A063528 Smallest number such that it and its successor are both divisible by an n-th power larger than 1.

2, 8, 80, 80, 1215, 16767, 76544, 636416, 3995648, 24151040, 36315135, 689278976, 1487503359, 1487503359, 155240824832, 785129144319, 4857090670592, 45922887663615, 157197025673216, 1375916505694208, 2280241934368767, 2280241934368767, 2280241934368767

Offset: 1

Author

Erich Friedman, Aug 01 2001

Keywords

• nonn

Comments

Lesser of the smallest pair of consecutive numbers divisible by an n-th power.

To get a(j), max exponent[=A051953(n)] of a(j) and 1+a(j) should exceed (j-1).

One can find a solution for primes p and q by solving p^n*i + 1 = q^n*j; then p^n*i is a solution. This solution will be less than (p*q)^n but greater than max(p,q)^n. Thus finding the solutions for 2, 3 (p=2,q=3 and p=3,q=2), one need at most also look at 2, 5 and 3, 5. It appears that the solution with 2, 3 is always optimal. - Franklin T. Adams-Watters, May 27 2011

Examples

			a(4) = 80 since 2^4 = 16 divides 80 and 3^4 = 81 divides 81.
		

References

• J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 242, p. 67, Ellipses, Paris 2008.

Links

• Franklin T. Adams-Watters, Table of n, a(n) for n = 1..100

Crossrefs

We need A051903(a[n]) > n-1 and A051903(a[n]+1) > n-1.

Cf. A068780, A068781, A068140, A068782, A068783, A068784.

Cf. A045330, A059737.

Programs

Mathematica

k = 4; Do[k = k - 2; a = b = 0; While[ b = Max[ Transpose[ FactorInteger[k]] [[2]]]; a <= n || b <= n, k++; a = b]; Print[k - 1], {n, 0, 19} ]

PARI

b(n,p=2,q=3)=local(i);i=Mod(p,q^n)^-n; min(p^n*lift(i)-1,p^n*lift(-i))
a(n)=local(r);r=b(n);if(r>5^n,r=min(r,min(b(n,2,5),b(n,3,5))));r /* Franklin T. Adams-Watters, May 27 2011 */

Extensions

More terms from Jud McCranie, Aug 06 2001