k log k 2 k (log k) 2 - f() k log log k)2 ' Further clearly by the itegral test Thus k 1 log k =log (, + 6)+ log log 8 log 1 +8' (loglogk) 2 _ < 5(log log.) 2-2 log (1 +6) log log 2 6 log log k(log k)2 (1 +6) log log ' log log k S log log -l k(log k) log, 6 (log log ) 2 1 log ( 1 -I-6)- ~g-(-k) 2(1-1-6) log log (1 + S) log log -1() 6 log log. (1 +S)log l+d log Hece from (39) : r 1 ~ 1 l 8 (loglog ) = exp 2 ~L 1+ 9(k) 1 ~~g(k)j - (1+b) 2 log? (40) Thus if we put m= 1 +' i (38) we obtai from (40) : log (1 +d) log log log log log - 8 f() log (1 +8) log e+2 [log log +log (1 x-8)]2+[f()+c] [log log +log (1+6)] < + < [log + 2 (log log ) 2 + /() log log ] [ (o ólgo) 2 + log (1+ 6) log log log log I + log ~f() log which is easily see to be false because of the ucacelled term c log log o the left side of the iequality (sice the coefficiet of /() is greater o the left side tha o the right side). 5. The third term o f the asymptotic expasio o f a, (for b, = a,l ad ay 2 > 1) We ote that formula (27) was obtaied by usig oly step oe for the computatio of a, that is by usig formula (4) ad ot formula (6). It is ot possible to get the ext term without usig steps of all orders. To do this, we have to calculate successively for m=1, 2,... all the q,, occurrig i (6). q, is defied as the smallest iteger for which : m-1 m (ax-v - 1) < mq i -- qb. =o Because of (26) it ca be see that : q =k- mlogk +o (logkj'

4 ad (6) becomes : 1 1 a k - ~ = [ fl ( ai - a. )] [k+~ ~...+m )lo k k i < m log k +0( 109k) i g %i with 0<0<1. Writig 0m istead of 0 ad usig (26) this becomes : ak -~ -~-~ ( ai k l k ai m-1 m log k (41) ai< mlogk )] [1+( m)1 with 0 < dm < 1. We ow use the fact that, because of (26) : (42) IT (2i k k (ai- 1) m log k < "i < (m. 1) log k M logm-1 I 1 + log k + 0 (log k) ( k k + log k)]' Rewritig (41) for (m -1) istead of m ad comparig with (41), we fid, usig (42) : m) 1 2 m- m-1 m-1 log 1 k - 1+( log ad so : =1 }(-1 } 1 I 2 I 0m M 6"' (43) -1 + log = + 1 +o(')- M-1 am-1 m m-1 l l + em)logk 1 0 (log J' Rewritig (43) for (m-1), (m-2),..., 2 istead of m, summig up ad cacellig we fid (44) 1 + log m = Bm m 11 +0(1). 1+0 (log k)1' But 0 < 0,m < 1 ad for large k ad m (44) ca oly hold if Thus : (45) 01 =Y+0(1). lim 01 =Y. k--> oo Formula (41) ow becomes (46) k- [T (aiati 1 )] I (-l+y+ log m)+o(1). k ai<- m ad, for m=1 : (47) k - [ ( aa ai 1)] +(-1+Y)+0(1).

5 1 2 8 We ote that (45) ad (26) together yield (46), so that (47) cotais all the iformatio that results from the use of formula (6) with the estimate qm = k - m log k + o (log k) of qm. Ay improvemet of the o(1) term i (47) ca oly result from a improvemet of these estimates of qm. Put ow k = log k + 2 (log log k) 2 ;- (2-y) log log k+ f (k). The we ca show by the same method as was used i provig (30), but by more laborious computatios that f(k) = o(log log k) : we supress all details. This completes the proof of the result stated at the ed of the itroductio. Several further questios ca be asked about the ak all of which have bee ivestigated for the sequece of primes e.g. Is it true that lim if (ak+, - a k ) < oo? Is it true that lim sup (a k+1 -ak )/log k=oo, 1 ) we do ot kow the aswer to ay of these questios. After writig our paper we fid that the quadruple paper of V. GARDINER, R. LAZARUS, N. METROPOLIS ad S. ULAM deals with a slight variat of our case bk =ak, they make a table of these umbers up to (Math. Magazie 29 (1956), ). They further cojecture ak /p -~ 1. HAWKINS proved this cojecture ad CHOWLA proved a k =k log k+(2+o(1)) k(log log k) 2, the proofs of HAWKINS ad CHOWLA are ot yet published. Added March VIGGO BRUN asked the followig questio : Put = l,,+1= t - [l/ l]. Determie the smallest iteger k for which k+l=k (i.e. for which k+ 1 > k ). By the methods used i i dealig with the case b k =k+ 1 we ca prove that k=(i+0(1)) (.~C 2 /8) 7/a. DAVID, i a paper to appear i Riveo le Matematika, vol_ 11, cosiders the sequece ul =, uk =k [ Uk 1 ] ad asks whe uk = 0. This reduces to our problem for bk = k + 1. Israel Istitute of Techology, Haifa 1 ) The fact that lim sup (p7. 1-Pk)/iogPk=O is due to WESTZYNTHIUS, See P. ERDÖS, Quarterly Joural of Math. 6, (1934). "M '11 f (Pk+1- Pk)< has ever bee roved.