A342260
a(n)^2 is the least square that, when written in base n, has exactly n digits n-1.
Original entry on oeis.org
3, 31, 217, 268, 8399, 29110, 711243, 4676815, 31622764, 376863606, 12638826343, 38121744938, 1511790122972, 8648472039419, 243625577528103
Offset: 2
a(2) = 3: 3^2 = 9 is the least square with 2 binary ones: 1001;
a(3) = 31: 31^2 = 961 is the least square with 3 ternary digits 2: 1022121;
a(4) = 217: 217^2 = 47089 = 23133301_4;
a(5) = 268: 268^2 = 71824 = 4244244_5.
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isok(k, n) = #select(x->(x==n-1), digits(k^2, n)) == n;
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Apr 05 2021
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from sympy.ntheory.factor_ import digits
def A342260(n):
k = 1
while digits(k**2,n).count(n-1) != n:
k += 1
return k # Chai Wah Wu, Apr 05 2021
A342545
a(n)^2 is the least square that has exactly n 0's in base n.
Original entry on oeis.org
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
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
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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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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
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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
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