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%I #9 Mar 31 2012 10:26:36
%S 2,8,80,80,1215,16767,76544,636416,3995648,24151040,36315135,
%T 689278976,1487503359,1487503359,155240824832,785129144319,
%U 4857090670592,45922887663615,157197025673216,1375916505694208
%N Smallest number such that it and its successor are both divisible by an n-th power larger than 1.
%C Lesser of the smallest pair of consecutive numbers divisible by an n-th power.
%C To get a(j), max exponent[=A051953(n)] of a(j) and 1+a(j) should exceed (j-1).
%C 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.
%D J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 242, p. 67, Ellipses, Paris 2008.
%H Franklin T. Adams-Watters, <a href="/A063528/b063528.txt">Table of n, a(n) for n = 1..100</a>
%e a[4]=80 since 2^4=16 divides 80 and 3^4=81 divides 81
%t 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} ]
%o (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))
%o 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 */
%Y We need A051903(a[n]) > n-1 and A051903(a[n]+1) > n-1.
%Y Cf. A068780, A068781, A068140, A068782, A068783, A068784.
%Y Cf. A045330, A059737.
%K nonn
%O 1,1
%A _Erich Friedman_, Aug 01 2001
%E More terms from _Jud McCranie_, Aug 06 2001