On the smallest abundant number not divisible by the first k primes

Transcription

1 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 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 Communicated by M. Van den Bergh Mathematics Subject Classification : 11A32, 11Y70. Key words and hrases : abundant numbers, rimes. Bull. Belg. Math. Soc. 12(2005), 39 44

2 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 ) = 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:

3 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) Table 1. The values A(k) for 1 k 7.

4 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 > ǫ 1 = 2 2ǫ ǫ 1 > 2 ǫ. Now take 0 < ǫ 4 < ( ǫ)/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 ǫ < 2 + ǫ. 5 Therefore if k max{k 1, k 2 } then ln u(k) /ln k 2 < ǫ.

5 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 < 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.

6 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), ; University of the Virgin Islands 2 John Brewers Bay St. Thomas VI USA diannuc@uvi.edu