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A063528 Smallest number such that it and its successor are both divisible by an n-th power larger than 1.

%I A063528 #12 May 25 2025 20:48:12
%S A063528 2,8,80,80,1215,16767,76544,636416,3995648,24151040,36315135,
%T A063528 689278976,1487503359,1487503359,155240824832,785129144319,
%U A063528 4857090670592,45922887663615,157197025673216,1375916505694208,2280241934368767,2280241934368767,2280241934368767
%N A063528 Smallest number such that it and its successor are both divisible by an n-th power larger than 1.
%C A063528 Lesser of the smallest pair of consecutive numbers divisible by an n-th power.
%C A063528 To get a(j), max exponent[=A051953(n)] of a(j) and 1+a(j) should exceed (j-1).
%C A063528 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 A063528 J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 242, p. 67, Ellipses, Paris 2008.
%H A063528 Franklin T. Adams-Watters, <a href="/A063528/b063528.txt">Table of n, a(n) for n = 1..100</a>
%e A063528 a(4) = 80 since 2^4 = 16 divides 80 and 3^4 = 81 divides 81.
%t A063528 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 A063528 (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 A063528 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 A063528 We need A051903(a[n]) > n-1 and A051903(a[n]+1) > n-1.
%Y A063528 Cf. A068780, A068781, A068140, A068782, A068783, A068784.
%Y A063528 Cf. A045330, A059737.
%K A063528 nonn
%O A063528 1,1
%A A063528 _Erich Friedman_, Aug 01 2001
%E A063528 More terms from _Jud McCranie_, Aug 06 2001