A094268 Starting term of smallest consecutive n-tuples of abundant numbers.

12, 5775, 171078830

Offset: 1

Author

Lekraj Beedassy, Jun 02 2004

Keywords

• hard
• bref
• more
• nonn

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

• 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.