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%I #21 Nov 05 2023 14:34:54
%S 2,2,2,118,382,580408,178758,12,1254514,53067715,51773525,314537797,
%T 110242999
%N Least number > 1 such that the nonzero product of the digits of its n-th power is also an n-th power.
%C 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.
%C If a(14) exists, it is greater than 4*10^8. - _Derek Orr_, Feb 17 2014
%C If a(14) exists, it is greater than 5.6*10^10. - _Sean A. Irvine_, Nov 05 2023
%e 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.
%t Do[k = 2; While[a = Apply[Times, IntegerDigits[k^n]]; a == 0 || !IntegerQ[a^(1/n)], k++ ]; Print[k], {n, 1, 10} ]
%o (Python)
%o import sympy
%o from sympy import factorint
%o def DigitProd(x):
%o ..total = 1
%o ..for i in str(x):
%o ....total *= int(i)
%o ..return total
%o def Prod(x):
%o ..n = 2
%o ..while n < 4*(10**8):
%o ....if DigitProd(n**x) != 0 and DigitProd(n**x) != 1:
%o ......count = 0
%o ......for i in list(factorint(DigitProd(n**x)).values()):
%o ........if (int(i)/x) % 1 == 0:
%o ..........count += 1
%o ........else:
%o ..........break
%o ......if count == len(list(factorint(DigitProd(n**x)).values())):
%o ........return n
%o ......else:
%o ........n += 1
%o ....else:
%o ......n += 1
%o x = 1
%o while x < 100:
%o ..print(Prod(x))
%o ..x += 1 # _Derek Orr_, Feb 13 2014
%K nonn,base
%O 1,1
%A _Robert G. Wilson v_, Jan 15 2002
%E a(10)-a(13) from _Derek Orr_, Feb 13 2014
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