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Number of solutions for a 10-digit number whose n-th power contains each digit (0-9) exactly n times.
2

%I #25 Jan 14 2024 20:33:38

%S 3265920,468372,65663,15487,5020,1930,855,417,246,114,97,45,33,24,20,

%T 18,7,6,1,3,2,3,0,1,0,2,0,0,0,0,0,0,0,0,1,0,0,1

%N Number of solutions for a 10-digit number whose n-th power contains each digit (0-9) exactly n times.

%C A number with 10*n digits may have all ten digits (0-9) repeated n times. The probability of this is (10n)!/((n!)^10 * 10^((10*n)-10^(10*n-1)). There are 10^10-10^(10-1/n)) numbers which are n-th powers of 10-digit numbers. So there may exist Count=(10n)!*(10^10-10^(10-1/n)))/((n!)^10 * 10^((10*n)-10^(2*n-1)) numbers with the desired property.

%C No solutions were found for n = 39 to 1000.

%e a(20) = 3 because there are 3 10-digit numbers (8951993472, 9921107394, and 9985819785) whose 20th power contains each digit (0-9) 20 times.

%o (Python)

%o def flag(p, n):

%o return all(p.count(d) == n for d in "0123456789")

%o def a(n):

%o num=0

%o for i in range(10**10-1, 3*int(10**(10-1/n)/3), -3):

%o if flag(str(i**n), n):

%o num+=1

%o return(num)

%Y Cf. A010784, A078255, A154532, A154566.

%K nonn,base,more

%O 1,1

%A _Zhining Yang_, Nov 26 2022