proposed

approved

editing

proposed

If n is in A033845, a(n) = 2^(n/2); otherwise a(n) = gpf(n)^(n/gpf(n)). - Franklin T. Adams-Watters, Jun 15 2014

Cf. A002581, A092975 (maximum instead of minimum), A033845.

proposed

editing

editing

proposed

Minimum among the numbers p^(n/p), where p is any a prime divisor factor of n.

The setting a(1)=1 is conventional.

a(1)=1 is conventional. Upper bound (for any n): a(n) <= (3^(1/3))^n = A002581^n.

For prime p, a(p)=p.

For n>1: When gpf(n)>3 then a(n)=gpf(n)^(n/gpf(n)); otherwise if n is even then a(n)=2^(n/2); otherwise a(n)=3^(n/3).

p = factor(n); m = 2^n;

for(k=1, #p[, 1], q=p[k, 1]; q=q^(n\qp[k, 1]); if(q<m, m=q));

Cf. A002581, A092975 (maximum instead of minimum).

allocated for Stanislav SykoraMinimum among the numbers p^(n/p), where p is any prime divisor of n.

1, 2, 3, 4, 5, 8, 7, 16, 27, 25, 11, 64, 13, 49, 125, 256, 17, 512, 19, 625, 343, 121, 23, 4096, 3125, 169, 19683, 2401, 29, 15625, 31, 65536, 1331, 289, 16807, 262144, 37, 361, 2197, 390625, 41, 117649, 43, 14641, 1953125, 529, 47, 16777216, 823543, 9765625, 4913, 28561, 53

1,2

a(1)=1 is conventional. Upper bound: a(n) <= (3^(1/3))^n = A002581^n.

Stanislav Sykora, <a href="/A243405/b243405.txt">Table of n, a(n) for n = 1..2000</a>

a(12)=64 because 2^(12/2)=64 is smaller than 3^(12/3)=81.

(PARI) A243405(n)= {my(m, k, p, q); if(n==1, return(1));

p = factor(n); m = 2^n;

for(k=1, #p[, 1], q=p[k, 1]; q=q^(n\q); if(q<m, m=q));

return (m); }

allocated

nonn

Stanislav Sykora, Jun 04 2014

approved

editing

allocating

allocated

allocated for Stanislav Sykora

allocating

approved