A342545 a(n)^2 is the least square that has exactly n 0's in base n.

2, 24, 16, 280, 216, 3430, 4096, 19683, 100000, 4348377, 2985984, 154457888, 105413504, 4442343750, 4294967296, 313909084845, 198359290368, 8712567840033, 10240000000000, 500396429346030, 584318301411328, 38112390316557080, 36520347436056576, 298023223876953125

Offset: 2

Author

Hugo Pfoertner, Apr 07 2021

Keywords

• nonn
• base

Examples

			   n    a(n)         a(n)^2   in base n
   2       2              4   100
   3      24            576   210100
   4      16            256   10000
   5     280          78400   10002100
   6     216          46656   1000000
   7    3430       11764900   202000000
   8    4096       16777216   100000000
   9   19683      387420489   1000000000
  10  100000    10000000000   10000000000
  11 4348377 18908382534129   6030000000000
  12 2985984  8916100448256   1000000000000
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 2..702

Crossrefs

Cf. A000290, A062971, A193678, A342260, A342546.

Programs

PARI

for(b=2,12,for(k=1,oo,my(s=k^2,v=digits(s,b));if(sum(k=1,#v,v[k]==0)==b,print1(k,", ");break)))

Python

from numba import njit
@njit # works with 64 bits through a(14)
def digits0(n, b):
  count0 = 0
  while n >= b:
    n, r = divmod(n, b)
    count0 += (r==0)
  return count0 + (n==0)
from sympy import integer_nthroot
def a(n):
  an = integer_nthroot(n**n, 2)[0]
  while digits0(an*an, n) != n: an += 1
  return an
print([a(n) for n in range(2, 13)]) # Michael S. Branicky, Apr 07 2021

Python

from itertools import product
from functools import reduce
from sympy.utilities.iterables import multiset_permutations
from sympy import integer_nthroot
def A342545(n):
    for a in range(1,n):
        p, q = integer_nthroot(a*n**n,2)
        if q: return p
    l = 1
    while True:
        cmax = n**(l+n+1)
        for a in range(1,n):
            c = cmax
            for b in product(range(1,n),repeat=l):
                for d in multiset_permutations((0,)*n+b):
                    p, q = integer_nthroot(reduce(lambda c, y: c*n+y, [a]+d),2)
                    if q: c = min(c,p)
            if c < cmax:
                return c
        l += 1 # Chai Wah Wu, Apr 07 2021

Formula

a(2*n) = A062971(n) = 2*A193678(n).

Extensions

More terms from Chai Wah Wu, Apr 07 2021