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

1,1

Comments

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.

No solutions were found for n = 39 to 1000.

Links

Table of n, a(n) for n=1..38.

Example

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

Prog

(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)

Crossrefs

Cf. A010784, A078255, A154532, A154566.

Sequence in context: A206083 A199634 A203987 * A217110 A217111 A036471

Adjacent sequences: A357752 A357753 A357754 * A357756 A357757 A357758

Keyword

nonn,base,more

Author

Zhining Yang, Nov 26 2022

Status

approved