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A066736
Least number > 1 such that the nonzero product of the digits of its n-th power is also an n-th power.
0
2, 2, 2, 118, 382, 580408, 178758, 12, 1254514, 53067715, 51773525, 314537797, 110242999
OFFSET
1,1
COMMENTS
The n-th roots of the product of the digits of the n-th power of a(n) are 2, 2, 2, 42, 24, 1440, 2520, 6, 10080. Because the numbers get larger quicker, the available candidates decreases. See A020665. Therefore this sequence might be finite or have a preponderance of blank entries.
If a(14) exists, it is greater than 4*10^8. - Derek Orr, Feb 17 2014
If a(14) exists, it is greater than 5.6*10^10. - Sean A. Irvine, Nov 05 2023
EXAMPLE
382^5 = 8134236862432 and the product of these digits is 7962624 = 24^5 (another fifth power). Since 382 is the smallest number with this property, a(5) = 382.
MATHEMATICA
Do[k = 2; While[a = Apply[Times, IntegerDigits[k^n]]; a == 0 || !IntegerQ[a^(1/n)], k++ ]; Print[k], {n, 1, 10} ]
PROG
(Python)
import sympy
from sympy import factorint
def DigitProd(x):
..total = 1
..for i in str(x):
....total *= int(i)
..return total
def Prod(x):
..n = 2
..while n < 4*(10**8):
....if DigitProd(n**x) != 0 and DigitProd(n**x) != 1:
......count = 0
......for i in list(factorint(DigitProd(n**x)).values()):
........if (int(i)/x) % 1 == 0:
..........count += 1
........else:
..........break
......if count == len(list(factorint(DigitProd(n**x)).values())):
........return n
......else:
........n += 1
....else:
......n += 1
x = 1
while x < 100:
..print(Prod(x))
..x += 1 # Derek Orr, Feb 13 2014
CROSSREFS
Sequence in context: A102384 A275533 A304463 * A324302 A216668 A057330
KEYWORD
nonn,base
AUTHOR
Robert G. Wilson v, Jan 15 2002
EXTENSIONS
a(10)-a(13) from Derek Orr, Feb 13 2014
STATUS
approved