A069648 a(1) = 1, otherwise smallest m > 1 such that the sum of digits of m^n is k^n for some k > 1.

1, 2, 2, 11, 47, 46, 983, 193534, 676644395

Offset: 1

Author

Amarnath Murthy, Apr 04 2002

Keywords

• more
• nonn
• base

Comments

n-th root of A069647(n).

Probably k = 2 in all cases. - Charles R Greathouse IV, Feb 26 2014

Crossrefs

Cf. A069647.

Programs

PARI

a237992(maxn, maxm) = {
  print1("1, ");
  for(n=1, maxn,
    for(m=2, maxm,
      t=eval(Vec(Str(m^n)));
      d=sum(i=1, #t, t[i]);
      if(d>1 && ispower(d, n), print1(m, ", "); break())
    )
  )
}
a237992(8, 1000000) \\ Colin Barker, Feb 23 2014

PARI

a(n)=if(n==1,return(1)); my(t=2^n,t3=3^n,k,s); while((s=sumdigits(k++^n))=t3 && ispower(s,n)),); k \\ Charles R Greathouse IV, Feb 26 2014

Python

import sympy
from sympy import factorint
def DigitSum(x):
    return sum(int(i) for i in str(x))
def PowExp(p):
    n = 2
    while n < 10000*(10**(int(2**p/9)/p)):
        if DigitSum(n**p) != 1:
            count = 0
            for i in list(factorint(DigitSum(n**p)).values()):
                if (int(i)/p) % 1 == 0:
                    count += 1
            if count == len(list(factorint(DigitSum(n**p)).values())):
                return n
            else:
                n += 1
        else:
            n += 1
print(1)
x = 2
while x < 20:
    print(PowExp(x))
    x += 1
# Derek Orr, Feb 16 2014

Python

from sympy import factorint
def A069648(n):
    if (n == 1):
        return 1
    else:
        m = 2
        while True:
            x = sum(int(d) for d in str(m**n))
            if x > 1 and not any(map(lambda x:x%n,factorint(x).values())):
                return m
            m += 1 # Chai Wah Wu, Aug 11 2014

Formula

a(n) > c^2^n for n > 1 where c = 10^(1/81) = 1.0288.... - Charles R Greathouse IV, Feb 26 2014

Extensions

Corrected and extended by David Wasserman, Apr 23 2003