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
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
Programs
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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)))
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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
Extensions
More terms from Chai Wah Wu, Apr 07 2021