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 A094268 Starting term of smallest consecutive n-tuples of abundant numbers. 4
 12, 5775, 171078830 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The triple 171078830, 171078831, 171078832 was apparently found by Laurent Hodges and Michael Reid in 1995. The starting term of the smallest consecutive 4-tuple of abundant numbers is at most 141363708067871564084949719820472453374 - Bruno Mishutka (bruno.mishutka(AT)googlemail.com), Nov 01 2007 Paul Erdős showed that there are two absolute constants c1, c2 such that for all large n there are at least c1 log log log n but not more than c2 log log log n consecutive abundant numbers less than n. - Bruno Mishutka (bruno.mishutka(AT)googlemail.com), Nov 01 2007 From Jianing Song, Apr 10 2021: (Start) a(n) exists for all n. Proof: since the infinite product Product_{p prime} (1 + 1/p) diverges, we can find a strictly increasing sequence {b(m)} such that b(0) = 0, Product_{k=b(m)+1..b(m+1)} (1 + 1/prime(k)) > 2 for all m. Given n, by Chinese Remainder Theorem, we can find N such that N + m divides Product_{k=b(m)+1..b(m+1)} prime(k) for m = 0..n-1, then sigma(N + m)/(N + m) >= Product_{k=b(m)+1..b(m+1)} (1 + 1/prime(k)) > 2. For example, if N is divisible by 2*3*5, N+1 is divisible by 7*11*...*73, N+2 is divisible by 79*83*...*7499, N+3 is divisible by 7507*7517*...*57081677, N+4 is divisible by 57081679*57081697*...*(some very large prime), then N through N+4 are consecutive abundant numbers. Of course, the number N found using this method will be extremely large, since Product_{k=1..K} (1 + 1/prime(k)) ~ log(log(K)). (End) REFERENCES J.-M. De Koninck and A. Mercier, 1001 Problemes en Theorie Classique Des Nombres, Problem 771, pp. 98, 327, Ellipses, Paris, 2004. S. Kravitz, Three Consecutive Abundant Numbers, Journal of Recreational Mathematics, 26:2 (1994), 149. Solution by L. Hodges and M. Reid, JRM, 27:2 (1995), 156-157. LINKS Table of n, a(n) for n=1..3. Paul Erdős, Note on consecutive abundant numbers, J. London Math. Soc. 10, 128-131 (1935). Carlos Rivera, Puzzle 878. Consecutive abundant integers CROSSREFS Cf. A005101, A005231, A096399. Sequence in context: A230749 A003793 A171669 * A208865 A012607 A167072 Adjacent sequences: A094265 A094266 A094267 * A094269 A094270 A094271 KEYWORD hard,bref,more,nonn AUTHOR Lekraj Beedassy, Jun 02 2004 STATUS approved

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Last modified July 24 01:47 EDT 2024. Contains 374575 sequences. (Running on oeis4.)