A154532 a(n) is the largest 10-digit number whose n-th power contains each digit (0-9) n times, or -1 no such number exists.

9876543210, 9994363488, 9999257781, 9999112926, 9995722269, 9999409158, 9998033316, 9993870774, 9986053188, 9964052493, 9975246786, 9966918135, 9938689137, 9998781633, 9813743148, 9970902252, 9740383767, 9829440591, 9873773268, 9985819785, 9766102146, 9863817738

Offset: 1

Author

Zhining Yang, Jan 11 2009

Keywords

• nonn
• base

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.

From a(23) to a(110) the only terms which exist are a(24)=9793730157, a(26)=9347769564, a(35)=9959167017, and a(38)=9501874278. (The other values of a(n) are -1.) - Zhining Yang, Oct 05 2022

Examples

			a(18) = 9829440591, so each digit (0-9) appears 18 times in the decimal expansion of 9829440591^18.
		

Links

• northwolves et al., Ten Digit Numbers
• Zhining Yang, Largest Ten Digit Powers

Crossrefs

Cf. A010784, A078255, A154566.

Programs

Python

def flag(p, n):
    b = True
    for i in range(10):
        if not p.count(str(i)) == n:
            b = False
            break
    return b
def a(n):
    for i in range(10 ** 10 - 1, 3 * int(10 ** (10 - 1 / n) / 3), -3):
        p = str(i ** n)
        if flag(p, n) == True:
            return i
            break
for i in range(1, 23):
    print(i, a(i))  # Zhining Yang, Oct 10 2022

Python

def flag(p, n):
    return all(p.count(d) == n for d in "0123456789")
def a(n):
    return next(i for i in range(10**10-1,3*int(10**(10-1/n)/3), -3) if flag(str(i**n), n))
for i in range(2, 23):
    print(i,a(i))  # Michael_S._Branicky, Oct 10 2022

Extensions

Edited by N. J. A. Sloane, Jan 12 2009

a(19)-a(22) from Zhining Yang, Oct 05 2022

Definition revised by N. J. A. Sloane, Nov 22 2022