On the smallest abundant number not divisible by the first k rimes Douglas E. Iannucci Abstract We say a ositive integer n is abundant if σ(n) > 2n, where σ(n) denotes the sum of the ositive divisors of n. Number the rimes in ascending order: 1 = 2, 2 = 3, and so forth. Let A(k) denote the smallest abundant number not divisible by 1, 2,..., k. In this aer we find A(k) for 1 k 7, and we show that for all ǫ > 0, (1 ǫ)(k ln k) 2 ǫ < ln A(k) < (1 + ǫ)(k ln k) 2+ǫ for all sufficiently large k. 1 Introduction We say a ositive integer n is abundant if σ(n) > 2n, where σ(n) denotes the sum of the ositive divisors of n. The smallest abundant number is 12, and the smallest odd abundant number is 945. With a comuter search, Whalen and Miller [3] found 5 2 7 11 13 17 19 23 29 to be an odd abundant number not divisible by 3, and they raised the general question of how one goes about finding the smallest abundant number not divisible by the first k rimes. We number the rimes in ascending order: 1 = 2, 2 = 3, and so forth. Let A(k) denote the smallest abundant number not divisible by 1, 2,..., k. Note that A(1) = 945. In this aer we devise an algorithm to find A(k), and we aly it to find A(k) for 1 k 7. We shall also rove Theorem 1. For every ǫ > 0 we have whenever k is sufficiently large. (1 ǫ)(k ln k) 2 ǫ < ln A(k) < (1 + ǫ)(k ln k) 2+ǫ Received by the editors February 2002. Communicated by M. Van den Bergh. 1991 Mathematics Subject Classification : 11A32, 11Y70. Key words and hrases : abundant numbers, rimes. Bull. Belg. Math. Soc. 12(2005), 39 44
40 D. E. Iannucci 2 Preliminaries For a ositive integer n we define the index of n to be σ 1 (n) = σ(n) n. Thus n is abundant if σ 1 (n) > 2. The function σ 1 is multilicative, and for rime and integer a 1 we have σ 1 ( a ) = 1 + 1 + 1 2 + + 1 a. Therefore σ 1 ( a ) increases with a, and in fact + 1 σ 1 ( a ) < 1. (1) If < q are rimes then q/(q 1) < ( + 1)/ and so for all integers a 1, b 1, we have σ 1 (q b ) < σ 1 ( a ). (2) For each integer k 1 let us define V t (k) = t j=k+1 j + 1 j for integers t > k. By Theorem 19 in [1] and Theorem 3 of 28, Chater VII in [2], V t (k) increases without bound as t increases, and therefore we may define v(k) = min { t : V t (k) > 2 }. Since V t (k) = σ 1 ( k+1 k+2 t ), we have A(k) k+1 k+2 v(k). (3) We may also obtain a lower bound for A(k). For each integer k 1 we define U t (k) = t j=k+1 j j 1 for integers t > k. Since /( 1) > ( + 1)/, we have U t (k) > V t (k) and so we may define u(k) = min { t : U t (k) > 2 }. (4) Note that u(k) v(k). We can show that A(k) k+1 k+2 u(k) ; in fact we can show more:
On the smallest abundant number 41 Lemma 1. A(k) is divisible by k+1 k+2 u(k). Proof. Let M = k+1 k+2 u(k) and suose M A(k). Let A(k) have the unique rime factorization given by A(k) = t i=1 q a i i for distinct rimes q 1 < q 2 < < q t, and ositive integers a i, 1 i t. Note that q 1 > k. Hence q i k+i for all i, 1 i t. We have t u(k) k. For, otherwise by (1), (2), σ 1 (A(k)) < k+1 k+1 1 k+2 k+2 1 u(k) 1 u(k) 1 1, which imlies σ 1 (A(k)) 2 by (4); this contradicts the abundance of A(k). Since M A(k), we have j A(k) for some j such that k + 1 j u(k). Therefore, since t u(k) k, at least one of the rimes q i dividing A(k) must be greater than u(k). Without loss of generality we may assume q 1 > u(k). Then by (2), σ 1 ( j q a 2 2 qat t ) > σ 1(q a 1 1 qa 2 2 qat t ) > 2. But then, j q a 2 2 qat t < q a 1 1 qa 2 2 qat t = A(k), which contradicts the minimality of A(k). 3 An Algorithm From Lemma 1, we may devise an algorithm for finding A(k): (1) Find u(k), as given by (4). (2) Let P k = 1 2 k. Let m run through the ositive integers which are relatively rime to P k until we find It follows that M(k) = min { m : σ 1(m k+1 k+2 u(k) ) > 2 }. (m,p k )=1 A(k) = M(k) k+1 k+2 u(k). Note that by (3) we have M(k) u(k)+1 u(k)+2 v(k). Using the UBASIC software ackage, a comuter search emloying the algorithm was conducted to find A(k) for 1 k 7. In Table 1 is given the values for M(k) and A(k), along with those of u(k) and v(k), for 1 k 7. k u(k) v(k) M(k) A(k) 1 7 13 3 2 3 3 5 7 2 23 31 5 29 5 2 7 11 13 17 19 23 29 3 61 73 7 11 67 7 2 11 2 13 17 59 61 67 4 127 149 11 13 131 137 11 2 13 2 17 19 131 137 5 199 233 13 17 211 223 227 13 2 17 2 19 23 223 227 6 337 367 17 19 23 347 349 17 2 19 2 23 2 29 31 347 349 7 479 521 19 23 29 487 491 499 19 2 23 2 29 2 31 37 491 499 Table 1. The values A(k) for 1 k 7.
42 D. E. Iannucci 4 Behavior of A(k) In this section we estimate the growth of A(k) by roving Theorem 1. We begin by stating a result due to Mertens (Theorem 429 in [1]), e γ lim x ln x x 1 = 1, (5) where the roduct is taken over rimes and where γ denotes Euler s constant. We now rove Lemma 2. ln u(k) lim = 2. x ln k Proof. Let 0 < ǫ < 2 be given. Take 0 < ǫ 1 < ǫ/(2 ǫ) (so that 2ǫ 1 /(1 + ǫ 1 ) < ǫ), and take 0 < ǫ 2 < ǫ 1 /(2 + ǫ 1 ) (so that (1 + ǫ 2 )/(1 ǫ 2 ) < 1 + ǫ 1 ). By (5), there exists an integer k 1 such that for all x k1 we have (1 ǫ 2 )e γ ln x < x 1 < (1 + ǫ 2)e γ ln x. Note that by (4) we have 2 < Thus for all k k 1 we have hence k < u(k) 1 = u(k) 1. k 1 2 < (1 + ǫ 2)e γ ln u(k) (1 ǫ 2 )e γ ln k < (1 + ǫ 1 ) ln u(k) ln k, ln u(k) ln k > 2 1 + ǫ 1 = 2 2ǫ 1 1 + ǫ 1 > 2 ǫ. Now take 0 < ǫ 4 < ( 3 + 9 + 4ǫ)/2 (so that 3ǫ 4 + ǫ 2 4 < ǫ), take 0 < ǫ 5 < ǫ 4 /(2 + ǫ 4 ) (so that 2/(1 ǫ 5 ) < 2 + ǫ 4 ), and take 0 < ǫ 6 < ǫ 5 /(2 ǫ 5 ) (so that (1 ǫ 6 )/(1 + ǫ 6 ) < 1 ǫ 5 ). By (5) there exists an integer k 2 such that for all k k 2 we have 1/( k 1) < ǫ 4 and such that for all x k2 we have (1 ǫ 6 )e γ ln x < x 1 < (1 + ǫ 6)e γ ln x. By (4) we have and so for k k 2 hence 2 u(k) u(k) 1 k < u(k) 1 = u(k) 1, k 1 2 u(k) u(k) 1 (1 ǫ 6)e γ ln u(k) (1 + ǫ 6 )e γ ln k > (1 ǫ 5 ) ln u(k) ln k, ln u(k) < u(k) ln k u(k) 1 2 < (1 + ǫ 4 )(2 + ǫ 4 ) = 2 + 3ǫ 4 + ǫ 2 4 1 ǫ < 2 + ǫ. 5 Therefore if k max{k 1, k 2 } then ln u(k) /ln k 2 < ǫ.
On the smallest abundant number 43 An almost identical roof (omitted here) gives Lemma 3. ln v(k) lim = 2. x ln k The Prime Number Theorem (Theorem 8 in [1]) states that lim n An equivalent result (Theorem 420 in [1]) is where θ denotes the function, defined for x > 0, given by n n ln n = 1. (6) θ(x) lim x x = 1, (7) θ(x) = ln, x the sum being taken over rimes. We may now begin roving Theorem 1. Let ǫ > 0 be given. Take 0 < ǫ 1 < 4 1 + ǫ 1 (so that (1+ǫ1 ) 4 < 1+ǫ), take 0 < ǫ 2 < ǫ 1, and take 0 < ǫ 3 < min{1, ǫ}. By (7), there exists an integer k 1 such that for all k k 1 we have θ( v(k) ) < (1 + ǫ 1 ) v(k). By (6) there exists an integer k 2 such that for all k k 2 we have k < (1 + ǫ 2 )k ln k. By Lemma 3 there exists an integer k 3 such that for all k k 3 we have v(k) < 2+ǫ 3 k. Then by (3), if k max{k 1, k 2, k 3 }, we have hence ln A(k) v(k) j=k+1 ln j < θ( v(k) ), ln A(k) < (1 + ǫ 1 ) v(k) < (1 + ǫ 1 ) 2+ǫ 3 k < (1 + ǫ 1 )(1 + ǫ 2 ) 2+ǫ 3 (k lnk) 2+ǫ 3 < (1 + ǫ 1 ) 4 (k ln k) 2+ǫ 3 < (1 + ǫ)(k ln k) 2+ǫ. A similar roof (omitted here) shows that for sufficiently large k we have ln A(k) > (1 ǫ)(k lnk) 2 ǫ, and hence the roof of Theorem 1 is comlete.
44 D. E. Iannucci Bibliograhy [1] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, Oxford, 1979; [2] K. Kno, Theory and Alication of Infinite Series, Dover Publications, New York, 1990; [3] M. T. Whalen and C. L. Miller, Odd abundant numbers: some interesting observations, Jour. Rec. Math. 22 (1990), 257 261; University of the Virgin Islands 2 John Brewers Bay St. Thomas VI 00802 USA email: diannuc@uvi.edu