A357755 Number of solutions for a 10-digit number whose n-th power contains each digit (0-9) exactly n times.

3265920, 468372, 65663, 15487, 5020, 1930, 855, 417, 246, 114, 97, 45, 33, 24, 20, 18, 7, 6, 1, 3, 2, 3, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1

Offset: 1

Author

Zhining Yang, Nov 26 2022

Keywords

• nonn
• base
• more

Comments

A number with 10*n digits may have all ten digits (0-9) repeated n times. The probability of this is (10*n)!/((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 = (10*n)!*(10^10-10^(10-1/n))/((n!)^10 * (10^(10*n)-10^(10*n-1))) numbers with the desired property.

No solutions were found for n = 39 to 1000.

Examples

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

Crossrefs

Cf. A010784, A078255, A154532, A154566.

Programs

Python

def flag(p, n):
    return all(p.count(d) == n for d in "0123456789")
def a(n):
    num=0
    for i in range(10**10-1, 3*int(10**(10-1/n)/3), -3):
        if flag(str(i**n), n):
            num+=1
    return(num)