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A063528 Smallest number such that it and its successor are both divisible by an n-th power larger than 1. 13
2, 8, 80, 80, 1215, 16767, 76544, 636416, 3995648, 24151040, 36315135, 689278976, 1487503359, 1487503359, 155240824832, 785129144319, 4857090670592, 45922887663615, 157197025673216, 1375916505694208 (list; graph; refs; listen; history; text; internal format)
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: A057984 A215741 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

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Last modified October 17 13:57 EDT 2021. Contains 348049 sequences. (Running on oeis4.)