KONINKL. NEDERL. AKADEMIE VAN WETENSCHAPPEN AMSTERDAM Reprinted from Proceedings, Series A, 61, No. 1 and Indag. Math., 20, No.

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1 KONINKL. NEDERL. AKADEMIE VAN WETENSCHAPPEN AMSTERDAM Reprinted fro Proceedings, Series A, 6, No. and Indag. Math., 20, No., 95 8 MATHEMATIC S ON SEQUENCES OF INTEGERS GENERATED BY A SIEVIN G PROCES S B Y PAUL ERDÖS AND ERI JABOTINSK Y (Counicated by Prof. J. POPKEN at the eeting of June 29, 957 ) 0. Introduction PART I The Sieve of Erathostenes is an algorith which yields the sequence of all pries. This paper deals with a faily of soewhat siilar algorith s for creating sequences of integers. These algoriths depend on an initial integer 2 and on an auxilary sequence B of integers b ( =, 2,...) with b > 2. A faily of interediary sequences Xi ) (i =, 2,...) is fored, A(i) consisting of the integers ai >( =, 2,...). The sequence A'> is defined by : a > = 2+. The sequence A' i+l> is obtained fro the sequence A(i ) by striing out all the ters of the for ( = 0,,...) and by renaing th e reaining ters : a ±'>, a2 +>,.... Finally, the sequence A consisting o f integers a (=, 2,...) is defined by : a al. Two exaples of sieves will be considered : in the first the sequence B will be given in advance while in the second it will be deterined by the sieving process itself. In the first exaple we will tae b = +. If we choose 2=0, the first ters of the first several sequences A ' i> are (the nubers in bold are those to be struc out in the next sequence) : A'>(b=2) : Ac2>(b2=3) : A c8> (b 3 =4) : A(4)(b4=5) : A(5)(b5=6) : Therefore, in this case a =, a2 = 2, a3 = 4, a4 = 6, a, 0. Also a 6 =2, a, 8, a s = 22 and a9 = 30 because these nubers will no longer be struc out in the following sequences until each one of the reaches the hea d of the line.

2 6 In the second exaple we shall tae b,i = a. first ters of the first several sequences are : If we choose 2=, the A > (b =2) : A(2)(b2=3) : A(3)(63=5) : A(4)(b4=7) : Therefore, in this case : a = 2, a2 = 3, a3 = 5, a4 = 7 and also as =, as =3, a,=7, a5 =23, a9 =25 and a 0 =29. The following are the principal results obtained.. General explicit forulas giving a in ters of the b2(i =, 2,..., ) are found and various ore or less precise estiates of a are given under different restrictive assuptions as to the nature of the sequence B. The various estiates of a will be nown respectively as the zero-step, the one-step and the ulti-step estiate. 2. For b = + it is shown using the ore precise ulti-step estiate s that : a = For b = a it is shown that a Ic log Ic. The a are in this cas e (for every 2) asyptotic to the pries. The proof has soe siilarit y to that of the prie nuber theore and aes use of the Tauberia n self regulation of the sieving process. However, because of the greate r regularity of the present process as copared to the Erathostenes ethod, the asyptotic forula for a is obtained uch ore easily than tha t for the pries. 4. Again for b= a, using the one-step forulas, it is shown that : a2 a=l~<llc. ati a +0(), and hence it will follow that a=c log + z lc (log log ) 2 ± o(lc log log ) 2. Thus it is seen that for large we have a > p. It was surised by ER' JABOTINSKY in a paper read to the 953-eeting of the Israel Matheatica l Society, on heuristic grounds, that a p2 and that a oscillates aroun d p. The second surise is thus proven to be wrong. 5. Again for b = a using the ore precise ulti-step estiates fo r a it is shown that (y Euler 's constant ) `= H ai< a a7 ( y)+o(]). Fro this it is deduced tha t a =c log c -F 2c (log log ) 2 +(2 y) Ic log log Ic+o(Ic log log ).

3 7. General Explicit Forula s We shall denote by TX] the sallest integer > X. Considering the generation of the sequence A( i+l ) fro the sequence A, we see that : (i+) (i) bibi ~ Repeating this reasoning i ties we find that : a)+)=a~~) with '= Ib vl Ib zb2 I... b /cl... [bi. Using the fact that ag, = A+ ' and a(, i+) = ai+ we get, putting i + = : `a =íl+, () a=íl+ibóib r...rbb l l..., which is the first of the announced explicit forulas for a. We have : b, + bi. b' r b7 v' Applying this to () we deduce the following estiate for a(lc> 2) : 2 + rr-( b ) + rr( b )+ ' l i= bi ~~ i= bi i s= Li rr( b = bi ) ' or, because bvb 2 > : (2). 7r b. A+7( bz )~ ai~< A +( )fi(b i ). i= a i= This forula will be called the zero-step estiate for a. Forula () can be used for actual coputation of the a. Thus, in the case b = +, A= 0, let us for exaple copute a9. We have : Now : [ =2, r8. 2 =3, r6.3 a9 I r2 r3 r4 r5 r6 r7 r8.. 4, r5. 4 =5, r 4.5 =7, r3.7 0, r2.0 = 5 and 5 = 30. ri Therefore a9 = 30. We see that a9 is the last of the sequence of integers : 2, 3, 4, 5, 7, 0, 5, 30, which are the values of the successive bracets FT. We note that these integers first increase by, then by 2, then by ore and ore. If we new how any bracets produce an increase by, how an y by 2 and so on, we would get uch shorter expressions for a. In the general case, let Q be the sallest integer such that : b_q <Q.

4 7. 8 Then the first Q bracets give an increase by and we have fo r every q< Q : A+ I blbl I b 2 b (3 ) a = C r... [ b,cb? g Forula (3) taes into account the bracets which increase by step s of - qi... and can be called, "the explicit forula for the one-step ". By the sae toen forula () is "the explicit forula for the zero-step ". Forula (3) with q = Q leads to a second estiate for a. Naely : (4 ) Q a = A + [ -0 ( Q) TI ( b2 ), with 0 < 0, i=i which is the one-step estiate for a. Explicit forulas for steps 2, 3 and so on are conveniently established under the restrictive assuption that B is a non-decreasing sequence (b+l b for all ). This assuption happens to hold in the two particular exaples considered by us. Then the explicit forula for -step is (for =, 2,...) : (5 ) a = A + [b- [b2 b2 [... [ITb'`gg i q lqi,}..., i= o where qo = 0 and q is the sallest integer for which : (b-a,,,. ) < q q i= o The proof by induction is iediate. We note that is the nuber of bracets which give increases of n. For = forula (5 ) becoes forula (3) with Q= q. Forula (5) leads to the -step estiat e of a. Indeed, fro (5) : A+[ with : 4,,, an Q c (biba ) (q qi) at.<a+ II (bi v ) (q qi) ± R z i=o i= i= o R= (-js 4 s s= i= b-_i) ( q ['Fr(b2 bi ). and therefore : Tc 4», (6) a=2+ [ H (b.b?' z )]. [gg, o qi+0 (lc q)], with 0 < 0 <, which is the ulti-step estiate for a. Another result of a different ind which will be needed holds in the

5 particular case when li Co all is not required), naely if b~' -- oo then : 9 = oo (but the assuption that b +l > b for be (7 ) = [+ 0 ( )] ~ ( biv I). Indeed, in this case the sallest Q for which b _ Q <Q + is such that -Q=o() and inequality (4) becoes : All that has to be proven is that : - ( bi bi ) = +o() i= Q+ This will follow if we prove that : (7' ) i= Q+ ' = 0(). Choose any sall e> 0 and write : We have : [Is ] i Q+ bi i--q+ bi i=[ej+l b i [e] <LQ< s <2s, i= Q+ b i 8 because by definition, Q is the sallest integer such that b_q <Q+ and so b _ i > j for j < Q and here bi > Ic - i > Q -I-. Also fro %-~ oo we have : i=[ej+ bi which copletes the proof of (7') : 2. The case b =+, 2= 0 < =o ( ), e<i< We shall need an auxiliary sequence ai (i =, 2',...) of nubers define d by the recurrence relation : (8) ( - a'i) = i a i i= i (for =. 2,...). By a suitable cobination of relation (8) for, ( + ) and (+2), one sees readily that : ( 9 ) ( ' -) '

6 and : (0 ) It is well nown that : 2 0 Ki = 22 (2 r) i= = () 2 2 (2 ) + ' ~z) We shall now use forula (5) which becoes : [T2 r r (2) a= 2 I... n a- [7c,C q q {q i=o qi }..., where the q are defined as follows : qo = 0 and q,, is the sallest intege r for which : na ( q) < q qi. i= 0 Therefore is an integer for which : q (3) + < 2g qi < +2. i= o We now use the sequence ai and define the nubers Oa) by the relation : (4 ) qi q2 _ =ai -I-O( ) Fro (4) we deduce that : (5) and : q = (.G a i) + Oi ) x= (6) q qi= ( Z (x 20" i) + ) i= x=i i= or, using (3), (5), (6) and (8) we obtain by a siple coputation : (7) < (+i) O)'` ) < 2. i= We note that the su in inequality (7) is an integer because th e iddle ter in (3) is an integer. We shall not use this fact. Forula (2) now leads to the following estiate for a analogous to (6) : a = ( q + ) [q qi+ O (ic q )], with 0 < 0 <, i= o or, using (5), (6), (8) and (0) : ai = r (2 ) 22 B2) L i= J L -Li i [ ( 2 ) 22 ~ io )] Oa ( gna) 2, i=

7 or again by (5) : (8) \/ s r a = zz (2) 2 (2) ) 792 ) (rn) 22 ( i) 0i + + 0B( b.) ( o()) M (2) +0' [IC na i= i= /f \ 22 i= B t )] 2 with 0<0 ' < 2. na We now need estiates for 0?), for i Oa) and for ( i) 0 a ). i i= i Inequality (7) yields easily : Indeed, the first inequality of (9) holds for = and by inductio n for all and the two other inequalities follow fro this and (7). (All these are not the best possible inequalities.) Using () and (9 ) in (8) we find : 2 a= - [+0(»+O(v) O()+0(~n) + 'n +O(). Vn Choosing = [" ] we find : 7 00?)<2, i= (9) < i Bi) < 2, i= 2 < ( i) 0 c) < 4. i = Z (20) 2 a = ± 0( 'is). A sharpening of the inequalities (9) should lead to the reduction o f the index 3. Nuerical evidence indicates that : a= a2 +0(). 3. The ease b ig = a (and any > ) We shall show that in this case a/ oo so that forula (7), which now becoes : (2 ) ~~` = [ + 0 ( )] n (ai a2 ), i= holds. Indeed, a > and a is an unbounded and increasing function of. Let Q be the sallest integer such that a_q Q+. Then Q is such that Q 2<Q+ and : Q R

8 2'2 Forula (4) now becoes, by using first 0=0 and then 0= : iaii) a(22) (2 2 (ai / When lc --> oo both Q and ( Q) --> oo. If the product in (22) converged, a7c/ would be bounded, but then the product would diverge. Therefore the product diverges, so that a/ tends to 0o and Q/ to. Fro (2) we easily see that : a+l _ a7c_ o a9 ± and thus that a/ changes slowly. We now want to show that : (23) liinf log <. Indeed we would otherwise have : and so fro (2) : Since : a > ( + 6) log lc for > o with 6> 0, i l - (+6) Jog < cl~( (+6)i log = c 2 ei= we would have : = c3 +log log ±o(), 2 Jog 2 logloglc (+6) log < c4 e +6 = c4 (log )+h which is wrong. Siilarly it can be shown that : (24) lisup log,c >. Suppose now that : li sup og = + c (c > 0). ~ C (+ö)ilo g i Then, because of the slowness of change of a/ and because of (23), for a suitable choice of cl and c 2 so that 0 < cl < c2 < c, there exist two nu - bers l and 2 such that : and : a' l log < + cl and a7 > + lo c g 7c cl for l < < 2, (IN 2 log ~ > + c2 and a < + c 2 for l < < Ice.

9 We have : (25 ) 2 3 a.,, aa (l---ca) log /c2 2 (+c ) log i ' But fro (2) we have : Ic,~ c [ i 0 ( )] (aia z z ) = [ ±0( )] ei=, Q G i= y [+0()] ei 7 (+c,)ilogí = o()] (log 2 ) + [+ g ci which contradicts (25). Therefore : Siilarly it can be shown that : li sup l ac; =. 7c roe and hence : li inf log ' -,00 (26) li log ' c o as stated in the introduction.