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

Offset

1,1

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.

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

Example

a[4]=80 since 2^4=16 divides 80 and 3^4=81 divides 81

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} ]

Prog

(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 */

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.

Sequence in context: A215741 A365360 A071254 * A259705 A073561 A258970

Adjacent sequences: A063525 A063526 A063527 * A063529 A063530 A063531

Keyword

nonn

Author

Erich Friedman, Aug 01 2001

Extensions

More terms from Jud McCranie, Aug 06 2001

Status

approved