A072103 Sorted perfect powers a^b for a, b > 1 with duplication.

4, 8, 9, 16, 16, 25, 27, 32, 36, 49, 64, 64, 64, 81, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 256, 256, 289, 324, 343, 361, 400, 441, 484, 512, 512, 529, 576, 625, 625, 676, 729, 729, 729, 784, 841, 900, 961, 1000, 1024, 1024, 1024, 1089

Offset: 1

Author

Eric W. Weisstein, Jun 18 2002

Keywords

• nonn

Comments

If b is the largest integer such that n=a^b for some a > 1, then n occurs d(b)-1 times in this sequence (where d = A000005 is the number of divisors function). (This includes the case where b=1 and n does not occur in the sequence.) - M. F. Hasler, Jan 25 2015

Examples

			(a,b) = (2,4) and (4,2) both yield 2^4 = 4^2 = 16, therefore 16 is listed twice.
Similarly, 64 is listed 3 times since (a,b) = (2,6), (4,3) and (8,2) all yield 64.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..9999, recomputed with new offset by _M. F. Hasler_, Jan 25 2015
• Eric Weisstein's World of Mathematics, Perfect Power

Crossrefs

Cf. A001597, A000005, A253641, A253642.

Programs

Haskell

import Data.Set (singleton, findMin, deleteMin, insert)
a072103 n = a072103_list !! (n-1)
a072103_list = f 9 3 $ Set.singleton (4,2) where
   f zz z s
     | xx < zz   = xx : f zz z (Set.insert (x*xx, x) $ Set.deleteMin s)
     | otherwise = zz : f (zz+2*z+1) (z+1) (Set.insert (z*zz, z) s)
     where (xx, x) = Set.findMin s
-- Reinhard Zumkeller, Oct 04 2012

Maple

N:= 2000: # to get all entries <= N
sort([seq(seq(a^b, b = 2 .. floor(log[a](N))), a = 2 .. floor(sqrt(N)))]); # Robert Israel, Jan 25 2015

Mathematica

nn=60;Take[Sort[#[[1]]^#[[2]]&/@Tuples[Range[2,nn],2]],nn] (* Harvey P. Dale, Oct 03 2012 *)

PARI

is_A072103(n)=ispower(n)
for(n=1,999,(e=ispower(n))||next;fordiv(e,d,d>1 && print1(n","))) \\ M. F. Hasler, Jan 25 2015

Python

import numpy
from math import isqrt
upto = 1090
A072103 = []
for m in range(2,isqrt(upto)+1):
    k = 2
    while m**k < upto:
        A072103.append(m**k)
        k += 1
print(sorted(A072103)) # Karl-Heinz Hofmann, Sep 16 2023

Formula

Sum_{i>=2} Sum_{j>=2} 1/i^j = 1.

Extensions

Offset corrected and examples added by M. F. Hasler, Jan 25 2015