A076745 a(n) = the least positive integer k such that b(k) = n, where b(k) (A076526) is defined by b(k) = r * max{e_1,...,e_r} if k = p_1^e_1 *...* p_r^e_r is the canonical prime factorization of k.
2, 4, 8, 12, 32, 24, 128, 48, 120, 96, 2048, 192, 8192, 384, 480, 768, 131072, 960, 524288, 3072, 1920, 6144, 8388608, 3840, 36960, 24576, 7680, 13440, 536870912, 15360, 2147483648, 26880, 30720, 393216, 147840, 53760, 137438953472, 1572864, 122880, 107520, 2199023255552
Offset: 1
Examples
a(12) = 2 * max{1,2} = 4 since 12 = 2^2 * 3^1 and 12 is the least k for which b(k) = 4. Hence a(4) = 12.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..1000
- Carlos Rivera, Puzzle 201: The Arithmetic Function A(n), The Prime Puzzles and Problems Connection.
Crossrefs
Cf. A076526.
Programs
-
Mathematica
a[n_] := Min[Table[2^d*Times @@ Prime[Range[2, n/d]], {d, Divisors[n]}]]; Array[a, 50] (* Amiram Eldar, Sep 08 2024 *) -
PARI
a(n) = {my(f = factor(n), nd = numdiv(f), v = vector(nd), k = 0); fordiv(f, d, k++; v[k] = 2^d * prod(i = 1, n/d-1, prime(i+1))); vecmin(v);} \\ Amiram Eldar, Sep 08 2024
Formula
From Amiram Eldar, Sep 08 2024: (Start)
a(n) = Min_{d|n} (2^d * Product_{i=1..n/d-1} prime(i+1)).
a(p) = 2^p for a prime p.
a(2*p) = 3*2^p for a prime p.
a(3*p) = 15*2^p for a prime p > 2. (End)
Extensions
More terms from Amiram Eldar, Sep 08 2024