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• W2021561567 abstract "An aliquot sequence n: k, k = 0, 1, 2, . . ., is defined by n: 0 = n, n: k + 1 = a(n: k) n k, and a driver of an aliquot sequence is a number 2Av with A > 0, v odd, v12 +11 and 2A-1 It(v). Pollard has noted some errors in a proof in [1] that the drivers comprise the even perfect numbers and a finite set. These are now corrected in a revised proof. John Pollard has observed two inaccuracies and some obscurities in a proof in [1] for which we wish to substitute the following. THEOREM 2. The only drivers are 2, 233, 233.5, 253.7, 293.11.31 and the even perfect numbers. Proof A driver is 2A v with A > O, v odd, v 2A + 1 -1 and 2A-1 a(v). If v = 1, 2A-1 j 1, A = 1 and we have the downdriver 2. If v = 2A+1 1 is a Mersenne prime, the driver is an even perfect number. Henceforth, we assume that v > 1 and that 2A + 1 1 is composite. If pa 112A + 1 1, p prime, a > 0, define the deficiency, 6(p), of p to be 2d/pa, where 2d || q(pb) and pb || V, 0 2A+ -i = pa>412d, p d 2A-1 > Hl 2d and 2A 1 would not divide H1 (pb) = a(V). The power of 2 dividing J(pb) depends only on how many factors of the product (p + 1)(p2 + 1)(P4 + 1) ... divide J(pb), each factor other than p + 1 contributing a single 2. Hence, d = O if b is even and d = t + k I if b is odd, where 2t lIp + 1, there are k such factors, and thus 2k jib + 1. It then follows that 8(p) 1. Otherwise, 6(p) 5, and 6(p) 2. If we denote by l 8(p) the product of the deficiencies of the Mersenne prime factors of 2A + 1 1, it is not difficult to see that 4 8 32 128 4 8 32 64 8 ll 6 3 * 7 W 31 * 127 3 7 31 63 5 We now note that 2A + 1 1 contains at most one non-Mersenne prime factor. Received December 1, 1977; revised March 29, 1978 and March 9, 1979. AMS (MOS) subject classifications (1970). Primary 1OA20, Secondary 1OL1O. ? 1980 American Mathematical Society 0025-571 8/80/0000-0022/$01 .75 319 This content downloaded from 157.55.39.220 on Fri, 02 Sep 2016 05:01:58 UTC All use subject to http://about.jstor.org/terms 320 RICHARD K. GUY AND J. L. SELFRIDGE For having two such prime factors would imply that the product of the deficiencies would be less than 8(P1)5(P2) II 8() 7, a > 1 would imply 6(2q 1) 1 would imply p*7 772 5 4 For p = 3, a > 3 would imply 6(3) 5, 2A + 1 1 contains a non-Mersenne prime factor p1 and 6(3)6 (p) n (p) 3. We know that u = 3 or 5, since u > 7 would imply 8(2cu 1) H8(p) 7, there is a prime p, p 12A +1 1, p 1 (mod A + 1), giving a second non-Mersenne prime divisor of 2A+1 1. So we have c > 2, q1 > 2, u = 3 or 5 and -1 (2ql 1)(-1) . . . (-1)(2'u 1) (mod 2min(c, q1)+1) -1 (1)k1 (2q1 2Cu + 1), k is even and q1 = c. Now 2A +1 3, this would imply 2.3.q3 < qlq2q3 < 2q, + q2 + q3 + 3 < 4q3 + 3, a contradiction. So k = 2, q1q2 < 2q, + q2 + 3, (q 1)(q2 -2) < 5 q 2 =c and q2 = 3 or 5. Only the latter gives a solution; u = 3 and 293.11.31 is a driver. This content downloaded from 157.55.39.220 on Fri, 02 Sep 2016 05:01:58 UTC All use subject to http://about.jstor.org/terms CORRIGENDUM TO WHAT DRIVES AN ALIQUOT SEQUENCE? 321 Department of Mathematics and Statistics The University of Calgary Calgary, Alberta, Canada T2N 1N4 Department of Mathematical Sciences Northern Illinois University DeKalb, Illinois 60115 1. RICHARD K. GUY & J. L. SELFRIDGE, WVhat drives an aliquot sequence?, Math. Comp., v. 29, 1975, pp. 101-107. MR 52 #5542. This content downloaded from 157.55.39.220 on Fri, 02 Sep 2016 05:01:58 UTC All use subject to http://about.jstor.org/terms" @default.
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• W2021561567 title "Corrigendum to What Drives an Aliquot Sequence?" @default.
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• W2021561567 doi "https://doi.org/10.2307/2006239" @default.
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