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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
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.
Sequence in context: A215741 A365360 A071254 * A259705 A073561 A258970
KEYWORD
nonn
AUTHOR
Erich Friedman, Aug 01 2001
EXTENSIONS
More terms from Jud McCranie, Aug 06 2001
STATUS
approved