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
KEYWORD
nonn
AUTHOR
Erich Friedman, Aug 01 2001
EXTENSIONS
More terms from Jud McCranie, Aug 06 2001
STATUS
approved